{
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    "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT",
    "title": "Nudge content generation",
    "agent": "nudge-generation",
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    {
      "info": {
        "role": "user",
        "time": {
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        "agent": "nudge-generation",
        "model": {
          "providerID": "openai",
          "modelID": "gpt-6-astra",
          "variant": "low"
        },
        "system": "You create Codeforces learning content for Nudge. Your job is to understand the best approach for the problem and teach it clearly in your own words, using existing solutions to inform your reasoning.\n\nStart with a brief, targeted web search for existing solutions so relevant editorials are in context before substantial independent problem-solving. Use any supplied tutorial links first; otherwise search by contest ID and problem index. Prefer the official tutorial/editorial, and read relevant explanations rather than relying on search snippets. Look at accessible alternatives when useful, especially if the official explanation is missing, unclear, or unnecessarily complicated. Accepted submissions can also help clarify an approach. Respect supplied source lookup statuses and avoid repeatedly retrying blocked sources. If no usable source is accessible, proceed from the supplied statement.\n\nTreat existing solutions as a starting point, not a constraint. If an approach is clear and suitable, build on it without unnecessary re-derivation. If an explanation has gaps, seems incorrect, or uses a harder method than necessary, reason through it and derive a better or simpler approach when worthwhile. Choose a correct approach that is easy to teach and fits the constraints. Verify it against the supplied statement and edge cases before writing the hints, editorial, and C++ solution.\n\nThe editorial must be self-contained: explain the key insight, algorithm, why it works, complexity, and implementation details a learner needs, filling in steps the source skips. Keep research and verification concise; spend the output on teaching the solution, not narrating the research process. Follow the user prompt and output schema exactly.",
        "format": {
          "type": "json_schema",
          "schema": {
            "type": "object",
            "properties": {
              "status": {
                "type": "string",
                "enum": [
                  "success",
                  "unsolvable"
                ],
                "description": "Set to 'success' when the supplied statement defines a solvable programming problem. Set to 'unsolvable' only if the statement is fundamentally incomplete, contradictory, or depends on an inaccessible required resource."
              },
              "reason": {
                "anyOf": [
                  {
                    "type": "string",
                    "pattern": "^[^\\u0000]*$",
                    "minLength": 1,
                    "maxLength": 2000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'unsolvable', provide a detailed failure message to show users. Name the concrete statement/resource blocker and any relevant source lookup/access status, such as missing official tutorial, 403/Cloudflare challenge, 404, or unavailable accepted submissions. Otherwise null."
              },
              "hints": {
                "anyOf": [
                  {
                    "type": "array",
                    "minItems": 5,
                    "maxItems": 5,
                    "description": "Exactly 5 progressive hints, each building on the last.",
                    "items": {
                      "type": "object",
                      "properties": {
                        "order": {
                          "type": "integer",
                          "minimum": 1,
                          "maximum": 5,
                          "description": "Hint number, 1 through 5."
                        },
                        "content": {
                          "type": "string",
                          "pattern": "^[^\\u0000]*$",
                          "minLength": 1,
                          "maxLength": 10000,
                          "description": "Markdown hint text."
                        }
                      },
                      "required": [
                        "order",
                        "content"
                      ],
                      "additionalProperties": false
                    }
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', provide exactly 5 progressive hints. Otherwise null."
              },
              "editorial": {
                "anyOf": [
                  {
                    "type": "string",
                    "pattern": "^[^\\u0000]*$",
                    "minLength": 1,
                    "maxLength": 100000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', a prose editorial explaining the solution in Nudge's own voice, with no links, citations, or source/research notes. Otherwise null."
              },
              "solution": {
                "anyOf": [
                  {
                    "type": "string",
                    "pattern": "^[^\\u0000]*$",
                    "minLength": 1,
                    "maxLength": 200000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', a complete C++ solution. Otherwise null."
              }
            },
            "required": [
              "status",
              "reason",
              "hints",
              "editorial",
              "solution"
            ],
            "additionalProperties": false,
            "title": "problem_response",
            "description": "Submit the generated content for the problem, or report if it is unsolvable."
          },
          "retryCount": 2
        },
        "id": "msg_0adf26219001y10ecLXnzSGaGT",
        "sessionID": "ses_f520d9df5ffeeRxgcTgdKVVPLP"
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      "parts": [
        {
          "type": "text",
          "text": "Generate Codeforces learning content: five progressive hints, an original editorial, and AC-quality C++.\n\nFollow the editorial-first research workflow in the system instructions. Use the supplied statement to verify that the source solves this exact problem.\n\nWrite clean Markdown with LaTeX as needed. Use quick and clever humor when appropriate. Tell it like it is (don't sugar-coat responses), and use very casual language. You are fully allowed to swear, just don't overdo it like a sailor (be natural). Deconstruct any false assumptions.\n\nGenerate content for Codeforces problem 2262E: \"Paired Bracket Sequences\".\n\nKnown problem metadata:\n- Contest ID: 2262\n- Index: E\n- Name: Paired Bracket Sequences\n- Rating: unknown / unrated\n- Tags: combinatorics, math\n- Problem URL: https://codeforces.com/contest/2262/problem/E\n\nUse the rating and tags as weak signals only. The supplied statement is the source of truth.\n\nProblem Statement:\n<problem-statement>\n<div class=\"header\"><div class=\"title\">E. Paired Bracket Sequences</div><div class=\"time-limit\"><div class=\"property-title\">time limit per test</div>2 seconds</div><div class=\"memory-limit\"><div class=\"property-title\">memory limit per test</div>256 megabytes</div><div class=\"input-file input-standard\"><div class=\"property-title\">input</div>standard input</div><div class=\"output-file output-standard\"><div class=\"property-title\">output</div>standard output</div></div><div><p> </p><p>Farmer John is interested in bracket sequences with common pairings.</p><p>A balanced bracket sequence is one in which every opening bracket is matched with a later closing bracket, and no prefix contains more closing brackets than opening brackets.</p><p>For a balanced bracket sequence $$$s$$$, a pairing is a pair of positions $$$(i,j)$$$ with $$$i \\lt j$$$ such that $$$s_i$$$ is an opening bracket, $$$s_j$$$ is a closing bracket, and the substring strictly between them is balanced. </p><p>Consider an ordered pair $$$(s,t)$$$ of balanced bracket sequences, where both $$$s$$$ and $$$t$$$ have length $$$2n$$$. A pairing $$$(i,j)$$$ is common to $$$s$$$ and $$$t$$$ if $$$(i,j)$$$ is a pairing in both sequences. For example, the bracket sequences $$$\\texttt{(()())}$$$ and $$$\\texttt{((()))}$$$ have one common pairing, namely the outer pair $$$(1,6)$$$.</p><p>Two ordered pairs $$$(s,t)$$$ and $$$(s',t')$$$ are considered different if $$$s \\ne s'$$$ or $$$t \\ne t'$$$.</p><p>For each $$$0 \\le k \\le n$$$, Farmer John wants to know the number of ordered pairs $$$(s,t)$$$ of balanced bracket sequences of length $$$2n$$$ with exactly $$$k$$$ common pairings. Since these numbers may be large, output them modulo $$$M$$$.</p></div><div class=\"input-specification\"><div class=\"section-title\">Input</div><p>Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 500$$$). The description of the test cases follows.</p><p>The first line of each testcase will contain two integers, $$$n$$$ and $$$M$$$ ($$$1 \\leq n \\leq 500$$$, $$$10^8 \\leq M \\leq 10^9$$$) — half the length of the bracket sequence and the modulo to output the answer in. It is guaranteed that $$$M$$$ is prime.</p><p>It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$. </p></div><div class=\"output-specification\"><div class=\"section-title\">Output</div><p>For each testcase, output $$$n + 1$$$ integers, the answer modulo $$$M$$$ for each $$$0 \\leq k \\leq n$$$.</p></div><div class=\"sample-tests\"><div class=\"section-title\">Example</div><div class=\"sample-test\"><div class=\"input\"><div class=\"title\">Input</div><pre><div class=\"test-example-line test-example-line-even test-example-line-0\">5</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">1 998244353</div><div class=\"test-example-line test-example-line-even test-example-line-2\">2 998244353</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">3 998244353</div><div class=\"test-example-line test-example-line-even test-example-line-4\">4 998244353</div><div class=\"test-example-line test-example-line-odd test-example-line-5\">8 998244353</div></pre></div><div class=\"output\"><div class=\"title\">Output</div><pre><div class=\"test-example-line test-example-line-odd test-example-line-1\">0 1 </div><div class=\"test-example-line test-example-line-even test-example-line-2\">2 0 2 </div><div class=\"test-example-line test-example-line-odd test-example-line-3\">8 12 0 5 </div><div class=\"test-example-line test-example-line-even test-example-line-4\">62 64 56 0 14 </div><div class=\"test-example-line test-example-line-odd test-example-line-5\">378790 646560 542960 296800 127400 34944 16016 0 1430 </div></pre></div></div></div><div class=\"note\"><div class=\"section-title\">Note</div><p>For the first test case, there is only one balanced bracket sequence with $$$n=1$$$: $$$\\texttt{()}$$$. It has exactly one pairing, namely $$$(1,2)$$$. Therefore, the only ordered pair of bracket sequences has exactly one common pairing, so the answer is $$$[0,1]$$$.</p><p>For the second test case, there are exactly two balanced bracket sequences with $$$n=2$$$: $$$a=\\texttt{(())}$$$ and $$$b=\\texttt{()()}$$$. The pairings of $$$a$$$ are $$$(1,4)$$$ and $$$(2,3)$$$, while the pairings of $$$b$$$ are $$$(1,2)$$$ and $$$(3,4)$$$.</p><p>Thus, the ordered pairs $$$(a,b)$$$ and $$$(b,a)$$$ have $$$0$$$ common pairings, while the ordered pairs $$$(a,a)$$$ and $$$(b,b)$$$ have $$$2$$$ common pairings. Therefore, the answer is $$$[2,0,2]$$$.</p></div>\n</problem-statement>\n\n\nSource lookup status:\n<source-lookup-status>\n- Tutorial (en): https://codeforces.com/blog/entry/156688 loaded (200 OK; title: \"Codeforces Round 1120 (Div 1, Div 2) Editorial - Codeforces\"; mentions 2262E)\n</source-lookup-status>\n\nGenerate:\n1. Five progressive hints, from a gentle nudge to the key insight.\n2. A self-contained editorial that teaches the solution clearly, including the key insight, algorithm, correctness argument, complexity, and necessary implementation details. Explain non-obvious steps without padding or a discovery-process transcript.\n3. A complete C++26 solution that gets AC on Codeforces.\n\nFormatting & Style Rules:\n- Hints and editorial: valid Markdown with LaTeX as needed (e.g., $dp[i]$, $$\\sum_{i=1}^{n} a_i$$).\n- Hints and editorial must read like Nudge's own explanation. No research notes, source notes, citations, Markdown links, URLs, or references to editorials/submissions/posts.\n- Solution: raw C++ only, no Markdown fences.\n- Keep C++ short, clean, standard, and single-file. Comments are fine when they clarify the idea.\n\nOutput strictness:\n- Return JSON matching the provided schema exactly.\n- Never include NUL (U+0000) characters in any field. Use literal $ or $$ for math delimiters, never control characters.\n- If the problem is solvable, return `status: \"success\"`, `reason: null`, and fill in `hints`, `editorial`, and `solution`.\n- Return `status: \"unsolvable\"`, a short `reason`, and set `hints`, `editorial`, and `solution` to null only when the supplied statement is fundamentally incomplete, contradictory, or dependent on an inaccessible resource needed to define the task. Missing research, high rating, uncertainty, or \"can't guarantee AC\" is not a reason to return unsolvable.\n- If you return `status: \"unsolvable\"`, the `reason` is shown directly to users. Be concrete: say what exact statement/resource blocker stopped you, and include relevant source lookup status if it mattered (for example: official tutorial missing, Codeforces returned 403/Cloudflare challenge, tutorial page 404, accepted submissions unavailable).\n- Each hint must be an object with `order` (1 through 5) and `content` (the Markdown hint text), not a plain string. The `content` must contain just the hint text, with no \"Hint 1:\" or subtitles. The UI adds those automatically.\n- Do not start the editorial with an \"# Editorial\" heading. The UI already adds that section. Feel free to add other headers as needed though.\n- Use styling in the editorial if needed! Just that first header is no good but any subheaders or bolding or whatever is good!\n\nFor the C++ solution, you MUST use this template and work around it:\n\n```cpp\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n}\n```",
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        "finish": "tool-calls",
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            "output": "---\nimage: https://codeforces.com/codeforces.org/s/85295/images/codeforces-telegram-square-1024x1024.png\ntitle: Codeforces Round 1120 (Div 1, Div 2) Editorial - Codeforces\ndescription: Codeforces. Programming competitions and contests, programming community\n---\n\n<!DOCTYPE HTML PUBLIC \"-//W3C//DTD HTML 4.01//EN\"> \n\n  \n[![Codeforces](//codeforces.com/codeforces.org/s/85295/images/codeforces-logo-with-telegram-555x130.png \"Codeforces\")](https://codeforces.com/) \n\n[![In English](//codeforces.com/codeforces.org/s/85295/images/flags/24/gb.png \"In English\")](https://codeforces.com/blog/entry/156688?locale=en) [![По-русски](//codeforces.com/codeforces.org/s/85295/images/flags/24/ru.png \"По-русски\")](https://codeforces.com/blog/entry/156688?locale=ru) \n\n[Enter](https://codeforces.com/enter?back=%2Fblog%2Fentry%2F156688)| [Register](https://codeforces.com/register) \n\n  \n* [Home](https://codeforces.com/)\n* [Top](https://codeforces.com/top)\n* [Catalog](https://codeforces.com/catalog)\n* [Contests](https://codeforces.com/contests)\n* [Gym](https://codeforces.com/gyms)\n* [Problemset](https://codeforces.com/problemset)\n* [Groups](https://codeforces.com/groups)\n* 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Errichto\") | 139      |\n| 6  | [adamant](https://codeforces.com/profile/adamant \"Master adamant\")                       | 136      |\n| 7  | [AmShZ](https://codeforces.com/profile/AmShZ \"Grandmaster AmShZ\")                        | 134      |\n| 8  | [maroonrk](https://codeforces.com/profile/maroonrk \"Legendary Grandmaster maroonrk\")     | 133      |\n| 9  | [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\") | 132      |\n| 10 | [qwexd](https://codeforces.com/profile/qwexd \"Master qwexd\")                             | 129      |\n\n|  | [View all →](https://codeforces.com/top-contributed) |\n|  | ---------------------------------------------------- |\n\n→ Find user \n\nHandle: \n\n→ Recent actions \n\n* [Kerolloz](https://codeforces.com/profile/Kerolloz \"Newbie Kerolloz\") → [Codeforces JSON-Problem-Scraper API](https://codeforces.com/blog/entry/75074) ![New 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Round 1119 (Div. 3) Editorial](https://codeforces.com/blog/entry/156457) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/85295/images/icons/comment-12x12.png \"New comment(s)\")\n* [MikeMirzayanov](https://codeforces.com/profile/MikeMirzayanov \"Headquarters, MikeMirzayanov\") → [Telegram Returns as Title Sponsor of Codeforces!](https://codeforces.com/blog/entry/156620) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/85295/images/icons/comment-12x12.png \"New comment(s)\")\n* [AmShZ](https://codeforces.com/profile/AmShZ \"Grandmaster AmShZ\") → [Repovive Premier Round 8 — $300 Prize Pool](https://codeforces.com/blog/entry/156827) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/85295/images/icons/comment-12x12.png \"New comment(s)\")\n* [DuyMinh3005](https://codeforces.com/profile/DuyMinh3005 \"Pupil DuyMinh3005\") → [Is this acceptable?](https://codeforces.com/blog/entry/155682) ![Text created or updated](//codeforces.com/codeforces.org/s/85295/images/icons/x-update-12x12.png \"Text created or updated\")\n* [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") → [Codeforces Round 1120 (Div. 1, Div. 2)](https://codeforces.com/blog/entry/156576) ![Text created or updated](//codeforces.com/codeforces.org/s/85295/images/icons/x-update-12x12.png \"Text created or updated\")\n* [Down\\_bad\\_4\\_haiender288](https://codeforces.com/profile/Down%5Fbad%5F4%5Fhaiender288 \"Pupil Down_bad_4_haiender288\") → [new manmade horrors (barely within comprehension) just dropped](https://codeforces.com/blog/entry/156825) ![Text created or updated](//codeforces.com/codeforces.org/s/85295/images/icons/x-update-12x12.png \"Text created or updated\")\n* [qwexd](https://codeforces.com/profile/qwexd \"Master qwexd\") → [Codeforces Round 1121 (Div. 2)](https://codeforces.com/blog/entry/156653) ![Text created or updated](//codeforces.com/codeforces.org/s/85295/images/icons/x-update-12x12.png \"Text created or updated\")\n* [lumibons](https://codeforces.com/profile/lumibons \"International Grandmaster lumibons\") → [IOI 2027 Call for Tasks](https://codeforces.com/blog/entry/156793) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/85295/images/icons/comment-12x12.png \"New comment(s)\")\n* [OmarAfterMidnight](https://codeforces.com/profile/OmarAfterMidnight \"Pupil OmarAfterMidnight\") → [VoidHub Update: Relive CPC Contests as If You Were There](https://codeforces.com/blog/entry/156807) ![Text created or updated](//codeforces.com/codeforces.org/s/85295/images/icons/x-update-12x12.png \"Text created or updated\")\n* [Grizoo](https://codeforces.com/profile/Grizoo \"Candidate Master Grizoo\") → [\\[GYM\\] Aleppo Collegiate Programming Contest 2026](https://codeforces.com/blog/entry/156707) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/85295/images/icons/comment-12x12.png \"New comment(s)\")\n* [pllk](https://codeforces.com/profile/pllk \"Grandmaster pllk\") → [Competitive Programmer's Handbook — a new book on competitive programming](https://codeforces.com/blog/entry/50728) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/85295/images/icons/comment-12x12.png \"New comment(s)\") ![Necropost](//codeforces.com/codeforces.org/s/85295/images/icons/hourglass.png \"Necropost\")\n* [sadwqdqEF213](https://codeforces.com/profile/sadwqdqEF213 \"Expert sadwqdqEF213\") → [Introducing SoSOI 2027 - The First Solar System Olympiad in Informatics](https://codeforces.com/blog/entry/156550) ![Text created or updated](//codeforces.com/codeforces.org/s/85295/images/icons/x-update-12x12.png \"Text created or updated\")\n* [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") → [Codeforces Round 1120 (Div 1, Div 2) Editorial](https://codeforces.com/blog/entry/156688) ![Text created or updated](//codeforces.com/codeforces.org/s/85295/images/icons/x-update-12x12.png \"Text created or updated\")\n* [adarshsolanki2004](https://codeforces.com/profile/adarshsolanki2004 \"Newbie adarshsolanki2004\") → [Cloudflare anti-bot verification not appearing](https://codeforces.com/blog/entry/151688) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/85295/images/icons/comment-12x12.png \"New comment(s)\")\n* [123gjweq2](https://codeforces.com/profile/123gjweq2 \"Master 123gjweq2\") → [Is this a good problem + can you please help me gauge its difficulty?](https://codeforces.com/blog/entry/156760) ![Text created or updated](//codeforces.com/codeforces.org/s/85295/images/icons/x-update-12x12.png \"Text created or updated\")\n\n|  | [Detailed →](https://codeforces.com/recent-actions) |\n|  | --------------------------------------------------- |\n\n* [kondasujay2](https://codeforces.com/profile/kondasujay2)\n* [Blog](https://codeforces.com/blog/kondasujay2)\n* [Teams](https://codeforces.com/teams/with/kondasujay2)\n* [Submissions](https://codeforces.com/submissions/kondasujay2)\n* [Groups](https://codeforces.com/groups/with/kondasujay2)\n* [Contests](https://codeforces.com/contests/with/kondasujay2)\n* [Problemsetting](https://codeforces.com/contests/writer/kondasujay2)\n\n### [kondasujay2's blog](https://codeforces.com/blog/kondasujay2)\n\n[ Codeforces Round 1120 (Div 1, Div 2) Editorial ](https://codeforces.com/blog/entry/156688) \n\nBy [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\"), [history](https://codeforces.com/topic/157332/en14), 4 days ago, ![In English](//codeforces.com/codeforces.org/s/85295/images/flags/24/gb.png \"In English\") \n\nThank you to everyone who participated! \n\nWe are sorry for the poor testcases on C1, along with C1 and C2's solutions being so similar to the recent Div3 E. \n\n[2263A - Min Max Game](https://codeforces.com/contest/2263/problem/A \"Codeforces Round 1120 (Div. 2)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nvoid tc() {\n    int n; cin >> n;\n    int cnt = 0;\n    vector<int> a(n);\n    for(int i = 0; i < n; i++) {\n        cin >> a[i];\n        cnt += a[i];\n    }\n    if(cnt >= n - cnt) {\n        cout << \"Bessie\\n\";\n    } else {\n        cout << \"Elsie\\n\";\n    }\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263B - Min Matrices](https://codeforces.com/contest/2263/problem/B \"Codeforces Round 1120 (Div. 2)\")\n\nAuthor: [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include<bits/stdc++.h>\n \nusing namespace std;\n \nvoid solve() {\n    int n, k; cin >> n >> k;\n    int x = k-n+1;\n    if(!(1 <= x && x <= n)) { cout << -1; return; }\n    vector<vector<int>> ans(n, vector<int>(n));\n    ans[0][0] = 1;\n    int on = 2;\n    for(int i = 1; i < x; i++) {\n        ans[0][i] = on; on++;\n        ans[i][0] = on; on++;\n    }\n    for(int i = x; i < n; i++) {\n        ans[i][i] = on; on++;\n    }\n\n    for(int i = 0; i < n; i++) {\n        for(int j = 0; j < n; j++) {\n            if(!ans[i][j]) {\n                ans[i][j] = on; on++;\n            }\n        }\n    }\n\n    for(int j = 0; j < n; j++) {\n        if(j) cout << \"\\n\";\n        for(auto &i : ans[j]) cout << i << \" \";\n    }\n}\n\nint main() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) {\n        solve();\n        if(t)   cout << \"\\n\";\n    }\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263C1 - Floor of MEX (Easy Version)](https://codeforces.com/contest/2263/problem/C1 \"Codeforces Round 1120 (Div. 2)\") / [2262A1 - Floor of MEX (Easy Version)](https://codeforces.com/contest/2262/problem/A1 \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [CutSandstone](https://codeforces.com/profile/CutSandstone \"Candidate Master CutSandstone\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nvoid tc() {\n    int n; cin >> n;\n    vector<int> b(n + 1);\n    for(int i = 1; i <= n; i++) cin >> b[i];\n    vector<pair<int, int>> goodivs;\n    vector<pair<int, int>> badivs;\n    for(int i = 1; i <= n; i++) {\n        badivs.push_back({b[i] * i, (b[i] + 1) * i - 1});\n        for(int j = 0; j < b[i]; j++) {\n            goodivs.push_back({j * i, (j + 1) * i - 1});\n        }\n    }\n    vector<int> d(n + 1);\n    for(auto [l, r] : badivs) {\n        d[min(n, l)]++;\n        d[min(n, r + 1)]--;\n    }\n    int csm = 0;\n    vector<int> gnums;\n    for(int i = 0; i < n; i++) {\n        csm += d[i];\n        if(csm == 0) {\n            gnums.push_back(i);\n        }\n    }\n    cout << gnums.size() << endl;\n    for(int i : gnums) cout << i << \" \";\n    cout << endl;\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263C2 - Floor of MEX (Hard Version)](https://codeforces.com/contest/2263/problem/C2 \"Codeforces Round 1120 (Div. 2)\") / [2262A2 - Floor of MEX (Hard Version)](https://codeforces.com/contest/2262/problem/A2 \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nconst int MOD = 1e9 + 7;\n\nvoid tc() {\n    int n; cin >> n;\n    vector<int> b(n + 1);\n    for(int i = 1; i <= n; i++) cin >> b[i];\n    vector<pair<int, int>> goodivs;\n    vector<pair<int, int>> badivs;\n    for(int i = 1; i <= n; i++) {\n        if(b[i] > (n - 1) / i + 1) {\n            cout << 0 << endl;\n            return;\n        }\n        badivs.push_back({b[i] * i, (b[i] + 1) * i - 1});\n        for(int j = 0; j < b[i]; j++) {\n            goodivs.push_back({j * i, (j + 1) * i - 1});\n        }\n    }\n    vector<int> d(n + 1);\n    for(auto [l, r] : badivs) {\n        d[min(n, l)]++;\n        d[min(n, r + 1)]--;\n    }\n    int csm = 0;\n    vector<int> gnums;\n    for(int i = 0; i < n; i++) {\n        csm += d[i];\n        if(csm == 0) {\n            gnums.push_back(i);\n        }\n    }\n    set<pair<int, int>> st;\n    sort(goodivs.begin(), goodivs.end(), [&] (pair<int, int> iv, pair<int, int> iv2) {\n        return iv.second - iv.first < iv2.second - iv2.first;\n    });\n    for(auto [l, r] : goodivs) {\n        auto it = st.lower_bound({l, 0});\n        if(it != st.end() && it->second <= r) {\n            continue;\n        }\n        st.insert({l, r});\n    }\n\n    vector<pair<int, int>> impivs(st.begin(), st.end());\n    int l = 0, r = 0;\n    vector<int> dp(impivs.size() + 1);\n    dp[0] = 1;\n    int poss = 1;\n    for(int i : gnums) {\n        while(l < impivs.size() && impivs[l].second < i) {\n            poss += MOD - dp[l];\n            poss %= MOD;\n            l++;\n        }\n        while(r < impivs.size() && impivs[r].first <= i) r++;\n        dp[r] += poss;\n        dp[r] %= MOD;\n        poss += poss;\n        poss %= MOD;\n    }\n    cout << dp[impivs.size()] << endl;\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263D - Culling Game](https://codeforces.com/contest/2263/problem/D \"Codeforces Round 1120 (Div. 2)\") / [2262B - Culling Game](https://codeforces.com/contest/2262/problem/B \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nstruct BIT {\n    vector<ll> tr;\n    BIT(int n) : tr(n + 1) {}\n    void add(int k, ll x) {\n        while(k < tr.size()) {\n            tr[k] += x;\n            k += k&-k;\n        }\n    }\n    ll sum(int k) {\n        ll res = 0;\n        while(k > 0) {\n            res += tr[k];\n            k -= k&-k;\n        }\n        return res;\n    }\n};\n\nvoid tc() {\n    int n; cin >> n;\n    vector<int> a(n), p(n);\n    for(int i = 0; i < n; i++) {\n        cin >> a[i];\n    }\n    for(int i = 0; i < n; i++) {\n        cin >> p[i]; p[i]--;\n    }\n    reverse(p.begin(), p.end());\n    BIT bit(n);\n    set<int> imp; \n    vector<int> ans(n);\n    for(int i = 0; i < n; i++) {\n        int k = p[i];\n        auto it = imp.upper_bound(k);\n        int pimp;\n        if(it == imp.begin() || bit.sum(k) - bit.sum(*prev(it)) < a[k]) {\n            imp.insert(k);\n            pimp = k;\n        } else {\n            pimp = *prev(it);\n        }\n        bit.add(k + 1, a[k]);\n        while(imp.upper_bound(k) != imp.end()) {\n            int j = *imp.upper_bound(k);\n            if(bit.sum(j) - bit.sum(pimp) >= a[j]) {\n                imp.erase(j);\n            } else {\n                break;\n            }\n        }\n        ans[i] = imp.size() - 1;\n    }\n    reverse(ans.begin(), ans.end());\n    for(int i : ans) cout << i << \" \";\n    cout << endl;\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263E - Traveling the World](https://codeforces.com/contest/2263/problem/E \"Codeforces Round 1120 (Div. 2)\") / [2262C - Traveling the World](https://codeforces.com/contest/2262/problem/C \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nconst int MOD = 1e9 + 7;\n\nvoid tc() {\n    int n; cin >> n;\n    vector<ll> a(n);\n    for(int i = 0; i < n; i++) cin >> a[i];\n    int k = n / 2;\n\n    int val = 1;\n    for(int i = 1; i <= k; i++) {\n        val = (ll)i * val % MOD;\n    }\n    for(int i = 1; i <= k - (n % 2 == 0 ? 2 : 1); i++) {\n        val = (ll)i * val % MOD;\n    }\n\n    set<ll> xposs;\n    xposs.insert(a[n - 1] - a[n - 2]);\n    xposs.insert(a[n - 1] - a[n - 3]);\n\n    int ans = 0;\n    for(ll x : xposs) {\n        vector<ll> b;\n        for(int i = 0; i < k; i++) {\n            b.push_back(i * x);\n        }\n        for(int i = 0; i < n - k; i++) {\n            b.push_back(a[n - 1] - i * x);\n        }\n        sort(b.begin(), b.end());\n        if(a == b) {\n            ans += val;\n            if(ans >= MOD) ans -= MOD;\n        }\n    }\n    cout << ans << endl;\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263F - PLUSworld](https://codeforces.com/contest/2263/problem/F \"Codeforces Round 1120 (Div. 2)\") / [2262D - PLUSworld](https://codeforces.com/contest/2262/problem/D \"Codeforces Round 1120 (Div. 1)\")\n\nAuthors: [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") & [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nvoid tc() {\n    int n; cin >> n;\n\n    vector<int> a(n), b(n);\n    for(int i = 0; i < n; i++) cin >> a[i], a[i]--;\n    for(int i = 0; i < n; i++) cin >> b[i], b[i]--;\n\n    int badcnt = 0;\n    for(int i = 0; i < n; i++) {\n        badcnt += a[i] != b[i];\n    }\n    if(badcnt == 0) {\n        cout << \"0 1\\n\";\n        return;\n    }\n\n    // mark everything which will ever be reached\n    vector<bool> useful(n);\n    for(int i = 0; i < n; i++) {\n        if(a[i] == b[i]) continue;\n        int x = i;\n        while(!useful[x]) {\n            useful[x] = true;\n            x = b[x];\n        }\n    }\n\n    // anything with b[i] == i once we reach we can't reach anything else, \n    // at most one of them can exit\n    vector<int> ends;\n    for(int i = 0; i < n; i++) {\n        if(useful[i] && b[i] == i) ends.push_back(i);\n    }\n\n    if(ends.size() > 1) {\n        cout << \"-1\\n\";\n        return;\n    }\n\n    vector<vector<int>> adj(n), radj(n);\n    for(int i = 0; i < n; i++) {\n        if(a[i] > b[i]) {\n            cout << \"-1\\n\";\n            return;\n        }\n        for(int j = a[i]; j <= b[i]; j++) {\n            adj[i].push_back(j);\n            radj[j].push_back(i);\n        }\n    }\n\n    // tarjan algorithm, from cp-algorithms\n    vector<int> st;    // - stack holding the unclaimed vertices\n    vector<int> roots(n, -1); // - keeps track of the SCC roots of the vertices\n    int timer = 0;         // - dfs timestamp counter\n    vector<int> t_in(n, -1);  // - keeps track of the dfs timestamp of the vertices\n    vector<int> t_low(n, -1); // - keeps track of the lowest t_in of unclaimed vertices\n    // reachable in the subtree\n\n    // implements the tarjan algorithm for strongly connected components\n    auto dfs = [&](auto&& self, int s) -> void {\n        t_low[s] = t_in[s] = timer++;\n        st.push_back(s);\n\n        for (auto u : adj[s]) {\n            if (t_in[u] == -1) { // tree-edge\n                self(self, u);\n                t_low[s] = min(t_low[s], t_low[u]);\n            } else if (roots[u] == -1) { // back-edge, cross-edge or forward-edge to an unclaimed vertex\n                t_low[s] = min(t_low[s], t_in[u]);\n            }\n        }\n\n        if (t_low[s] == t_in[s]) { // vertex is a root\n            while (true) {\n                int u = st.back();\n                st.pop_back();\n                roots[u] = s; // claims the vertex\n                if (u == s)\n                    break;\n            }\n        }\n    };\n\n    // applies the tarjan algorithm to all the vertices\n    for (int v = 0; v < n; v++) {\n        if (t_in[v] == -1) {\n            dfs(dfs, v);\n        }\n    }\n    vector<vector<int>> comps(n);\n    for(int v = 0; v < n; v++) {\n        comps[roots[v]].push_back(v);\n    }\n\n    vector<int> p(n, -1);\n    vector<int> sz(n);\n    for(int i = 0; i < n; i++) {\n        if(a[i] != b[i]) {\n            sz[roots[i]]++;\n        }\n        if(useful[i] && roots[i] != roots[b[i]]) {\n            if(p[roots[b[i]]] != -1) {\n                cout << \"-1\\n\";\n                return;\n            }\n            p[roots[b[i]]] = roots[i];\n        }\n    }\n    vector<int> chain;\n    int x = roots[ends[0]];\n    int csz = 0;\n    while(x != -1) {\n        chain.push_back(x);\n        csz += sz[x];\n        x = p[x];\n    }\n    if(csz != badcnt) {\n        cout << \"-1\\n\";\n        return;\n    }\n    reverse(chain.begin(), chain.end());\n\n    {\n        x = comps[chain[0]][0];\n        vector<int> ops;\n        auto op1 = [&] () {\n            ops.push_back(1);\n            a[x]++;\n        };\n        auto op2 = [&] () {\n            ops.push_back(2);\n            x = a[x];\n        };\n        auto solve_scc = [&] () {\n            int rt = roots[x];\n            vector<int> comp = comps[rt];\n\n            vector<int> p2(n, -1);\n            // only one thing in comp case\n            auto rdfs = [&] (auto&& self, int s) -> void {\n                for(int u : radj[s]) {\n                    if(roots[u] != rt) continue;\n                    if(p2[u] == -1) {\n                        p2[u] = s;\n                        self(self, u);\n                    }\n                }\n            };\n\n            // root a tree\n            rdfs(rdfs, comp[0]);\n\n            for(int i = 0; i < comp.size(); i++) {\n                while(x != comp[i]) {\n                    while(a[x] < max(p2[x], comp[i])) op1();\n                    op2();\n                }\n                while(a[x] < b[x]) op1();\n                op2();\n            }\n        };\n        for(int i = 0; i < chain.size(); i++) solve_scc();\n\n        cout << ops.size() << \" \" << comps[chain[0]][0] + 1 << endl;\n        for(int i : ops) cout << i << \" \";\n        cout << endl;\n\n        for(int i = 0; i < n; i++) assert(a[i] == b[i]);\n    }\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Magic**\n\nWith this, you can prove a significantly simpler solution. We will add this proof later.\n\n**Magic Solution**\n\n```\n// Note: AI generated solution\n#include <bits/stdc++.h>\nusing namespace std;\n\nint main() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n\n    int t;\n    cin >> t;\n\n    while (t--) {\n        int n;\n        cin >> n;\n\n        vector<int> a(n + 1), b(n + 1);\n        for (int i = 1; i <= n; i++) {\n            cin >> a[i];\n        }\n        for (int i = 1; i <= n; i++) {\n            cin >> b[i];\n        }\n\n        int start = 1, remaining = 0;\n        bool possible = true;\n\n        for (int i = 1; i <= n; i++) {\n            if (a[i] > b[i]) {\n                possible = false;\n            }\n            if (a[i] < b[i]) {\n                if (remaining == 0) {\n                    start = i;\n                }\n                remaining++;\n            }\n        }\n\n        if (!possible) {\n            cout << -1 << '\\n';\n            continue;\n        }\n\n        vector<int> operations;\n        int position = start;\n\n        while (remaining > 0) {\n            if (a[position] < b[position]) {\n                while (a[position] < b[position] &&\n                       (a[position] >= position ||\n                        a[a[position]] == b[a[position]])) {\n                    a[position]++;\n                    operations.push_back(1);\n                }\n\n                if (a[position] == b[position]) {\n                    remaining--;\n                }\n            }\n\n            if (remaining == 0 || a[position] == position) {\n                break;\n            }\n\n            operations.push_back(2);\n            position = a[position];\n        }\n\n        if (remaining > 0) {\n            cout << -1 << '\\n';\n            continue;\n        }\n\n        cout << operations.size() << ' ' << start << '\\n';\n        for (int operation : operations) {\n            cout << operation << ' ';\n        }\n        cout << '\\n';\n    }\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2262E - Paired Bracket Sequences](https://codeforces.com/contest/2262/problem/E \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nint MOD;\n\nvoid madd(int& a, int b) {\n    a += b;\n    if(a >= MOD) a -= MOD;\n}\n\nvoid msub(int& a, int b) {\n    a -= b;\n    if(a < 0) a += MOD;\n}\n\nint mmul(int a, int b) {\n    return 1LL * a * b % MOD;\n}\n\nint bpow(int x, int y) {\n    return y == 0 ? 1 : mmul((y % 2) ? x : 1, bpow(mmul(x, x), y / 2));\n}\n\nvoid tc() {\n    int n; cin >> n >> MOD;\n    vector<int> f(2 * n + 1), invf(2 * n + 1);\n    f[0] = 1;\n    for(int i = 1; i <= 2 * n; i++) {\n        f[i] = mmul(f[i - 1], i);\n    }\n    invf[2 * n] = bpow(f[2 * n], MOD - 2);\n    for(int i = 2 * n; i >= 1; i--) {\n        invf[i - 1] = mmul(invf[i], i);\n    }\n    auto F = [&] (int n, int k) -> int {\n        return mmul(f[n], mmul(invf[k + 1], invf[n - k]));\n    };\n    auto choose = [&] (int n, int k) -> int {\n        return mmul(f[n], mmul(invf[k], invf[n - k]));\n    };\n    vector<vector<int>> p(n + 1, vector<int>(n + 1));\n    for(int i = 0; i <= n; i++) {\n        p[0][i] = mmul(F(2 * i, i), F(2 * i, i));\n    }\n    for(int i = 1; i <= n; i++) {\n        for(int j = 0; j <= n; j++) {\n            for(int k = 0; j + k <= n; k++) {\n                madd(p[i][j + k], mmul(p[i - 1][j], p[0][k]));\n            }\n        }\n    }\n    vector<int> ans(n + 1);\n    for(int i = 0; i <= n; i++) {\n        ans[i] = mmul(F(2 * n, i), p[i][n - i]);\n    }\n    for(int i = n - 1; i >= 0; i--) {\n        for(int j = i + 1; j <= n; j++) {\n            msub(ans[i], mmul(ans[j], choose(j, i)));\n        }\n    }\n    for(int i = 0; i <= n; i++) {\n        cout << ans[i] << \" \";\n    }\n    cout << endl;\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t; \n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2262F - Rank Removal](https://codeforces.com/contest/2262/problem/F \"Codeforces Round 1120 (Div. 1)\")\n\nAuthors: [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") & [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nconst int MXN = 400;\n\ntemplate<size_t SZ> int compute_inv(const vector<bitset<SZ>>& mat, vector<bitset<SZ>>& inv) {\n    int n = mat.size();\n    vector<bool> empty(n, true);\n    vector<bitset<SZ>> basis(n);\n\n    for(int i = 0; i < n; i++) {\n        bitset<SZ> cur = mat[i];\n        bitset<SZ> curinv;\n        curinv[i] = 1;\n        for(int j = 0; j < n; j++) {\n            if(cur[j]) {\n                if(empty[j]) {\n                    basis[j] = cur;\n                    inv[j] = curinv;\n                    empty[j] = false;\n                    break;\n                } else {\n                    cur ^= basis[j];\n                    curinv ^= inv[j];\n                }\n            }\n        }\n    }\n    for(int i = 0; i < n; i++) {\n        if(empty[i]) return 0;\n    }\n\n    for(int i = n - 1; i >= 1; i--) {\n        for(int j = 0; j < i; j++) {\n            if(basis[j][i]) {\n                basis[j] ^= basis[i];\n                inv[j] ^= inv[i];\n            }\n        }\n    }\n    return 1;\n}\n\nbool try_removal(vector<pair<int, int>> cells, vector<bitset<MXN>>& mat, vector<bitset<MXN>>& inv, vector<int>& row_cnt) {\n    int n = mat.size(), k = cells.size();\n    for(auto [x, y] : cells) {\n        if(mat[x][y] == 0) {\n            return false;\n        }\n    }\n    assert(k < 4);\n    vector<bitset<3>> K(k); \n    for(int i = 0; i < k; i++) {\n        for(int j = 0; j < k; j++) {\n            K[i][j] = (i == j) ^ inv[cells[i].second][cells[j].first];\n        }\n    }\n    vector<bitset<3>> invK(cells.size());\n    if(!compute_inv(K, invK)) {\n        return false;\n    }\n    for(auto [x, y] : cells) {\n        mat[x][y] = 0;\n        row_cnt[x]--;\n    }\n\n    auto old = inv;\n    for(int i = 0; i < n; i++) {\n        bitset<3> coef;\n        for(int alpha = 0; alpha < k; alpha++) {\n            int r = cells[alpha].first;\n\n            if(old[i][r]) {\n                coef ^= invK[alpha];\n            }\n        }\n\n        // delta = coef * (V^T * old)\n        for(int beta = 0; beta < k; beta++) {\n            if(coef[beta]) {\n                int c = cells[beta].second;\n                inv[i] ^= old[c];\n            }\n        }\n    }\n    return true;\n}\n\nvector<int> compute_matching(const vector<bitset<MXN>>& mat) {\n    int n = mat.size();\n    vector<vector<int>> adj(n);\n    for(int i = 0; i < n; i++) {\n        for(int j = 0; j < n; j++) {\n            if(mat[i][j]) adj[i].push_back(j);\n        }\n    }\n\n    vector<int> match(n, -1);\n    vector<bool> done(n);\n\n    function<bool(int)> dfs = [&] (int s) {\n        for(int u : adj[s]) {\n            if(!done[u]) {\n                done[u] = true;\n\n                if(match[u] == -1 || dfs(match[u])) {\n                    match[u] = s;\n                    return true;\n                }\n            }\n        }\n        return false;\n    };\n\n    for(int i = 0; i < n; i++) {\n        fill(done.begin(), done.end(), false);\n        dfs(i);\n    }\n\n    for(int i = 0; i < n; i++) {\n        assert(match[i] != -1);\n    }\n    return match;\n}\n\nvoid tc() {\n    int n, m; cin >> n >> m;\n\n    vector<bitset<MXN>> mat(n);\n    vector<int> row_cnt(n);\n    for(int i = 0; i < m; i++) {\n        int x, y; cin >> x >> y;\n        x--, y--;\n        mat[x][y] = 1;\n        row_cnt[x]++;\n    }\n\n    vector<bitset<MXN>> inv(n);\n    compute_inv(mat, inv);\n\n    int sm = m;\n    int left = (sm + n - 1) / n;\n    cout << left << '\\n';\n\n    while(sm > 0) {\n        vector<bitset<MXN>> omat = mat;\n        if(sm <= n) {\n            for(int i = 0; i < n; i++) {\n                for(int j = 0; j < n; j++) {\n                    mat[i][j] = 0;\n                }\n            }\n            row_cnt.assign(n, 0);\n        } else if(sm <= 2 * n) {\n            vector<int> matching = compute_matching(mat);\n            for(int i = 0; i < n; i++) {\n                for(int j = 0; j < n; j++) {\n                    mat[i][j] = 0;\n                }\n            }\n            row_cnt.assign(n, 0);\n            for(int i = 0; i < sm - n; i++) {\n                mat[matching[i]][i] = 1;\n                row_cnt[matching[i]]++;\n            }\n        } else if(sm <= 3 * n) {\n            vector<int> matching = compute_matching(mat);\n            int above = 0;\n            for(int i = 0; i < n; i++) {\n                for(int j = matching[i] + 1; j < n; j++) {\n                    above += mat[j][i];\n                }\n            }\n\n            vector<bitset<MXN>> nmat(n);\n            vector<int> nrow_cnt(n);\n            int ones_left = sm - n;\n            for(int i = 0; i < n; i++) {\n                nmat[matching[i]][i] = 1;\n                nrow_cnt[matching[i]]++;\n                ones_left--;\n            }\n            // make matrix either upper or lower triangular, whichever one\n            // removes a sufficient number of ones\n            if(above >= ones_left) {\n                for(int i = 0; i < n; i++) {\n                    for(int j = matching[i] + 1; j < n; j++) {\n                        if(mat[j][i]) {\n                            ones_left--;\n                            nmat[j][i] = 1;\n                            nrow_cnt[j]++;\n                            if(ones_left == 0) break;\n                        }\n                    }\n                    if(ones_left == 0) break;\n                }\n            } else {\n                for(int i = 0; i < n; i++) {\n                    for(int j = 0; j < matching[i]; j++) {\n                        if(mat[j][i]) {\n                            ones_left--;\n                            nmat[j][i] = 1;\n                            nrow_cnt[j]++;\n                            if(ones_left == 0) break;\n                        }\n                    }\n                    if(ones_left == 0) break;\n                }\n            }\n            mat = nmat;\n            row_cnt = nrow_cnt;\n        } else {\n            int csm = sm;\n            if(n % 2) {\n                // odd case (remove either 1 or 3)\n                int dup_row = -1;\n                for(int i = 0; i < n; i++) {\n                    if(row_cnt[i] >= 2) {\n                        dup_row = i;\n                        break;\n                    }\n                }\n                assert(dup_row != -1);\n\n                vector<int> cols;\n                for(int i = 0; i < n; i++) {\n                    if(mat[dup_row][i]) cols.push_back(i);\n                }\n\n                assert(cols.size() >= 2);\n\n                for(int i = 0; i < n; i++) {\n                    if(try_removal({{i, cols[0]}}, mat, inv, row_cnt)) {\n                        csm--;\n                        break;\n                    }\n                    if(try_removal({{i, cols[1]}}, mat, inv, row_cnt)) {\n                        csm--;\n                        break;\n                    }\n                }\n\n                if(csm == sm) {\n                    for(int i = 0; i < n; i++) {\n                        if(i != dup_row) {\n                            // remove L shape\n                            if(try_removal(\n                                    {{i, cols[0]}, \n                                     {dup_row, cols[0]}, \n                                     {dup_row, cols[1]}},\n                                mat, inv, row_cnt)) {\n                                csm -= 3;\n                                break;\n                            }\n                        }\n                    }\n                    assert(sm != csm);\n                }\n            }\n\n            while(csm > sm - n) {\n                // remove 2 at a time\n                int dup_row = -1;\n                for(int i = 0; i < n; i++) {\n                    if(row_cnt[i] >= 3) {\n                        dup_row = i;\n                        break;\n                    }\n                }\n\n                vector<int> cols;\n                for(int i = 0; i < n; i++) {\n                    if(mat[dup_row][i]) cols.push_back(i);\n                }\n\n                assert(cols.size() >= 3);\n\n                if(!try_removal({{dup_row, cols[0]}, {dup_row, cols[1]}}, mat, inv, row_cnt)) {\n                    if(!try_removal({{dup_row, cols[1]}, {dup_row, cols[2]}}, mat, inv, row_cnt)) {\n                        if(!try_removal({{dup_row, cols[0]}, {dup_row, cols[2]}}, mat, inv, row_cnt)) {\n                            assert(false);\n                        }\n                    }\n                }\n\n                csm -= 2;\n            }\n            assert(csm == sm - n);\n        }\n\n        vector<pair<int, int>> move;\n        for(int i = 0; i < n; i++) {\n            for(int j = 0; j < n; j++) {\n                if(omat[i][j] == 1 && mat[i][j] == 0) {\n                    move.push_back({i, j});\n                }\n            }\n        }\n        cout << move.size() << \"\\n\";\n        for(auto [x, y] : move) cout << x + 1 << \" \" << y + 1 << \"\\n\";\n        sm -= n;\n\n        int csm = 0;\n        for(int i = 0; i < n; i++) {\n            csm += row_cnt[i];\n        }\n        assert(csm == max(0, sm));\n    }\n}\n\nint main() {\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Tutorial 2**\n\nCredit to [BurnedChicken](https://codeforces.com/profile/BurnedChicken \"International Grandmaster BurnedChicken\") for finding this in testing!\n\nWe will first prove that for any row/column of a rank $$$n$$$ matrix, we can zero out everything in that row/column except one, and maintain rank $$$n$$$. This is because rank $$$n$$$ means the determinant is $$$1$$$, so there are exactly an odd number of perfect matchings in total (read Solution 1 for more understanding). Therefore at least $$$1$$$ of the ones in the row/column must be in an odd number of perfect matchings (since if they were all even then the total number would be even). This means if we delete all other ones but that one, then we would keep rank $$$n$$$.\n\nThis actually all you need to prove to solve the problem. Note that after we clear out one row except one number, we can clear out that column except that one number (even partially) and the determinant will not change (and the same works when clearing out a column with the leftover row). \n\nFor every turn, we want to try to find $$$n$$$ things to remove without changing the rank. Therefore, we begin by finding the one in the biggest row/column we can keep while removing everything else. Then we remove everything in the column/row of that one, and the everything in the row/column into the queue. If the queue + the current removals is big enough to reach $$$n$$$ removals, then we can just remove the ones in the queue. If the queue is too small, then we repeat one the next biggest row/column. Because the queue contains the row/column with the most ones, when the current removals will never exceed $$$n$$$ if the the queue + current removals is less than $$$n$$$. Once we run out of row/columns to clear, we can start removing the ones we excluded removing, since we have to reduce the rank at this point.\n\nTo find the one we must keep, you have to utilize the inverse, since we can clear out the row/column on $$$M\\_{i, j}$$$ if $$$M^{-1}\\_{j, i} = 1$$$ (proof is by Cramer's Rule). We can keep track of the inverse after this clear out operation in $$$O(n^2 / w)$$$ if done properly (using something similar to Woodbury Matrix Identity, look at Solution 1 for details). If implemented properly, this results in a $$$O(n^3 / w)$$$ solution.\n\n**Solution 2**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n \nconst int SZ = 400;\n \n \ntemplate<size_t SZ> int compute_inv(const vector<bitset<SZ>>& mat, vector<bitset<SZ>>& inv) {\n    int n = mat.size();\n    vector<bool> empty(n, true);\n    vector<bitset<SZ>> basis(n);\n \n    for(int i = 0; i < n; i++) {\n        bitset<SZ> cur = mat[i];\n        bitset<SZ> curinv;\n        curinv[i] = 1;\n        for(int j = 0; j < n; j++) {\n            if(cur[j]) {\n                if(empty[j]) {\n                    basis[j] = cur;\n                    inv[j] = curinv;\n                    empty[j] = false;\n                    break;\n                } else {\n                    cur ^= basis[j];\n                    curinv ^= inv[j];\n                }\n            }\n        }\n    }\n    for(int i = 0; i < n; i++) {\n        if(empty[i]) return 0;\n    }\n \n    for(int i = n - 1; i >= 1; i--) {\n        for(int j = 0; j < i; j++) {\n            if(basis[j][i]) {\n                basis[j] ^= basis[i];\n                inv[j] ^= inv[i];\n            }\n        }\n    }\n    return 1;\n}\n \nvoid tc() {\n    int n, m; cin >> n >> m;\n    vector<bitset<SZ>> mat(n);\n    for(int i = 0; i < m; i++) {\n        int x, y; cin >> x >> y;\n        x--, y--;\n        mat[x][y] = 1;\n    }\n \n    vector<bitset<SZ>> inv(n);\n    compute_inv(mat, inv);\n    vector<int> cnts(2 * n);\n    for(int i = 0; i < n; i++) {\n        for(int j = 0; j < n; j++) {\n            cnts[i] += mat[i][j];\n            cnts[n + j] += mat[i][j];\n        }\n    }\n    queue<pair<int, int>> mustq, freeq;\n    vector<pair<int, int>> centers;\n    vector<vector<pair<int, int>>> moves;\n    while(true) {\n        while(mustq.size() + freeq.size() < n) {\n            int best_rc = 0;\n            for(int rc = 0; rc < 2 * n; rc++) {\n                if(cnts[rc] >= cnts[best_rc]) best_rc = rc;\n            }\n            if(cnts[best_rc] == 0) {\n                break;\n            }\n \n            int x = -1, y = -1;\n            if(best_rc < n) {\n                x = best_rc;\n                for(int j = 0; j < n; j++) {\n                    if(mat[x][j] && inv[j][x]) {\n                        y = j;\n                        break;\n                    }\n                }\n            } else {\n                y = best_rc - n;\n                for(int j = 0; j < n; j++) {\n                    if(mat[j][y] && inv[y][j]) {\n                        x = j;\n                        break;\n                    }\n                }\n            }\n            assert(x != -1 && y != -1);\n \n            centers.push_back({x, y});\n            vector<pair<int, int>> must, free;\n            auto erase = [&] (int a, int b) {\n                mat[a][b] = 0;\n                cnts[a]--;\n                cnts[b + n]--;\n            };\n            for(int j = 0; j < n; j++) {\n                if(mat[x][j] && j != y) {\n                    must.push_back({x, j});\n                    erase(x, j);\n                }\n            }\n            for(int j = 0; j < n; j++) {\n                if(mat[j][y] && j != x) {\n                    free.push_back({j, y});\n                    erase(j, y);\n                }\n            }\n            erase(x, y);\n            assert(cnts[x] == 0 && cnts[n + y] == 0);\n \n            if(best_rc < n) swap(must, free);\n            for(auto u : must) mustq.push(u);\n            for(auto u : free) freeq.push(u);\n \n            bitset<SZ> row = inv[y];\n            for(int j = 0; j < n; j++) {\n                if(cnts[j + n] > 0 && inv[j][x]) {\n                    inv[j] ^= row;\n                }\n            }\n        }\n        int left = n;\n        vector<pair<int, int>> move;\n        while(left > 0 && mustq.size() > 0) {\n            left--;\n            move.push_back(mustq.front());\n            mustq.pop();\n        }\n        while(left > 0 && freeq.size() > 0) {\n            left--;\n            move.push_back(freeq.front());\n            freeq.pop();\n        }\n        while(left > 0 && centers.size() > 0) {\n            left--;\n            move.push_back(centers.back());\n            centers.pop_back();\n        }\n        moves.push_back(move);\n        if(centers.size() == 0) break;\n    }\n    cout << moves.size() << \"\\n\";\n    for(auto move : moves) {\n        cout << move.size() << \"\\n\";\n        for(auto [u, v] : move) cout << u + 1 << \" \" << v + 1 << \"\\n\";\n    }\n}\n \nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n![](//codeforces.com/codeforces.org/s/85295/images/icons/paperclip-16x16.png) Tutorial of [Codeforces Round 1120 (Div. 1)](https://codeforces.com/contest/2262) \n\n\n...107981 bytes truncated...\n\nThe tool call succeeded but the output was truncated. 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              "path": "/Users/zayd/.local/share/opencode/tool-output/tool_0adf27ac4001aV221K6Nz3MgLR"
            },
            "output": "Found 100 matches (more matches available)\n/Users/zayd/.local/share/opencode/tool-output/tool_0acc2e452001c5MTVZp6cLP82T:\n  Line 96: * [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") → [Codeforces Round 1120 (Div 1, Div 2) Editorial](https://codeforces.com/blog/entry/156688) ![Text created or updated](//codeforces.com/codeforces.org/s/21069/images/icons/x-update-12x12.png \"Text created or updated\")\n\n  Line 830: [2262E - Paired Bracket Sequences](https://codeforces.com/contest/2262/problem/E \"Codeforces Round 1120 (Div. 1)\")\n\n\n/Users/zayd/.local/share/opencode/tool-output/tool_0acba1ffb001a0trOfQpBMs5wJ:\n  Line 96: * [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") → [Codeforces Round 1120 (Div 1, Div 2) Editorial](https://codeforces.com/blog/entry/156688) ![Text created or updated](//codeforces.com/codeforces.org/s/21069/images/icons/x-update-12x12.png \"Text created or updated\")\n\n\n/Users/zayd/.local/share/opencode/tool-output/tool_0acba132f001X5WsyrsszK4JAN:\n  Line 96: * [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") → [Codeforces Round 1120 (Div 1, Div 2) Editorial](https://codeforces.com/blog/entry/156688) ![Text created or updated](//codeforces.com/codeforces.org/s/30864/images/icons/x-update-12x12.png \"Text created or updated\")\n\n\n/Users/zayd/.local/share/opencode/tool-output/tool_0acb98e53001Pv9q24kB4837NW:\n  Line 96: * [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") → [Codeforces Round 1120 (Div 1, Div 2) Editorial](https://codeforces.com/blog/entry/156688) ![Text created or updated](//codeforces.com/codeforces.org/s/30864/images/icons/x-update-12x12.png \"Text created or updated\")\n\n\n/Users/zayd/.local/share/opencode/tool-output/tool_0acbdffb4001nW2Z0Ox3Ty4m2t:\n  Line 12: [![In English](//codeforces.com/codeforces.org/s/30864/images/flags/24/gb.png \"In English\")](https://codeforces.com/blog/entry/156688?locale=en) [![По-русски](//codeforces.com/codeforces.org/s/30864/images/flags/24/ru.png \"По-русски\")](https://codeforces.com/blog/entry/156688?locale=ru) \n\n  Line 14: [Enter](https://codeforces.com/enter?back=%2Fblog%2Fentry%2F156688)| [Register](https://codeforces.com/register) \n\n  Line 96: * [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") → [Codeforces Round 1120 (Div 1, Div 2) Editorial](https://codeforces.com/blog/entry/156688) ![Text created or updated](//codeforces.com/codeforces.org/s/30864/images/icons/x-update-12x12.png \"Text created or updated\")\n\n  Line 118: [ Codeforces Round 1120 (Div 1, Div 2) Editorial ](https://codeforces.com/blog/entry/156688) \n\n  Line 840: [2262E - Paired Bracket Sequences](https://codeforces.com/contest/2262/problem/E \"Codeforces Round 1120 (Div. 1)\")\n\n  Line 1457: * [![Comments](//codeforces.com/codeforces.org/s/30864/images/blog/comments_16x16.png \"Comments\")](https://codeforces.com/blog/entry/156688#comments) [55 ](https://codeforces.com/blog/entry/156688#comments)\n\n  Line 1467: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/kondasujay2) [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392138 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +1 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) _Auto comment: topic has been updated by [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") ([previous revision](https://codeforces.com/topic/157332/en6), [new revision](https://codeforces.com/topic/157332/en7), [compare](https://codeforces.com/topic/157332/diff/en6/en7))._ [→](#) [Reply](#) |\n\n  Line 1473: | » » [ ![](https://codeforces.com/userpic.codeforces.org/2572984/avatar/6e314e22b534732f.jpg) ](https://codeforces.com/profile/Askew%5Fbreadcrumb6) [Askew\\_breadcrumb6](https://codeforces.com/profile/Askew%5Fbreadcrumb6 \"Newbie Askew_breadcrumb6\") | 38 hours ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392736 \"Link to comment\") [^](#comment-1392138 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) In Problem C2's official solution, for(int i = 1; i <= n; i++) {        badivs.push\\_back({b\\[i\\] \\* i, (b\\[i\\] + 1) \\* i - 1});     for(int j = 0; j < b\\[i\\]; j++) {        goodivs.push\\_back({j \\* i, (j + 1) \\* i - 1});     } } How do you know this will not give a TLE? In the sense that the worst case time complexity of this loop is n^2 right? [→](#) [Reply](#) |\n\n  Line 1479: | » » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/chen%5Fweilian) [chen\\_weilian](https://codeforces.com/profile/chen%5Fweilian \"Master chen_weilian\") | 15 hours ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392862 \"Link to comment\") [^](#comment-1392736 \"Parent comment\") \\| [←](#)Rev. 3 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) The problem states that a solution must exist, which implies that a\\[i\\] <= (n-1) / i + 1 because numbers are at most n-1\\. Summing it up give a O(n\\*log n) complexity [→](#) [Reply](#) |\n\n  Line 1485: | » » » » [ ![](https://codeforces.com/userpic.codeforces.org/2572984/avatar/6e314e22b534732f.jpg) ](https://codeforces.com/profile/Askew%5Fbreadcrumb6) [Askew\\_breadcrumb6](https://codeforces.com/profile/Askew%5Fbreadcrumb6 \"Newbie Askew_breadcrumb6\") | 14 hours ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392865 \"Link to comment\") [^](#comment-1392862 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Oh that's right! Thank you!! [→](#) [Reply](#) |\n\n  Line 1491: | » [ ![](https://codeforces.com/userpic.codeforces.org/3913197/avatar/9ddf748287465df7.jpg) ](https://codeforces.com/profile/sukon) [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392142 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) _Auto comment: topic has been updated by [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") ([previous revision](https://codeforces.com/topic/157332/en7), [new revision](https://codeforces.com/topic/157332/en8), [compare](https://codeforces.com/topic/157332/diff/en7/en8))._ [→](#) [Reply](#) |\n\n  Line 1497: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/bakaasta) [bakaasta](https://codeforces.com/profile/bakaasta \"Newbie bakaasta\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392144 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +1 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Nice Contest [→](#) [Reply](#) |\n\n  Line 1503: | » [ ![](https://codeforces.com/userpic.codeforces.org/2948677/avatar/f15474cd5b82d030.jpg) ](https://codeforces.com/profile/AksLolCoding) [AksLolCoding](https://codeforces.com/profile/AksLolCoding \"Grandmaster AksLolCoding\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392148 \"Link to comment\") \\| [←](#)Rev. 2 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +16 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Div1A2/Div2C2 can be solved in $$$O(n\\\\log n)$$$ via counting sort if you store the smallest interval for each right endpoint. [→](#) [Reply](#) |\n\n  Line 1509: | » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/tfg) [tfg](https://codeforces.com/profile/tfg \"Master tfg\") | 3 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392327 \"Link to comment\") [^](#comment-1392148 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Remove intervals that have smaller intervals that are completely inside it and there remains O(N) intervals. I solved the DP part in O(N) using prefix sums and the only part is O(NlogN) to iterate through the initial intervals to find out for each L the smallest R of an interval starting at that L. [→](#) [Reply](#) |\n\n  Line 1515: | » [ ![](https://codeforces.com/userpic.codeforces.org/3591959/avatar/9d6d82e27c30d45d.jpg) ](https://codeforces.com/profile/mnnit.prakharg) [mnnit.prakharg](https://codeforces.com/profile/mnnit.prakharg \"Specialist mnnit.prakharg\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392152 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +20 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) For C2, my submission: [390474676](https://codeforces.com/contest/2263/submission/390474676 \"Submission 390474676 by mnnit.prakharg\"). It works in O(nlogn) time comp. and O(n) space comp.My DP: I have processed till i (0 <= i < n), and I want to find how many good sets exist with the largest element being i. Note till now, we are not considering intervals whose right endpoint is bigger that i.For transition, I will be summing up the dp\\[j\\] such that j < i, j represents what is the second largest element of the set, and there is no interval completely inside \\[j + 1, i — 1\\]. j will form a contiguous range, so we can use prefix sum on dp values.Thus, dp calc. part is O(n), but preprocessing intervals take O(nlogn) time.Final answer: let pos represents the maximum left endpoint of any interval in which at least one number should lie. ans = dp\\[pos\\] + dp\\[pos + 1\\] + ... + dp\\[n — 1\\]. [→](#) [Reply](#) |\n\n  Line 1521: | » » [ ![](https://codeforces.com/userpic.codeforces.org/5132546/avatar/14b4a73eb8b7dff1.jpg) ](https://codeforces.com/profile/fakeapple) [fakeapple](https://codeforces.com/profile/fakeapple \"Master fakeapple\") | 3 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392332 \"Link to comment\") [^](#comment-1392152 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I also solved it similarily :D<https://codeforces.com/contest/2262/submission/390455393> [→](#) [Reply](#) |\n\n  Line 1527: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/dooglius) [dooglius](https://codeforces.com/profile/dooglius \"Expert dooglius\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392153 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +8 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) \"Rate the problem\" appears to have all the problems sharing the same rating votes [→](#) [Reply](#) |\n\n  Line 1533: | » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/kondasujay2) [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392154 \"Link to comment\") [^](#comment-1392153 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +1 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) wow im stupid i will fix [→](#) [Reply](#) |\n\n  Line 1539: | » [ ![](https://codeforces.com/userpic.codeforces.org/3913197/avatar/9ddf748287465df7.jpg) ](https://codeforces.com/profile/sukon) [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392156 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) _Auto comment: topic has been updated by [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") ([previous revision](https://codeforces.com/topic/157332/en8), [new revision](https://codeforces.com/topic/157332/en9), [compare](https://codeforces.com/topic/157332/diff/en8/en9))._ [→](#) [Reply](#) |\n\n  Line 1545: | » [ ![](https://codeforces.com/userpic.codeforces.org/3913197/avatar/9ddf748287465df7.jpg) ](https://codeforces.com/profile/sukon) [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392157 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) _Auto comment: topic has been updated by [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") ([previous revision](https://codeforces.com/topic/157332/en11), [new revision](https://codeforces.com/topic/157332/en12), [compare](https://codeforces.com/topic/157332/diff/en11/en12))._ [→](#) [Reply](#) |\n\n  Line 1551: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/PiggyPump) [PiggyPump](https://codeforces.com/profile/PiggyPump \"Pupil PiggyPump\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392159 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Not a bad contest, but it was indeed tough, thx for the rating boost, but I mean it was like:A: 800 — 900 B: 1000 — 1100 C1: 1300, maybe 1400, because I had to use Seg Tree to quickly tell me which ranges of values needed to be excluded (but it wasn't necessary because you didn't have to do updates between queries)C2: 1700+, I read it and it seemed kinda cooked, like I don't know how to wrap my head around all the conditions that had to be satisifed -- and how to count the number of ways to satisfy all constraints. D: 2000+, read it and realized that your favorite pupil had never seen this kind of bullshit before.E, F: ??? [→](#) [Reply](#) |\n\n  Line 1557: | » » [ ![](https://codeforces.com/userpic.codeforces.org/3066219/avatar/fb18c426e88a104c.jpg) ](https://codeforces.com/profile/mrinmoy66sadhukhan) [mrinmoy66sadhukhan](https://codeforces.com/profile/mrinmoy66sadhukhan \"Expert mrinmoy66sadhukhan\") | 3 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392319 \"Link to comment\") [^](#comment-1392159 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) \"never seen this kind of bullshit..\" Lmao. [→](#) [Reply](#) |\n\n  Line 1562: | » » [ ![](https://codeforces.com/userpic.codeforces.org/4593468/avatar/1d2e2455af09c340.jpg) ](https://codeforces.com/profile/Geek%5FGamBit) [Geek\\_GamBit](https://codeforces.com/profile/Geek%5FGamBit \"Newbie Geek_GamBit\") | 20 hours ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392842 \"Link to comment\") [^](#comment-1392159 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Goated feedback LOL [→](#) [Reply](#) |\n\n  Line 1568: | » [ ![](https://codeforces.com/userpic.codeforces.org/3913197/avatar/9ddf748287465df7.jpg) ](https://codeforces.com/profile/sukon) [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392162 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) _Auto comment: topic has been updated by [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") ([previous revision](https://codeforces.com/topic/157332/en12), [new revision](https://codeforces.com/topic/157332/en13), [compare](https://codeforces.com/topic/157332/diff/en12/en13))._ [→](#) [Reply](#) |\n\n  Line 1574: | » [ ![](https://codeforces.com/userpic.codeforces.org/5347974/avatar/13eac3d99d6f3189.jpg) ](https://codeforces.com/profile/Sean222299) [Sean222299](https://codeforces.com/profile/Sean222299 \"Newbie Sean222299\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392163 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) C1 is hard for me :< [→](#) [Reply](#) |\n\n  Line 1580: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/misytnikita) [misytnikita](https://codeforces.com/profile/misytnikita \"Newbie misytnikita\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392173 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) B tricky math and realization, C1 really mathy [→](#) [Reply](#) |\n\n  Line 1586: | » » [ ![](https://codeforces.com/userpic.codeforces.org/5268131/avatar/fa8bb1ed65c58067.jpg) ](https://codeforces.com/profile/Krish-006-19) [Krish-006-19](https://codeforces.com/profile/Krish-006-19 \"Newbie Krish-006-19\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392199 \"Link to comment\") [^](#comment-1392173 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +1 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) i found out, in min k all diagonal things, and as k decreases we move minimum non diagonal elements to the diagonal and we keep on doing till k==n, and the max k happens when it goes, n=3, 1 2 3 4 5 6 7 8 9 here k is the max, whereas 1 5 9 4 2 6 7 8 3 here k=3hence it is bit observation, not that much math, the only thing u gotta do is to swap mat\\[i\\]\\[i\\] & mat\\[0\\]\\[i\\] and you gonna get ans [→](#) [Reply](#) |\n\n  Line 1592: | » » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/misytnikita) [misytnikita](https://codeforces.com/profile/misytnikita \"Newbie misytnikita\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392201 \"Link to comment\") [^](#comment-1392199 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) i have the same idea, but i get one RE with realization [→](#) [Reply](#) |\n\n  Line 1598: | » [ ![](https://codeforces.com/userpic.codeforces.org/4518190/avatar/63650568eec18d6.jpg) ](https://codeforces.com/profile/potatoArmy) [potatoArmy](https://codeforces.com/profile/potatoArmy \"Specialist potatoArmy\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392175 \"Link to comment\") \\| [←](#)Rev. 2 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) in C1 and C2, i think its difference array, not prefix sum right? [→](#) [Reply](#) |\n\n  Line 1604: | » » [ ![](https://codeforces.com/userpic.codeforces.org/5870632/avatar/b959e58fb1f45619.jpg) ](https://codeforces.com/profile/xiaolanjing) [xiaolanjing](https://codeforces.com/profile/xiaolanjing \"Newbie xiaolanjing\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392177 \"Link to comment\") [^](#comment-1392175 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Use a difference array to mark the intervals that are not allowed [→](#) [Reply](#) |\n\n  Line 1610: | » » » [ ![](https://codeforces.com/userpic.codeforces.org/4518190/avatar/63650568eec18d6.jpg) ](https://codeforces.com/profile/potatoArmy) [potatoArmy](https://codeforces.com/profile/potatoArmy \"Specialist potatoArmy\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392178 \"Link to comment\") [^](#comment-1392177 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) ik, but editorial says prefix sum, i think it should be difference array [→](#) [Reply](#) |\n\n  Line 1616: | » » » » [ ![](https://codeforces.com/userpic.codeforces.org/5870632/avatar/b959e58fb1f45619.jpg) ](https://codeforces.com/profile/xiaolanjing) [xiaolanjing](https://codeforces.com/profile/xiaolanjing \"Newbie xiaolanjing\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392181 \"Link to comment\") [^](#comment-1392178 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) After marking the forbidden intervals with a difference array, take prefix sums to determine which positions are covered.That's probably what the editorial meant^\\_^ [→](#) [Reply](#) |\n\n  Line 1622: | » » » » » [ ![](https://codeforces.com/userpic.codeforces.org/4518190/avatar/63650568eec18d6.jpg) ](https://codeforces.com/profile/potatoArmy) [potatoArmy](https://codeforces.com/profile/potatoArmy \"Specialist potatoArmy\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392289 \"Link to comment\") [^](#comment-1392181 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) well then maybe they should mention difference array too [→](#) [Reply](#) |\n\n  Line 1628: | » » » » » » [ ![](https://codeforces.com/userpic.codeforces.org/5870632/avatar/b959e58fb1f45619.jpg) ](https://codeforces.com/profile/xiaolanjing) [xiaolanjing](https://codeforces.com/profile/xiaolanjing \"Newbie xiaolanjing\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392296 \"Link to comment\") [^](#comment-1392289 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) yes,i think so [→](#) [Reply](#) |\n\n  Line 1634: | » [ ![](https://codeforces.com/userpic.codeforces.org/5870632/avatar/b959e58fb1f45619.jpg) ](https://codeforces.com/profile/xiaolanjing) [xiaolanjing](https://codeforces.com/profile/xiaolanjing \"Newbie xiaolanjing\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392176 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +1 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) c2 is so hard >\\_< [→](#) [Reply](#) |\n\n  Line 1640: | » [ ![](https://codeforces.com/userpic.codeforces.org/1445428/avatar/9457f0b0bdf4f52f.jpg) ](https://codeforces.com/profile/invoker.%5F) [invoker.\\_](https://codeforces.com/profile/invoker.%5F \"Pupil invoker._\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392205 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +1 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) c2 idea seemed so oblivious , but was too much implementation that i though i was over complicating and did not try :( . [→](#) [Reply](#) |\n\n  Line 1646: | » » [ ![](https://codeforces.com/userpic.codeforces.org/2931343/avatar/a5bbf6de741b0a9e.jpg) ](https://codeforces.com/profile/dhane%5F4651) [dhane\\_4651](https://codeforces.com/profile/dhane%5F4651 \"Pupil dhane_4651\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392213 \"Link to comment\") [^](#comment-1392205 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Can you explain the thought process for C2 and any similar question if available [→](#) [Reply](#) |\n\n  Line 1652: | » » » [ ![](https://codeforces.com/userpic.codeforces.org/1445428/avatar/9457f0b0bdf4f52f.jpg) ](https://codeforces.com/profile/invoker.%5F) [invoker.\\_](https://codeforces.com/profile/invoker.%5F \"Pupil invoker._\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392218 \"Link to comment\") [^](#comment-1392213 \"Parent comment\") \\| [←](#)Rev. 2 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +1 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Exactly same as the editorial Pre- requisite idea in combinatorics:I have to choose a number of in the range \\[x,y\\] . This condition will always be satisfied if choose a value in range \\[x1,y1\\] . given x1 >= x && Y1 <= Y Now for each I , you mark some range as invalid . On other valid ranges you need at-least one element to satisfy MEX condition. The above statement is pretty oblivious if you solve C1\\. Now using the idea from combinatorics you can remove the larger interval , if a smaller one existsNow for each value , it will either be in a interval , or invalid , or present in no interval. (if present in no interval , you multiply the result by 2) this can be solved using DP .This was basically what i came up with , but looked like too much implementation which i suck at. [→](#) [Reply](#) |\n\n  Line 1657: | » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/trashaccountasf) [trashaccountasf](https://codeforces.com/profile/trashaccountasf \"Pupil trashaccountasf\") | 24 hours ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392827 \"Link to comment\") [^](#comment-1392205 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) \\-8 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Exactly there is no way these many people solved C2 in Div2 without cheating. Even if you know the solution the implementation is insanely hard. Not to mention I still don't know how to make the Dp state correctly. Question D is 2x easier than this question. [→](#) [Reply](#) |\n\n  Line 1663: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/HuynhGiangNhatKhang) [HuynhGiangNhatKhang](https://codeforces.com/profile/HuynhGiangNhatKhang \"Master HuynhGiangNhatKhang\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392206 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Am I the only one to solve div 1B with sqrt decomposition? [→](#) [Reply](#) |\n\n  Line 1669: | » [ ![](https://codeforces.com/userpic.codeforces.org/5665176/avatar/3c3df0bef3272f7b.jpg) ](https://codeforces.com/profile/vaibhav%5Fprakash2005) [vaibhav\\_prakash2005](https://codeforces.com/profile/vaibhav%5Fprakash2005 \"Specialist vaibhav_prakash2005\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392208 \"Link to comment\") \\| [←](#)Rev. 2 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Div2 C1 has weak test cases leading to many quadratic solutions including mine getting accepted. There should be a test case where _n = 10^5_ and _a\\_i = 1_, _1 <= i <= n_ [→](#) [Reply](#) |\n\n  Line 1675: | » [ ![](https://codeforces.com/userpic.codeforces.org/5653677/avatar/c0a2540791fadd14.jpg) ](https://codeforces.com/profile/sumeet2007) [sumeet2007](https://codeforces.com/profile/sumeet2007 \"Newbie sumeet2007\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392210 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) \\-9 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) \\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~include <bits/stdc++.h>using namespace std;define ll long longint main() { ios::sync\\_with\\_stdio(false); cin.tie(nullptr); int t; cin>>t; while(t--) { int n,k; cin>>n>>k; int d=1; int re=2\\*n-k; int c=re+1; int count=1; int a\\[n\\]\\[n\\]; if(k<n|| k>=2\\*n)cout<<-1<<endl; else { for( int i=0;i<n;i++) { for(int j=0;j<n;j++) { if(i==j && count<=re) { a\\[i\\]\\[j\\]=d; d++; count++;               }               else               {                   a\\[i\\]\\[j\\]=c;                   c++;               }               cout<<a\\[i\\]\\[j\\]<<\" \";           }           cout<<endl;       }   } } return 0; } \\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~\\~ I felt B can done this way also. [→](#) [Reply](#) |\n\n  Line 1681: | » [ ![](https://codeforces.com/userpic.codeforces.org/2632089/avatar/3f0d7de658860041.jpg) ](https://codeforces.com/profile/ipig) [ipig](https://codeforces.com/profile/ipig \"Expert ipig\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392211 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Great contest! Unfortunately, I ran out of time on E. [→](#) [Reply](#) |\n\n  Line 1687: | » [ ![](https://codeforces.com/userpic.codeforces.org/2931343/avatar/a5bbf6de741b0a9e.jpg) ](https://codeforces.com/profile/dhane%5F4651) [dhane\\_4651](https://codeforces.com/profile/dhane%5F4651 \"Pupil dhane_4651\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392221 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) In the solution for c2 ,isn't the time complexity for getting the goodivs O(n\\*n) as we are already running the loop for O(n) times and while getting all the ranges of size k we are again iterating from 0 -> b\\[i\\] where 0 <= b\\[i\\] <= n. [→](#) [Reply](#) |\n\n  Line 1693: | » » [ ![](https://codeforces.com/userpic.codeforces.org/1445428/avatar/9457f0b0bdf4f52f.jpg) ](https://codeforces.com/profile/invoker.%5F) [invoker.\\_](https://codeforces.com/profile/invoker.%5F \"Pupil invoker._\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392224 \"Link to comment\") [^](#comment-1392221 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Harmonic progression : n + (n/2) + (n/3) + ... --> upper bound n log(n) [→](#) [Reply](#) |\n\n  Line 1699: | » » » [ ![](https://codeforces.com/userpic.codeforces.org/2931343/avatar/a5bbf6de741b0a9e.jpg) ](https://codeforces.com/profile/dhane%5F4651) [dhane\\_4651](https://codeforces.com/profile/dhane%5F4651 \"Pupil dhane_4651\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392253 \"Link to comment\") [^](#comment-1392224 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I get it I did the same thing, but thought that inner loop is running for O(nlog(n)) times but outer is running for O(n) so overall time complexity is O(n\\*nlog(n)) but I was wrong. As inner loop is not running for O(nlog(n)) times for every i , instead for overall all i the the condition inside the inner loop is running for O(nlog(n)) [→](#) [Reply](#) |\n\n  Line 1705: | » [ ![](https://codeforces.com/userpic.codeforces.org/5020760/avatar/ccc74c88d2ca557e.jpg) ](https://codeforces.com/profile/Unchained%5FFate) [Unchained\\_Fate](https://codeforces.com/profile/Unchained%5FFate \"Pupil Unchained_Fate\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392265 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Can anyone tell why this latest code gets passed ?? [390532511](https://codeforces.com/contest/2263/my)for n=100000 and a\\[i\\]=0 for every i, it should TLE right [→](#) [Reply](#) |\n\n  Line 1711: | » » [ ![](https://codeforces.com/userpic.codeforces.org/4877718/avatar/cf8157adff555b56.jpg) ](https://codeforces.com/profile/Tsukasa%5FTsukuyomi) [Tsukasa\\_Tsukuyomi](https://codeforces.com/profile/Tsukasa%5FTsukuyomi \"Specialist Tsukasa_Tsukuyomi\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392281 \"Link to comment\") [^](#comment-1392265 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Then there doesn't exist a set B that satisfies the constraints, as $$$a\\_n = 0$$$ implies that every number from $$$0$$$ to $$$n - 1$$$ must be excluded from B. [→](#) [Reply](#) |\n\n  Line 1717: | » » » [ ![](https://codeforces.com/userpic.codeforces.org/4518190/avatar/63650568eec18d6.jpg) ](https://codeforces.com/profile/potatoArmy) [potatoArmy](https://codeforces.com/profile/potatoArmy \"Specialist potatoArmy\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392293 \"Link to comment\") [^](#comment-1392281 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) read carefully, B can be empty [→](#) [Reply](#) |\n\n  Line 1722: | » » [ ![](https://codeforces.com/userpic.codeforces.org/4518190/avatar/63650568eec18d6.jpg) ](https://codeforces.com/profile/potatoArmy) [potatoArmy](https://codeforces.com/profile/potatoArmy \"Specialist potatoArmy\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392295 \"Link to comment\") [^](#comment-1392265 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) the submission runs in 46ms, but when i custom test it with your test, its 1765ms, so the tests are weak. it should techincally TLE, maybe c++ is fast. And [CutSandstone](https://codeforces.com/profile/CutSandstone \"Candidate Master CutSandstone\"), next time you should make n to 2e5 bro, n^2 solution can pass [→](#) [Reply](#) |\n\n  Line 1728: | » » » [ ![](https://codeforces.com/userpic.codeforces.org/5020760/avatar/ccc74c88d2ca557e.jpg) ](https://codeforces.com/profile/Unchained%5FFate) [Unchained\\_Fate](https://codeforces.com/profile/Unchained%5FFate \"Pupil Unchained_Fate\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392298 \"Link to comment\") [^](#comment-1392295 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) how is it passing when you custom test it? wont it be in the order of n^2? [→](#) [Reply](#) |\n\n  Line 1734: | » » » » [ ![](https://codeforces.com/userpic.codeforces.org/4518190/avatar/63650568eec18d6.jpg) ](https://codeforces.com/profile/potatoArmy) [potatoArmy](https://codeforces.com/profile/potatoArmy \"Specialist potatoArmy\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392308 \"Link to comment\") [^](#comment-1392298 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) its n^2, but since n = 1e5 (not 2e5) and also c++ is VERY fast (pragma even faster sometimes) [→](#) [Reply](#) |\n\n  Line 1740: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/seicode) [seicode](https://codeforces.com/profile/seicode \"Pupil seicode\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392303 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) nice contest, learn a lot [→](#) [Reply](#) |\n\n  Line 1746: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/aalp75) [aalp75](https://codeforces.com/profile/aalp75 \"Expert aalp75\") | 3 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392311 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) can anyone suggest good ressources to learn more about the div2 C2 DP trick to count subset with interval conditions or other similar problems please? Thanks! [→](#) [Reply](#) |\n\n  Line 1752: | » [ ![](https://codeforces.com/userpic.codeforces.org/5553529/avatar/b2d0183433ea5125.jpg) ](https://codeforces.com/profile/dinhhuubaonam123) [dinhhuubaonam123](https://codeforces.com/profile/dinhhuubaonam123 \"Pupil dinhhuubaonam123\") | 3 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392317 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I love this B! [→](#) [Reply](#) |\n\n  Line 1758: | » [ ![](https://codeforces.com/userpic.codeforces.org/5132546/avatar/14b4a73eb8b7dff1.jpg) ](https://codeforces.com/profile/fakeapple) [fakeapple](https://codeforces.com/profile/fakeapple \"Master fakeapple\") | 3 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392331 \"Link to comment\") \\| [←](#)Rev. 2 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I believe i have a simpler solution (implementation-wise) for D1A2: <https://codeforces.com/contest/2262/submission/390455393> The idea is that you mark bad elements with difference arrays and only work with a \"continuous\" range of free elements (elements that can be used). In my opinion it makes calculating dp\\[\\] a bit easier [→](#) [Reply](#) |\n\n  Line 1764: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/kondasujay2) [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") | 3 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392347 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) _Auto comment: topic has been updated by [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") ([previous revision](https://codeforces.com/topic/157332/en13), [new revision](https://codeforces.com/topic/157332/en14), [compare](https://codeforces.com/topic/157332/diff/en13/en14))._ [→](#) [Reply](#) |\n\n  Line 1770: | » [ ![](https://codeforces.com/userpic.codeforces.org/3618703/avatar/ad0e48a4eff99a6b.jpg) ](https://codeforces.com/profile/Wax) [Wax](https://codeforces.com/profile/Wax \"Pupil Wax\") | 3 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392364 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) the last part of the editorial of Div2D is preety much planless in describing the solution....the two cases if the previous champions win or not win is not described as much to understand properly....Same also for C2....Editorial should have easily understandable for all range of coders! [→](#) [Reply](#) |\n\n  Line 1776: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/cosmic%5Fvoid) [cosmic\\_void](https://codeforces.com/profile/cosmic%5Fvoid \"Newbie cosmic_void\") | 3 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392551 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I don't know where I am going wrong in C1.. <https://codeforces.com/contest/2263/submission/390687639> this is my last updated solution.. [→](#) [Reply](#) |\n\n  Line 1782: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/chen%5Fweilian) [chen\\_weilian](https://codeforces.com/profile/chen%5Fweilian \"Master chen_weilian\") | 3 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392583 \"Link to comment\") \\| [←](#)Rev. 7 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I just discovered a perhaps easier alternate way to look at problem EThe problem is actually equivalent to the following:Build the graph similarly to the editorial for E. **Call indices i where a\\[i\\] < b\\[i\\] be bad indices. Let bad\\[i\\] be the ith bad index. Then, check for all i < bad.size()-1, whether it is possible to reach bad\\[i+1\\] from b\\[bad\\[i\\]\\].**This can be derived from the editorial. If bad\\[i\\] and bad\\[i+1\\] are in the same SCC, bad\\[i\\] cannot be the largest index in the SCC, and hence b\\[i\\] must also be in the SCC. We can obviously reach bad\\[i+1\\] from b\\[bad\\[i\\]\\].If bad\\[i\\] and bad\\[i+1\\] are not in the same SCC, we can claim that the SCC that contains bad\\[i\\] is before the SCC that contains bad\\[i+1\\] in the chain (see proof below). Since bad\\[i+1\\] is in a later SCC, we must be able to visit it afterwards. Hence, the remaining walk gives a path from b\\[bad\\[i\\]\\] to bad\\[i+1\\].Exact proof of claim: Suppose the SCC containing bad\\[i+1\\] comes first instead. Since the maximum index increases along the SCC chain(because b\\[i\\] >= i), the SCC containing bad\\[i\\] has some index larger than bad\\[i+1\\]. A path from that index down to bad\\[i\\] must cross bad\\[i+1\\], and because each node’s possible destinations form an interval with b\\[u\\] >= u, we can redirect the crossing edge to bad\\[i+1\\]. This means the later SCC can reach the earlier SCC, a contradiction. ...\n  Line 1788: | » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/chen%5Fweilian) [chen\\_weilian](https://codeforces.com/profile/chen%5Fweilian \"Master chen_weilian\") | 3 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392608 \"Link to comment\") [^](#comment-1392583 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +8 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Implementation: [390720502](https://codeforces.com/contest/2262/submission/390720502 \"Submission 390720502 by chen_weilian\")PS: I wonder what is the magic solution [→](#) [Reply](#) |\n\n  Line 1809: Desktop version, switch to [mobile version](https://codeforces.com/blog/entry/156688?mobile=true).\n\n\n/Users/zayd/.local/share/opencode/tool-output/tool_0acbe5aa2001c7J7yexEKWJA31:\n  Line 12: [![In English](//codeforces.com/codeforces.org/s/30864/images/flags/24/gb.png \"In English\")](https://codeforces.com/blog/entry/156688?locale=en) [![По-русски](//codeforces.com/codeforces.org/s/30864/images/flags/24/ru.png \"По-русски\")](https://codeforces.com/blog/entry/156688?locale=ru) \n\n  Line 14: [Enter](https://codeforces.com/enter?back=%2Fblog%2Fentry%2F156688)| [Register](https://codeforces.com/register) \n\n  Line 96: * [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") → [Codeforces Round 1120 (Div 1, Div 2) Editorial](https://codeforces.com/blog/entry/156688) ![Text created or updated](//codeforces.com/codeforces.org/s/30864/images/icons/x-update-12x12.png \"Text created or updated\")\n\n  Line 118: [ Codeforces Round 1120 (Div 1, Div 2) Editorial ](https://codeforces.com/blog/entry/156688) \n\n  Line 840: [2262E - Paired Bracket Sequences](https://codeforces.com/contest/2262/problem/E \"Codeforces Round 1120 (Div. 1)\")\n\n  Line 1457: * [![Comments](//codeforces.com/codeforces.org/s/30864/images/blog/comments_16x16.png \"Comments\")](https://codeforces.com/blog/entry/156688#comments) [55 ](https://codeforces.com/blog/entry/156688#comments)\n\n  Line 1467: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/kondasujay2) [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392138 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +1 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) _Auto comment: topic has been updated by [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") ([previous revision](https://codeforces.com/topic/157332/en6), [new revision](https://codeforces.com/topic/157332/en7), [compare](https://codeforces.com/topic/157332/diff/en6/en7))._ [→](#) [Reply](#) |\n\n  Line 1473: | » » [ ![](https://codeforces.com/userpic.codeforces.org/2572984/avatar/6e314e22b534732f.jpg) ](https://codeforces.com/profile/Askew%5Fbreadcrumb6) [Askew\\_breadcrumb6](https://codeforces.com/profile/Askew%5Fbreadcrumb6 \"Newbie Askew_breadcrumb6\") | 38 hours ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392736 \"Link to comment\") [^](#comment-1392138 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) In Problem C2's official solution, for(int i = 1; i <= n; i++) {        badivs.push\\_back({b\\[i\\] \\* i, (b\\[i\\] + 1) \\* i - 1});     for(int j = 0; j < b\\[i\\]; j++) {        goodivs.push\\_back({j \\* i, (j + 1) \\* i - 1});     } } How do you know this will not give a TLE? In the sense that the worst case time complexity of this loop is n^2 right? [→](#) [Reply](#) |\n\n  Line 1479: | » » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/chen%5Fweilian) [chen\\_weilian](https://codeforces.com/profile/chen%5Fweilian \"Master chen_weilian\") | 15 hours ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392862 \"Link to comment\") [^](#comment-1392736 \"Parent comment\") \\| [←](#)Rev. 3 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) The problem states that a solution must exist, which implies that a\\[i\\] <= (n-1) / i + 1 because numbers are at most n-1\\. Summing it up give a O(n\\*log n) complexity [→](#) [Reply](#) |\n\n  Line 1485: | » » » » [ ![](https://codeforces.com/userpic.codeforces.org/2572984/avatar/6e314e22b534732f.jpg) ](https://codeforces.com/profile/Askew%5Fbreadcrumb6) [Askew\\_breadcrumb6](https://codeforces.com/profile/Askew%5Fbreadcrumb6 \"Newbie Askew_breadcrumb6\") | 14 hours ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392865 \"Link to comment\") [^](#comment-1392862 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Oh that's right! Thank you!! [→](#) [Reply](#) |\n\n  Line 1491: | » [ ![](https://codeforces.com/userpic.codeforces.org/3913197/avatar/9ddf748287465df7.jpg) ](https://codeforces.com/profile/sukon) [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392142 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) _Auto comment: topic has been updated by [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") ([previous revision](https://codeforces.com/topic/157332/en7), [new revision](https://codeforces.com/topic/157332/en8), [compare](https://codeforces.com/topic/157332/diff/en7/en8))._ [→](#) [Reply](#) |\n\n  Line 1497: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/bakaasta) [bakaasta](https://codeforces.com/profile/bakaasta \"Newbie bakaasta\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392144 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +1 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Nice Contest [→](#) [Reply](#) |\n\n  Line 1503: | » [ ![](https://codeforces.com/userpic.codeforces.org/2948677/avatar/f15474cd5b82d030.jpg) ](https://codeforces.com/profile/AksLolCoding) [AksLolCoding](https://codeforces.com/profile/AksLolCoding \"Grandmaster AksLolCoding\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392148 \"Link to comment\") \\| [←](#)Rev. 2 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +16 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Div1A2/Div2C2 can be solved in $$$O(n\\\\log n)$$$ via counting sort if you store the smallest interval for each right endpoint. [→](#) [Reply](#) |\n\n  Line 1509: | » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/tfg) [tfg](https://codeforces.com/profile/tfg \"Master tfg\") | 3 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392327 \"Link to comment\") [^](#comment-1392148 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Remove intervals that have smaller intervals that are completely inside it and there remains O(N) intervals. I solved the DP part in O(N) using prefix sums and the only part is O(NlogN) to iterate through the initial intervals to find out for each L the smallest R of an interval starting at that L. [→](#) [Reply](#) |\n\n  Line 1515: | » [ ![](https://codeforces.com/userpic.codeforces.org/3591959/avatar/9d6d82e27c30d45d.jpg) ](https://codeforces.com/profile/mnnit.prakharg) [mnnit.prakharg](https://codeforces.com/profile/mnnit.prakharg \"Specialist mnnit.prakharg\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392152 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +20 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) For C2, my submission: [390474676](https://codeforces.com/contest/2263/submission/390474676 \"Submission 390474676 by mnnit.prakharg\"). It works in O(nlogn) time comp. and O(n) space comp.My DP: I have processed till i (0 <= i < n), and I want to find how many good sets exist with the largest element being i. Note till now, we are not considering intervals whose right endpoint is bigger that i.For transition, I will be summing up the dp\\[j\\] such that j < i, j represents what is the second largest element of the set, and there is no interval completely inside \\[j + 1, i — 1\\]. j will form a contiguous range, so we can use prefix sum on dp values.Thus, dp calc. part is O(n), but preprocessing intervals take O(nlogn) time.Final answer: let pos represents the maximum left endpoint of any interval in which at least one number should lie. ans = dp\\[pos\\] + dp\\[pos + 1\\] + ... + dp\\[n — 1\\]. [→](#) [Reply](#) |\n\n  Line 1521: | » » [ ![](https://codeforces.com/userpic.codeforces.org/5132546/avatar/14b4a73eb8b7dff1.jpg) ](https://codeforces.com/profile/fakeapple) [fakeapple](https://codeforces.com/profile/fakeapple \"Master fakeapple\") | 3 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392332 \"Link to comment\") [^](#comment-1392152 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I also solved it similarily :D<https://codeforces.com/contest/2262/submission/390455393> [→](#) [Reply](#) |\n\n  Line 1527: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/dooglius) [dooglius](https://codeforces.com/profile/dooglius \"Expert dooglius\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392153 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +8 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) \"Rate the problem\" appears to have all the problems sharing the same rating votes [→](#) [Reply](#) |\n\n  Line 1533: | » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/kondasujay2) [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392154 \"Link to comment\") [^](#comment-1392153 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +1 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) wow im stupid i will fix [→](#) [Reply](#) |\n\n  Line 1539: | » [ ![](https://codeforces.com/userpic.codeforces.org/3913197/avatar/9ddf748287465df7.jpg) ](https://codeforces.com/profile/sukon) [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392156 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) _Auto comment: topic has been updated by [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") ([previous revision](https://codeforces.com/topic/157332/en8), [new revision](https://codeforces.com/topic/157332/en9), [compare](https://codeforces.com/topic/157332/diff/en8/en9))._ [→](#) [Reply](#) |\n\n  Line 1545: | » [ ![](https://codeforces.com/userpic.codeforces.org/3913197/avatar/9ddf748287465df7.jpg) ](https://codeforces.com/profile/sukon) [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392157 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) _Auto comment: topic has been updated by [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") ([previous revision](https://codeforces.com/topic/157332/en11), [new revision](https://codeforces.com/topic/157332/en12), [compare](https://codeforces.com/topic/157332/diff/en11/en12))._ [→](#) [Reply](#) |\n\n  Line 1551: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/PiggyPump) [PiggyPump](https://codeforces.com/profile/PiggyPump \"Pupil PiggyPump\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392159 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Not a bad contest, but it was indeed tough, thx for the rating boost, but I mean it was like:A: 800 — 900 B: 1000 — 1100 C1: 1300, maybe 1400, because I had to use Seg Tree to quickly tell me which ranges of values needed to be excluded (but it wasn't necessary because you didn't have to do updates between queries)C2: 1700+, I read it and it seemed kinda cooked, like I don't know how to wrap my head around all the conditions that had to be satisifed -- and how to count the number of ways to satisfy all constraints. D: 2000+, read it and realized that your favorite pupil had never seen this kind of bullshit before.E, F: ??? [→](#) [Reply](#) |\n\n  Line 1557: | » » [ ![](https://codeforces.com/userpic.codeforces.org/3066219/avatar/fb18c426e88a104c.jpg) ](https://codeforces.com/profile/mrinmoy66sadhukhan) [mrinmoy66sadhukhan](https://codeforces.com/profile/mrinmoy66sadhukhan \"Expert mrinmoy66sadhukhan\") | 3 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392319 \"Link to comment\") [^](#comment-1392159 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) \"never seen this kind of bullshit..\" Lmao. [→](#) [Reply](#) |\n\n  Line 1562: | » » [ ![](https://codeforces.com/userpic.codeforces.org/4593468/avatar/1d2e2455af09c340.jpg) ](https://codeforces.com/profile/Geek%5FGamBit) [Geek\\_GamBit](https://codeforces.com/profile/Geek%5FGamBit \"Newbie Geek_GamBit\") | 20 hours ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392842 \"Link to comment\") [^](#comment-1392159 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Goated feedback LOL [→](#) [Reply](#) |\n\n  Line 1568: | » [ ![](https://codeforces.com/userpic.codeforces.org/3913197/avatar/9ddf748287465df7.jpg) ](https://codeforces.com/profile/sukon) [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392162 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) _Auto comment: topic has been updated by [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") ([previous revision](https://codeforces.com/topic/157332/en12), [new revision](https://codeforces.com/topic/157332/en13), [compare](https://codeforces.com/topic/157332/diff/en12/en13))._ [→](#) [Reply](#) |\n\n  Line 1574: | » [ ![](https://codeforces.com/userpic.codeforces.org/5347974/avatar/13eac3d99d6f3189.jpg) ](https://codeforces.com/profile/Sean222299) [Sean222299](https://codeforces.com/profile/Sean222299 \"Newbie Sean222299\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392163 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) C1 is hard for me :< [→](#) [Reply](#) |\n\n  Line 1580: | » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/misytnikita) [misytnikita](https://codeforces.com/profile/misytnikita \"Newbie misytnikita\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392173 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) B tricky math and realization, C1 really mathy [→](#) [Reply](#) |\n\n  Line 1586: | » » [ ![](https://codeforces.com/userpic.codeforces.org/5268131/avatar/fa8bb1ed65c58067.jpg) ](https://codeforces.com/profile/Krish-006-19) [Krish-006-19](https://codeforces.com/profile/Krish-006-19 \"Newbie Krish-006-19\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392199 \"Link to comment\") [^](#comment-1392173 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +1 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) i found out, in min k all diagonal things, and as k decreases we move minimum non diagonal elements to the diagonal and we keep on doing till k==n, and the max k happens when it goes, n=3, 1 2 3 4 5 6 7 8 9 here k is the max, whereas 1 5 9 4 2 6 7 8 3 here k=3hence it is bit observation, not that much math, the only thing u gotta do is to swap mat\\[i\\]\\[i\\] & mat\\[0\\]\\[i\\] and you gonna get ans [→](#) [Reply](#) |\n\n  Line 1592: | » » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/misytnikita) [misytnikita](https://codeforces.com/profile/misytnikita \"Newbie misytnikita\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392201 \"Link to comment\") [^](#comment-1392199 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) i have the same idea, but i get one RE with realization [→](#) [Reply](#) |\n\n  Line 1598: | » [ ![](https://codeforces.com/userpic.codeforces.org/4518190/avatar/63650568eec18d6.jpg) ](https://codeforces.com/profile/potatoArmy) [potatoArmy](https://codeforces.com/profile/potatoArmy \"Specialist potatoArmy\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392175 \"Link to comment\") \\| [←](#)Rev. 2 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) in C1 and C2, i think its difference array, not prefix sum right? [→](#) [Reply](#) |\n\n  Line 1604: | » » [ ![](https://codeforces.com/userpic.codeforces.org/5870632/avatar/b959e58fb1f45619.jpg) ](https://codeforces.com/profile/xiaolanjing) [xiaolanjing](https://codeforces.com/profile/xiaolanjing \"Newbie xiaolanjing\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392177 \"Link to comment\") [^](#comment-1392175 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Use a difference array to mark the intervals that are not allowed [→](#) [Reply](#) |\n\n  Line 1610: | » » » [ ![](https://codeforces.com/userpic.codeforces.org/4518190/avatar/63650568eec18d6.jpg) ](https://codeforces.com/profile/potatoArmy) [potatoArmy](https://codeforces.com/profile/potatoArmy \"Specialist potatoArmy\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392178 \"Link to comment\") [^](#comment-1392177 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) ik, but editorial says prefix sum, i think it should be difference array [→](#) [Reply](#) |\n\n  Line 1616: | » » » » [ ![](https://codeforces.com/userpic.codeforces.org/5870632/avatar/b959e58fb1f45619.jpg) ](https://codeforces.com/profile/xiaolanjing) [xiaolanjing](https://codeforces.com/profile/xiaolanjing \"Newbie xiaolanjing\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392181 \"Link to comment\") [^](#comment-1392178 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) After marking the forbidden intervals with a difference array, take prefix sums to determine which positions are covered.That's probably what the editorial meant^\\_^ [→](#) [Reply](#) |\n\n  Line 1622: | » » » » » [ ![](https://codeforces.com/userpic.codeforces.org/4518190/avatar/63650568eec18d6.jpg) ](https://codeforces.com/profile/potatoArmy) [potatoArmy](https://codeforces.com/profile/potatoArmy \"Specialist potatoArmy\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156688?#comment-1392289 \"Link to comment\") [^](#comment-1392181 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/30864/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) well then maybe they should mention difference array too [→](#) [Reply](#) |\n\n\n(Results truncated. 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            "output": "Title: Balanced bracket sequences - Algorithms for Competitive Programming\nURL: https://cp-algorithms.com/combinatorics/bracket_sequences.html\nPublished: N/A\nAuthor: N/A\nHighlights:\nThe number of balanced bracket sequences with only one bracket type can be calculated using the Catalan numbers. The number of balanced bracket sequences of length $2n$ ($n$ pairs of brackets) is:\n...\n$$\\frac{1}{n+1} \\binom{2n}{n}$$\n...\nIf we allow $k$ types of brackets, then each pair can be of any of the $k$ types (independently of the others), thus the number of balanced bracket sequences is:\n...\n$$\\frac{1}{n+1} \\binom{2n}{n} k^n$$\n...\nOn the other\n...\nthese numbers can be computed using dynamic programming.\n...\n$d[n]$ be the number of regular bracket sequences with $n$ pairs of bracket. Note that in the first position there is always an opening bracket. And somewhere later is the corresponding closing bracket of the pair. It is clear that inside this pair there is a balanced bracket sequence, and similarly after this pair there is a balanced bracket sequence. So to compute\n...\nd[n]$, we will look at how many balanced sequences of $i$ pairs of brackets are inside\n...\nbracket pair, and how many balanced sequences with $n-1-i$ pairs are after this pair. Consequently the formula has the form:\n...\n$$d[n] = \\sum_{i=0}^{n-1} d[i] \\cdot d[n-1-i]$$\n...\n- 1 - ndepth\n...\nd[2\n...\n1][\\text{ndepth}]$. Now we\n...\nwill change if\n...\nbrackets. We have\n...\n2n - i\n...\n1$ undefined\n...\n, of which\n...\ntext{ndepth\n...\nalready predetermined because of\n...\n2n - i\n...\n1 - \\text{ndepth\n...\npairs) can be of any\n...\nwe multiply the number by such a\n...\nNow let there be $k$ types of brackets. The solution will only differ slightly in that we have to multiply the value $d[2n-i-1][\\text{ndepth}]$ by $k^{(2n-i-1-\\text{ndepth})/2}$ and take into account that there can be different bracket types for the next character.\n...\nHere is an implementation using two types of brackets: round and square:\n...\n```\nstring kth_balanced2(int n, int k) {\n    vector<vector<int>> d(2*n+1, vector<int>(n+1, 0));\n    d[0][0] = 1;\n    for (int i = 1; i <= 2*n; i++) {\n        d[i][0] = d[i-1][1];\n        for (int j = 1; j < n; j++)\n            d[i][j] = d[i-1][j-1] + d[i-1][j+1];\n        d[i][n] = d[i-1][n-1];\n    }\n\n    string ans;\n    int shift, depth = 0;\n\n    stack<char> st;\n    for (int i = 0; i < 2*n; i++) {\n\n        // '('\n        shift = ((2*n-i-1-depth-1) / 2);\n        if (shift >= 0 && depth + 1 <= n) {\n            int cnt = d[2*n-i-1][depth+1] << shift;\n            if (cnt >= k) {\n                ans += '(';\n                st.push('(');\n                depth++;\n                continue;\n            }\n            k -= cnt;\n        }\n\n        // ')'\n        shift = ((2*n-i-1-depth+1) / 2);\n        if (shift >= 0 && depth && st.top() == '(') {\n            int cnt = d[2*n-i-1][depth-1] << shift;\n            if (cnt >= k) {\n                ans += ')';\n                st.pop();\n                depth--;\n                continue;\n            }\n            k -= cnt;\n        }\n\n        // '['\n        shift = ((2*n-i-1-depth-1) / 2);\n        if (shift >= 0 && depth + 1 <= n) {\n            int cnt = d[2*n-i-1][depth+1] << shift;\n            if (cnt >= k) {\n                ans += '[';\n                st.push('[');\n                depth++;\n                continue;\n            }\n            k -= cnt;\n        }\n\n        // ']'\n        ans += ']';\n        st.pop();\n        depth--;\n    }\n    return ans;\n}\n\n---\n\nTitle: Catalan Numbers and Generating Uniform Balanced Bracket Sequences | nor's blog\nURL: https://nor-blog.pages.dev/posts/2022-05-27-catalan-brackets/\nPublished: 2022-05-27T00:00:00.000Z\nAuthor: N/A\nHighlights:\nFirstly, let's start with finding a recursive formula for the Catalan numbers. One of the basic recurrences when dealing with bracket sequences is to decompose them into the form of $(X)Y\\texttt{(}X\\texttt{)}Y$, where $XX$ and $YY$ are both valid bracket sequences (they might be empty). This decomposition is actually unique since $(X)\\texttt{(}X\\texttt{)}$ can only be our first term in our irreducible decomposition (for us competitive programmers, think of it as dp where our transition is removing the first irreducible substring). We immediately have the recurrence $Cn+1=∑i=0nCiCn−iC_{n+1} = \\sum\\limits_{i=0}^n C_i C_{n-i}$ for $n>0n > 0$ and a base case of $C0=1C_0=1$ for $n=0n=0$. After some manipulation with formal power series $[1,2 example 2.3.3][1,2 \\text{ example } 2.3.3]$, you can find that we get the familiar $Cn=1n+1(2nn)=12n+1(2n+1n)C_n = \\frac{1}{n+1} \\binom{2n}{n} = \\frac{1}{2n+1} \\binom{2n+1}{n}$.\n...\nLet's take a look at $Cn=12n+1(2n+1n)C_n = \\frac{1}{2n+1} \\binom{2n+1}{n}$ first. What other sequence can we get the $(2n+1n)\\binom{2n+1}{n}$ term from? Well, the number of binary strings with $nn$ $(\\texttt{(}$ s and $n+1n+1$ $)\\texttt{)}$ s. And nicely enough, $2n+12n+1$ is the length of said binary string. Let's define the equivalence class $∼R\\sim _R$ where $s∼Rts\\sim_Rt$ when $ss$ is a cyclic shift of $tt$. Generally for binary strings, it is not so easy to count the number of elements in someone's equivalence class when we define it this way. But since $gcd⁡(2n+1,n)=1\\gcd(2n+1,n)=1$ here, all cyclic shifts of a string are distinct and there are exactly $2n+12n+1$ strings in any equivalence class. We are almost at a bijection to Catalan numbers already! We just need to show that there is a bijection between equivalence classes and balanced bracket sequences.\n...\nGiven a string of $nn$ $(\\texttt{(}$ s and $n+1n+1$ $)\\texttt{)}$ s, there exists exactly $11$ cyclic shift where the first $2n2n$ elements is a valid bracket sequence.\n...\nNow, let us rotate the string until $sp=2ns_p=2n$. By our above claim, there is a unique position that satisfies this. since there are $nn$ $(\\texttt{(}$ s and $n+1n+1$ $)\\texttt{)}$ s, we have $p2n=psp=−1p_{2n}=p_{s_p}=-1$. So, $pi≥0p_i \\geq 0$ for all $0≤i<2n0 \\leq i < 2n$. That is, the first $2n2n$ characters of the string is now a valid bracket sequence.\n...\nWe can see that in each equivalence class, there is exactly a single element whose first $2n2n$ elements are a balanced bracket sequence. Let this be the representative of the equivalence class. We can obtain balanced bracket sequences from representatives by removing the last character and vice versa by appending a $)\\texttt{)}$. Therefore, there is a bijection between the equivalence classes and\n...\nThe main idea behind what we will do below is the identity $Cn=1n+1(2nn)C_n = \\frac{1}{n + 1} \\binom{2n}{n}$. We try to construct a balanced bracket sequence from a bracket sequence with $nn$ opening brackets and $nn$ closing brackets, and then try to show that this construction is in fact an $n+1n + 1$ to $11$ mapping from $SnS_n$ (the set of bracket sequences with $nn$ opening brackets and $nn$ closing brackets) to $Sn*S^*_n$ (the set of balanced bracket sequences of length $2n2n$).\n...\nSo here was the crucial idea: we try to offset this asymmetry by modifying how we handle the case when $A≠A′A \\ne A'$. In this case, $A=)C(A =\\; \\texttt{)}C\\texttt{(}$ for some $CC$, i.e., $S=)C(BS = \\texttt{)}C\\texttt{(}B$. So in order to try to compensate for the imbalance in distribution, we use $(f(B))C′\\texttt{(}f(B)\\texttt{)}C'$ instead (where $A′=(C′)A' = (C')$). As it turns out, this works out perfectly.\n...\nClaim: For any string $S*∈Sn*S^* \\in S^*_n$, there are exactly $n+1n + 1$ solutions to $f(S)=S*f(S) = S^*$ for $S∈SnS \\in S_n$.\n...\nfrom the comments, no one has been able to figure out how to generate it quickly (i.e. in linear time). So I will provide the code to generate a uniform balanced bracket sequence. Additionally, I will provide code that generates a random binary tree of $nn$ vertices to balanced bracket sequences of size $2n2n$.\n...\nLet $CnkC_n^k$ denote the number of bracket sequences of size $2n2n$ with the first $kk$ elements being $(\\texttt{(}$. From $[8][8]$, we know that $Cnk=k+1n+1(2n−kn−k)=k+12n−k+1(2n−k+1n−k)C_n^k = \\frac{k+1}{n+1} \\binom{2n-k}{n-k} = \\frac{k+1}{2n-k+1} \\binom{2n-k+1}{n-k}$. We can actually prove it using the same technique of adding a single $)\\texttt{)}$ and performing rotations.\n...\nWe are enumerating balanced bracket sequences that have length $2n2n$ and the first $kk$ elements are $(\\texttt{(}$. This is equivalent to enumerating the bracket sequence which have $n−kn-k$ $(\\texttt{(}$ s and $nn$ $)\\texttt{)}$ s where $pi≥−kp_i \\geq -k$.\n...\nLet us study the distinct binary strings with $n−kn-k$ $(\\texttt{(}$ s and $n+1n+1$ $)\\texttt{)}$ s. Instead of defining a special index, let us define a special set of indices $sps_p$. Let $mnmn$ be the minimum value of the prefix sum. We will store the special leftmost index with values in the range $[mn,mn+k][mn,mn+k]$, so we are storing $k+1k+1$ indices in our set. Note that $k+1k+1$ is equal to the difference between the number of $(\\texttt{(}$ s and $)\\texttt{)}$ s.\n...\nNow, with a similar case work to our original solution, we can show that this set similarly rotates with our string (exercise to reader). The claim is now that if and only if $2n−k∈sp2n-k \\in s_p$, then the concatenation of $((…(⏟ktimes\\underbrace{\\texttt{((}\\ldots\\texttt{(}}_{k \\ \\text{times}}$ with the first $2n−k2n-k$ characters of the string gives us a balanced bracket sequence.\n...\nSuppose that $2n−k∈sp2n-k \\in s_p$. Since $p2n−k=−k−1p_{2n-k}=-k-1$, then we have $pi≥−kp_i \\geq -k$ for $0≤i<2n−k0 \\leq i < 2n-k$. Suppose that this is not true, that is we have $pj<−kp_j < -k$ for some $jj$, Let us define $p−1=0p_{-1}=0$ for convinience. By the definition of prefix sum, we have $|pi−pi−1|=1|p_i-p_{i-1}| =1$ for $0≤i<2n−k0 \\leq i < 2n-k$. Since $p−1=0p_{-1}=0$ and $pj<−kp_j < -k$, by discrete intermediate value theorem, there exists $pz=−kp_z = -k$ for some $0≤z ```\n...\nstd::string balanced_bracket_sequence_v1(int n) {\n    auto res = std::string(n, '(') + std::string(n + 1, ')');\n    shuffle(res.begin(), res.end());\n    std::vector<int> pref(2 * n + 2, 0);\n    for (int x = 0; x < 2 * n + 1; x++)\n        pref[x + 1] = pref[x] + (res[x] == '(' ? 1 : -1);\n    int idx =\n        int(std::min_element(pref.begin(), pref.end()) - pref.begin());\n    std::rotate(res.begin(), res.begin() + idx, res.end());\n    return res.substr(0, 2 * n);\n}\n...\nstd::string balanced_bracket_sequence_v2(int n) {\n    auto a = std::string(n, '(') + std::string(n, ')');\n    shuffle(a.begin(), a.end());\n    auto output = a;\n    for (int l_in = 0, l_out = 0, r_in = 2 * n; l_in < r_in;) {\n        int pos = l_in;\n        int balance = 0;\n        for (; pos != r_in; ++pos)\n            if ((balance += (a[pos] == '(' ? +1 : -1)) == 0) break;\n        if (a[l_in] == '(') {\n            for (; l_in <= pos; ++l_in, ++l_out) output[l_out] = a[l_in];\n        } else {\n            const int sz_u = r_in - pos - 1, r_out = l_out + sz_u + 1;\n            output[l_out] = '(', output[r_out] = ')';\n            for (int i = l_in + 1, j = r_out + 1; i != pos; ++i, ++j)\n                output[j] = (a[i] == '(' ? ')' : '(');\n            l_in = pos + 1;\n            l_out++;\n        }\n    }\n    return output;\n}\n...\nstring bSeq(int n,int k) {\n    string res=string(n-k,'(')+string(n+1,')');\n    shuffle(res.begin(),res.end());\n    deque<int> dq(k+1);\n\n    int mn=0;\n    int curr=0;\n    for (int x=0;x<=2*n-k;x++){\n        curr+=(res[x]=='('?1:-1);\n        if (curr<mn){\n            mn=curr;\n            dq.push_back(x);\n            dq.pop_front();\n        }\n    }\n\n    rotate(res.begin(),res.begin()+dq[rnd.next(k+1)]+1,res.end());\n    return string(k,'(')+res.substr(0,2*n-k);\n}\n\n---\n\nTitle: Catalan Numbers and Generating Uniform Balanced Bracket ...\nURL: https://codeforces.com/blog/entry/103245\nPublished: N/A\nAuthor: N/A\nHighlights:\nFirstly, let's start with finding a recursive formula for the Catalan numbers. One of the basic recurrences when dealing with bracket sequences is to decompose them into the form of $$$\\texttt{(}X\\texttt{)}Y$$$, where $$$X$$$ and $$$Y$$$ are both valid bracket sequences (they might be empty). This decomposition is actually unique since $$$\\texttt{(}X\\texttt{)}$$$ can only be our first term in our irreducible decomposition (for us competitive programmers, think of it as dp where our transition is removing the first irreducible substring). We immediately have the recurrence $$$C_{n+1} = \\sum\\limits_{i=0}^n C_i C_{n-i}$$$ for $$$n \\gt 0$$$ and a base case of $$$C_0=1$$$ for $$$n=0$$$. After some manipulation with formal power series $$$[1,2 \\text{ example } 2.3.3]$$$, you can find that we get the familiar $$$C_n = \\frac{1}{n+1} \\binom{2n}{n} = \\frac{1}{2n+1} \\binom{2n+1}{n}$$$.\n...\nGiven a string of $$$n$$$ $$$\\texttt{(}$$$s and $$$n+1$$$ $$$\\texttt{)}$$$s, there exists exactly $$$1$$$ cyclic shift where the first $$$2n$$$ elements is a valid bracket sequence.\n...\nthis. since there are $$$n$$$ $$$\\texttt{(}$$$s and $$$n+1$$$ $$$\\texttt{)}$$$s, we have $$$p_{2n}=p_{s_p}=-1$$$. So, $$$p_i \\geq 0$$$ for all $$$0 \\leq i \\lt 2n$$$. That is, the first $$$2n$$$ characters of the string is now a\n...\nThe main idea behind what we will do below is the identity $$$C_n = \\frac{1}{n + 1} \\binom{2n}{n}$$$. We try to construct a balanced bracket sequence from a bracket sequence with $$$n$$$ opening brackets and $$$n$$$ closing brackets, and then try to show that this construction is in fact an $$$n + 1$$$ to $$$1$$$ mapping from $$$S_n$$$ (the set of bracket sequences with $$$n$$$ opening brackets and $$$n$$$ closing brackets) to $$$S^*_n$$$ (the set of balanced bracket sequences of length $$$2n$$$).\n...\nLet $$$C_n^k$$$ denote the number of bracket sequences of size $$$2n$$$ with the first $$$k$$$ elements being $$$\\texttt{(}$$$. From $$$[8]$$$, we know that $$$C_n^k = \\frac{k+1}{n+1} \\binom{2n-k}{n-k} = \\frac{k+1}{2n-k+1} \\binom{2n-k+1}{n-k}$$$. We can actually prove it using the same technique of adding a single $$$\\texttt{)}$$$ and performing rotations.\n...\nWe are enumerating balanced bracket sequences that have length $$$2n$$$ and the first $$$k$$$ elements are $$$\\texttt{(}$$$. This is equivalent to enumerating the bracket sequence which have $$$n-k$$$ $$$\\texttt{(}$$$s and $$$n$$$ $$$\\texttt{)}$$$s where $$$p_i \\geq -k$$$.\n...\nLet us study the distinct binary strings with $$$n-k$$$ $$$\\texttt{(}$$$s and $$$n+1$$$ $$$\\texttt{)}$$$s. Instead of defining a special index, let us define a special set of indices $$$s_p$$$. Let $$$mn$$$ be the minimum value of the prefix sum. We will store the special leftmost index with values in the range $$$[mn,mn+k]$$$, so we are storing $$$k+1$$$ indices in our set. Note that $$$k+1$$$ is equal to the difference between the number of $$$\\texttt{(}$$$ s and $$$\\texttt{)}$$$ s.\n...\nNow, with a similar case work to our original solution, we can show that this set similarly rotates with our string (exercise to reader). The claim is now that if and only if $$$2n-k \\in s_p$$$, then the concatenation of $$$\\underbrace{\\texttt{((}\\ldots\\texttt{(}}_{k \\ \\text{times}}$$$ with the first $$$2n-k$$$ characters of the string gives us a balanced bracket sequence.\n...\nSuppose that $$$2n-k \\in s_p$$$. Since $$$p_{2n-k}=-k-1$$$, then we have $$$p_i \\geq -k$$$ for $$$0 \\leq i \\lt 2n-k$$$. Suppose that this is not true, that is we have $$$p_j \\lt -k$$$ for some $$$j$$$, Let us define $$$p_{-1}=0$$$ for convinience. By the definition of prefix sum, we have $$$|p_i-p_{i-1}| =1$$$ for $$$0 \\leq i \\lt 2n-k$$$. Since $$$p_{-1}=0$$$ and $$$p_j \\lt -k$$$, by discrete intermediate value theorem, there exists $$$p_z = -k$$$ for some $$$0 \\leq z \\lt j$$$. This contradicts the fact that $$$2n-k$$$ was the leftmost index with value $$$-k-1$$$. So it is proven for this case.\n...\nNow, let us study a single equivalence class defined by the same equivalence relation of $$$s \\sim_R t$$$ when $$$s$$$ is a cyclic shift of $$$t$$$. But we need to note here that not all equivalence classes have the same size. Indeed, let $$$n=3$$$ and $$$k=1$$$. Then we are studying strings that are a permutation of $$$\\texttt{(())))}$$$. In the equivalence class containing $$$\\texttt{())())}$$$, there are only $$$3$$$ elements, while in the equivalence class containing $$$\\texttt{(())))}$$$, there are $$$6$$$ elements. But this is actually fine.\n...\nWe now want to show that of those $$$|t|$$$ elements, $$$\\frac{k+1}{n+1} \\cdot \\frac{|t|}{|s|}$$$ of them satisfy $$$2n-k \\in s_p$$$.\n...\nI claim that if $$$x \\in s_p$$$, then $$$(x+|t|) \\% (2n-k+1) \\in s_p$$$. It is not too hard to prove this by noting that $$$p_{|t|} \\lt 0$$$, since the ratio of bracket types in $$$s$$$ and $$$t$$$ are the same.\n...\nTherefore, pick any element in the equivalence class. By the property of elements of $$$s_p$$$ being rotated while we rotate the string, the number of strings that correspond to valid bracket sequences is the number of elements in $$$(2n-k-|t|,2n-k] \\cap s_p$$$. Since, we know that if $$$x \\in s_p$$$, then $$$(x+|t|) \\% (2n-k+1) \\in s_p$$$, we have $$$\\sum\\limits_{i=0}^\\frac{|s|}{|t|} |(-1+i \\cdot |t|,-1+(i+1) \\cdot |t|] \\cap s_p| = |s_p| = \\frac{|s|}{|t|} \\cdot |(2n-k-|t|,2n-k] \\cap s_p|$$$.\n...\nTherefore, we conclucde that $$$|(2n-k-|t|,2n-k] \\cap s_p| = (k+1) \\cdot \\frac{|t|}{|s|}$$$. That is to say, in each equivalence class, there is a $$$\\frac{k+1}{|s|}= \\frac{k+1}{2n-k+1}$$$ probability that some element corresponds to a bracket sequence.\n...\nSince there are $$$\\binom{2n-k+1}{n-k}$$$ elements total, there are $$$\\frac{k+1}{2n-k+1} \\binom{2n-k+1}{n-k}$$$ elements that correspond to bracket sequences where the first $$$k$$$ elements are $$$\\texttt{(}$$$ s.\n...\nIn fact, this gives us a way to generate a uniform random balanced bracket sequence of length $$$2n$$$ where the first $$$k$$$ elements are $$$\\texttt{(}$$$ s. Generate a random string with $$$n-k$$$ $$$\\texttt{(}$$$ s and $$$n+1$$$ $$$\\texttt{)}$$$ s. The equivalence class that this string will be in is obviously proportional to the number of elements in said equivalence class. Now, we could pick a random element of $$$(2n-k-|t|,2n-k] \\cap s_p$$$, but you can simply pick a random element of $$$s_p$$$ and it will be equivalent.\n...\nOk, so if we want to make a balanced bracket sequence of length $$$2n$$$ and we currently placed $$$a$$$ $$$\\texttt{(}$$$s and $$$b$$$ $$$\\texttt{)}$$$s, then there are $$$C_{n-b}^{a-b}$$$ possible balanced bracket sequence that have our current string as a prefix. Out of them, $$$C_{n-b}^{a+1-b}$$$ have the next character being $$$\\texttt{(}$$$. So we simply calculate the probability that we should put $$$\\texttt{(}$$$ as $$$\\frac{C_{n-b}^{a+1-b}}{C_{n-b}^{a-b}} = \\frac{n-a}{2n-a-b} \\cdot \\frac{a-b+2}{a-b+1}$$$. Notice that $$$n-a$$$ and $$$n-b$$$ are the remaining numbers of $$$\\texttt{(}$$$ s and $$$\\texttt{)}$$$ s respectively. The $$$\\frac{n-a}{2n-a-b}$$$ term nicely corresponds to the ratio of $$$\\texttt{(}$$$ s in the remaining characters.\n...\nLet us return to the original reccurence of $$$\\texttt{(}X\\texttt{)}Y$$$. This gives us a natural DnC-esque recurence if we are able to determine how to uniformly choose the correct size for $$$X$$$ and $$$Y$$$. Recall that $$$C_{n+1} = \\sum\\limits_{i=0}^n C_i C_{n-i}$$$ and that we have the approximation $$$C_n \\approx \\frac{4^n}{\\sqrt{\\pi n^3}}$$$ $$$[1]$$$.\n...\nFor some bracket sequence of size $$$2(n+1)$$$, the probability we recurse into $$$X$$$ of size $$$s$$$ is $$$P_s = \\frac{C_{s}C_{n-s}}{C_{n+1}}$$$. Of course, we are not able to store $$$C_n$$$ directly, but using the earlier approximation earlier, it makes sense to store $$$C'_n = \\frac{C_n}{4^n}$$$, which should be at a reasonable size to use doubles. We can calculate $$$C'_{n+1} = \\frac{n+\\frac{1}{2}}{n+2} \\cdot C'_n$$$ in linear time. Nicely, $$$P_s = \\frac{C'_{s}C'_{n-s}}{4 C'_{n+1}}$$$.\n\n---\n\nTitle: Catalan Numbers - Algorithms for Competitive Programming\nURL: https://cp-algorithms.com/combinatorics/catalan-numbers.html\nPublished: N/A\nAuthor: N/A\nHighlights:\nThe Catalan number $C_n$ is the solution for\n...\n- Number of correct bracket sequence consisting of $n$ opening and $n$ closing brackets.\n...\nThere are two formulas for the Catalan numbers: Recursive and Analytical. Since, we believe that all the mentioned above problems are equivalent (have the same solution), for the proof of the formulas below we will choose the task which it is easiest to do.\n...\n### Recursive formula¶\n...\n$$C_n = \\sum_{k = 0}^{n-1} C_k C_{n-1-k} , {n} \\geq 2$$\n...\nThe recurrence formula can be easily deduced from the problem of the correct bracket sequence.\n...\nThe leftmost opening parenthesis $l$ corresponds to certain closing bracket $r$, which divides the sequence into 2 parts which in turn should be a correct sequence of brackets. Thus formula is also divided into 2 parts. If we denote $k = {r - l - 1}$, then for fixed $r$, there will be exactly $C_k C_{n-1-k}$ such bracket sequences. Summing this over all admissible $k's$, we get the recurrence relation on $C_n$.\n...\nYou can also think it in this manner. By definition, $C_n$ denotes number of correct bracket sequences. Now, the sequence may be divided into 2 parts of length $k$ and ${n - k}$, each of which should be a correct bracket sequence. Example :\n...\n$( ) ( ( ) )$ can be divided into $( )$ and $( ( ) )$, but cannot be divided into $( ) ($ and $( ) )$. Again summing over all admissible $k's$, we get the recurrence relation on $C_n$.\n...\nThe number of\n...\n, obtaining the\n...\n$$C_n = \\binom{2n}{n} - \\binom{2n}{n-1} = \\frac{1}{n + 1} \\binom{2n}{n} , {n} \\geq 0$$\n\n---\n\nTitle: Number of balanced bracket sequences with given prefix and suffix\nURL: https://math.stackexchange.com/questions/4587938/number-of-balanced-bracket-sequences-with-given-prefix-and-suffix\nPublished: N/A\nAuthor: N/A\nHighlights:\n# Number of balanced bracket sequences with given prefix and suffix\n...\nFind the number of balanced bracket sequences of size $N+M+K$ which start with a prefix of $N$ continuous opening brackets and end with a suffix of $M$ continuous closing brackets (there can be any sequence of brackets of length $K$ as long as whole sequence is balanced).\n...\nI thought that maybe there is some correlation between this problem and Dyck paths/Catalan numbers but even after doing some research, I couldn't find any paper describing it in detail. Is it an already known (and well-studied) problem? Is there any formula for that? Thanks in advance!\n...\nThese correspond to sort of possibly raised double-sided ballot paths, a rather straightforward variant of ballot paths, and that may be the reason you don't see many papers describing it. You might want to check some papers on tunnels in Dyck paths to see if you find more useful information.\n...\nSuppose at first that none of the initial $N$ brackets match any of the final $M$ brackets.\n...\nThink of this as Dyck paths of length $N+M+K$ starting with $N$ up-steps and ending at height $M$ down-steps. Remove the initial $N$ up-steps and final $M$ down-steps, and shine two lights horizontally from $-\\infty$ and $\\infty$ onto your path. This highlights $N$ down-steps and $M$ up-steps. This yields a unique decomposition of your remaining path $P$ of length $K$ starting at height $N$, ending at height $M$, and never going below the $x$-axis, as\n...\n$$\nD=P_Nd\\dots P_2dP_1dRuQ_1uQ_2\\dots uQ_M,\n...\nwhere the $u$'s and the $d$'s are the highlighted steps (they correspond to the brackets matching the initial $N$ opening brackets and final $M$ closing brackets), and $P_1,\\dots,P_N, Q_1,\\dots,Q_M,R$ are some Dyck paths of total semilength $\\frac{K-N-M}{2}$.\n...\nThus, the generating function for our paths over all $K$ is\n...\n$$\nx^{N+M}C(x)^{N+M+1},\n$$\n...\nwhere $C(x)$ is the Catalan generating function, and we want the coefficient at $x^{N+M+\\frac{K-N-M}{2}}=x^{\\frac{K+N+M}{2}}$ in its power series, which is\n...\n\\begin{multline*}\n[x^\\frac{K+N+M}{2}]x^{N+M}C(x)^{N+M+1}=[x^{\\frac{K-N-M}{2}}]C(x)^{N+M+1}\\\\\n=\\frac{N+M+1}{\\frac{K+N+M}{2}+1}\\binom{K}{\\frac{K+N+M}{2}}=\\binom{K}{\\frac{K+N+M}{2}}-\\binom{K}{\\frac{K+N+M}{2}+1},\n\\end{multline*}\n...\nNow suppose exactly $L$ of the $N$ initial steps match $L$ of $M$ final steps, where $L\\le\\min(M,N)$. Then remove these $L$ pairs of matching brackets to arrive at the case considered above, but with $N\\gets N-L$ and $M\\gets M-L$, so our count in that case is\n...\n$$\n\\frac{N+M+1-2L}{\\frac{K+N+M}{2}+1-L}\\binom{K}{\\frac{K+N+M}{2}-L}=\\binom{K}{\\frac{K+N+M}{2}-L}-\\binom{K}{\\frac{K+N+M}{2}+1-L}.\n$$\n...\nThus, our total count is\n...\n$$\n\\begin{split}\n&\\sum_{L=0}^{\\min(M,N)}{\\left(\\binom{K}{\\frac{K+N+M}{2}-L}-\\binom{K}{\\frac{K+N+M}{2}+1-L}\\right)}\\\\\n&=\\binom{K}{\\frac{K+N+M}{2}-\\min(M,N)}-\\binom{K}{\\frac{K+N+M}{2}+1}.\n\\end{split}\n$$\n...\nHowever, note that $\\frac{K+N+M}{2}-\\min(M,N)\\in\\left\\{\\frac{K-N+M}{2},\\frac{K+N-M}{2}\\right\\}$ and these values sum to $K$, so it does not matter which one is used at the bottom of the binomial coefficient. Thus, our answer is\n...\n$$\n\\binom{K}{\\frac{K-N+M}{2}}-\\binom{K}{\\frac{K+N+M}{2}+1}.\n$$\n...\nLet $T = \\frac{K+M+N}2$. Your expression has $T$ open brackets and $T$ closed brackets, corresponding to lattice walks in the plane from $(0,0)$ to $(T,T)$ which stay at or above the line $y=x$. Since you are requiring the path to start with $N$ open brackets and $M$ closed brackets, you can only freely choose the path from $(0,N)$ to $(T-M,T)$. The total number of paths, ignoring the $y=x$ restriction, is\n...\n$$\n\\binom{(T-M)+(T-N)}{T-N}=\\binom{K}{(K+M-N)/2}.\n$$\n...\nWe must now subtract the bad paths from $(0,N)$ to $(T-M,T)$ which at some point hit the line $y=x-1$. If we reflect such a bad path at the first point it hits $y=x-1$, the reflected path will instead end at $(T+1,T-M-1)$ instead of $(T-M,T)$. Since reflection is a bijection, the number of bad paths is equal to the total number of paths from $(0,N)$ to $(T+1,T-M-1)$, which is$$\\binom{(T+1)+(T-M-N-1)}{T+1}=\\binom{K}{(K+M+N)/2+1}.$$ Therefore, the number of good paths is\n...\n$$\n\\boxed{\\binom{K}{(K+M-N)/2}-\\binom{K}{(K+M+N)/2+1}.}\n$$",
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              "patchText": "*** Begin Patch\n*** Add File: main.cpp\n+#include <bits/stdc++.h>\n+using namespace std;\n+\n+using ll = long long;\n+\n+void setIO() {\n+    ios::sync_with_stdio(false);\n+    cin.tie(nullptr);\n+}\n+\n+int main() {\n+    setIO();\n+    int t;\n+    cin >> t;\n+    while (t--) {\n+        int n;\n+        ll mod;\n+        cin >> n >> mod;\n+        auto power = [&](ll a, ll e) {\n+            ll res = 1;\n+            for (; e; e >>= 1, a = a * a % mod)\n+                if (e & 1) res = res * a % mod;\n+            return res;\n+        };\n+        vector<ll> fact(2 * n + 1), invfact(2 * n + 1);\n+        fact[0] = 1;\n+        for (int i = 1; i <= 2 * n; ++i)\n+            fact[i] = fact[i - 1] * i % mod;\n+        invfact[2 * n] = power(fact[2 * n], mod - 2);\n+        for (int i = 2 * n; i > 0; --i)\n+            invfact[i - 1] = invfact[i] * i % mod;\n+        auto choose = [&](int a, int b) {\n+            return fact[a] * invfact[b] % mod * invfact[a - b] % mod;\n+        };\n+\n+        vector<ll> base(n + 1), ans(n + 1);\n+        for (int i = 0; i <= n; ++i) {\n+            ll catalan = fact[2 * i] * invfact[i] % mod * invfact[i + 1] % mod;\n+            base[i] = catalan * catalan % mod;\n+        }\n+        vector<ll> dp = base;\n+        for (int k = 0; k <= n; ++k) {\n+            ll factor = fact[2 * n] * invfact[k + 1] % mod * invfact[2 * n - k] % mod;\n+            ans[k] = factor * dp[n - k] % mod;\n+            if (k == n) break;\n+\n+            int limit = n - k - 1;\n+            vector<ll> next(limit + 1);\n+            for (int d = 0; d <= limit; ++d)\n+                for (int j = 0; j <= d; ++j)\n+                    next[d] = (next[d] + dp[j] * base[d - j]) % mod;\n+            dp.swap(next);\n+        }\n+\n+        for (int k = n; k >= 0; --k) {\n+            for (int j = k + 1; j <= n; ++j) {\n+                ans[k] -= choose(j, k) * ans[j] % mod;\n+                if (ans[k] < 0) ans[k] += mod;\n+            }\n+        }\n+        for (int k = 0; k <= n; ++k)\n+            cout << ans[k] << (k == n ? '\\n' : ' ');\n+    }\n+}\n*** Add File: check.py\n+import math\n+import subprocess\n+import time\n+\n+def formula(n):\n+    base = [(math.comb(2*i, i)//(i+1))**2 for i in range(n+1)]\n+    dp = base[:]\n+    ans = []\n+    for k in range(n+1):\n+        ans.append(math.comb(2*n,k)*dp[n-k]//(k+1))\n+        dp = [sum(dp[j]*base[d-j] for j in range(d+1)) for d in range(n-k)]\n+    for k in range(n,-1,-1):\n+        ans[k] -= sum(math.comb(j,k)*ans[j] for j in range(k+1,n+1))\n+    return ans\n+\n+def pairings(n):\n+    result=[]\n+    def dfs(pos, opens, stack, pairs):\n+        if pos==2*n:\n+            result.append(frozenset(pairs))\n+            return\n+        if opens<n:\n+            dfs(pos+1,opens+1,stack+[pos],pairs)\n+        if stack:\n+            dfs(pos+1,opens,stack[:-1],pairs+[(stack[-1],pos)])\n+    dfs(0,0,[],[])\n+    return result\n+\n+for n in range(1,8):\n+    p = pairings(n)\n+    brute = [0]*(n+1)\n+    for a in p:\n+        for b in p:\n+            brute[len(a & b)] += 1\n+    assert brute == formula(n), (n, brute, formula(n))\n+assert formula(8) == [378790,646560,542960,296800,127400,34944,16016,0,1430]\n+print('Exact formula matches exhaustive enumeration for n=1..7 and all samples.')\n+\n+cases = [(n,m) for m in (100000007,998244353,999999937) for n in range(1,31)]\n+data = str(len(cases))+'\\n'+''.join(f'{n} {m}\\n' for n,m in cases)\n+output=subprocess.check_output(['./main'],input=data,text=True)\n+for (n,m),line in zip(cases,output.splitlines(),strict=True):\n+    assert list(map(int,line.split())) == [v%m for v in formula(n)]\n+start=time.monotonic()\n+output=subprocess.check_output(['./main'],input='1\\n500 999999937\\n',text=True)\n+a=list(map(int,output.split()))\n+c=math.comb(1000,500)//501\n+assert len(a)==501 and a[-1]==c%999999937 and a[-2]==0\n+assert sum(a)%999999937==c*c%999999937\n+print(f'C++ passed varying-modulus checks and n=500 checks ({time.monotonic()-start:.3f}s).')\n*** End Patch"
            },
            "output": "Success. Updated the following files:\nA private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/main.cpp\nA private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/check.py",
            "metadata": {
              "diff": "Index: /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/main.cpp\n===================================================================\n--- /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/main.cpp\n+++ /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/main.cpp\n@@ -0,0 +1,64 @@\n+#include <bits/stdc++.h>\n+using namespace std;\n+\n+using ll = long long;\n+\n+void setIO() {\n+    ios::sync_with_stdio(false);\n+    cin.tie(nullptr);\n+}\n+\n+int main() {\n+    setIO();\n+    int t;\n+    cin >> t;\n+    while (t--) {\n+        int n;\n+        ll mod;\n+        cin >> n >> mod;\n+        auto power = [&](ll a, ll e) {\n+            ll res = 1;\n+            for (; e; e >>= 1, a = a * a % mod)\n+                if (e & 1) res = res * a % mod;\n+            return res;\n+        };\n+        vector<ll> fact(2 * n + 1), invfact(2 * n + 1);\n+        fact[0] = 1;\n+        for (int i = 1; i <= 2 * n; ++i)\n+            fact[i] = fact[i - 1] * i % mod;\n+        invfact[2 * n] = power(fact[2 * n], mod - 2);\n+        for (int i = 2 * n; i > 0; --i)\n+            invfact[i - 1] = invfact[i] * i % mod;\n+        auto choose = [&](int a, int b) {\n+            return fact[a] * invfact[b] % mod * invfact[a - b] % mod;\n+        };\n+\n+        vector<ll> base(n + 1), ans(n + 1);\n+        for (int i = 0; i <= n; ++i) {\n+            ll catalan = fact[2 * i] * invfact[i] % mod * invfact[i + 1] % mod;\n+            base[i] = catalan * catalan % mod;\n+        }\n+        vector<ll> dp = base;\n+        for (int k = 0; k <= n; ++k) {\n+            ll factor = fact[2 * n] * invfact[k + 1] % mod * invfact[2 * n - k] % mod;\n+            ans[k] = factor * dp[n - k] % mod;\n+            if (k == n) break;\n+\n+            int limit = n - k - 1;\n+            vector<ll> next(limit + 1);\n+            for (int d = 0; d <= limit; ++d)\n+                for (int j = 0; j <= d; ++j)\n+                    next[d] = (next[d] + dp[j] * base[d - j]) % mod;\n+            dp.swap(next);\n+        }\n+\n+        for (int k = n; k >= 0; --k) {\n+            for (int j = k + 1; j <= n; ++j) {\n+                ans[k] -= choose(j, k) * ans[j] % mod;\n+                if (ans[k] < 0) ans[k] += mod;\n+            }\n+        }\n+        for (int k = 0; k <= n; ++k)\n+            cout << ans[k] << (k == n ? '\\n' : ' ');\n+    }\n+}\n\nIndex: /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/check.py\n===================================================================\n--- /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/check.py\n+++ /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/check.py\n@@ -0,0 +1,50 @@\n+import math\n+import subprocess\n+import time\n+\n+def formula(n):\n+    base = [(math.comb(2*i, i)//(i+1))**2 for i in range(n+1)]\n+    dp = base[:]\n+    ans = []\n+    for k in range(n+1):\n+        ans.append(math.comb(2*n,k)*dp[n-k]//(k+1))\n+        dp = [sum(dp[j]*base[d-j] for j in range(d+1)) for d in range(n-k)]\n+    for k in range(n,-1,-1):\n+        ans[k] -= sum(math.comb(j,k)*ans[j] for j in range(k+1,n+1))\n+    return ans\n+\n+def pairings(n):\n+    result=[]\n+    def dfs(pos, opens, stack, pairs):\n+        if pos==2*n:\n+            result.append(frozenset(pairs))\n+            return\n+        if opens<n:\n+            dfs(pos+1,opens+1,stack+[pos],pairs)\n+        if stack:\n+            dfs(pos+1,opens,stack[:-1],pairs+[(stack[-1],pos)])\n+    dfs(0,0,[],[])\n+    return result\n+\n+for n in range(1,8):\n+    p = pairings(n)\n+    brute = [0]*(n+1)\n+    for a in p:\n+        for b in p:\n+            brute[len(a & b)] += 1\n+    assert brute == formula(n), (n, brute, formula(n))\n+assert formula(8) == [378790,646560,542960,296800,127400,34944,16016,0,1430]\n+print('Exact formula matches exhaustive enumeration for n=1..7 and all samples.')\n+\n+cases = [(n,m) for m in (100000007,998244353,999999937) for n in range(1,31)]\n+data = str(len(cases))+'\\n'+''.join(f'{n} {m}\\n' for n,m in cases)\n+output=subprocess.check_output(['./main'],input=data,text=True)\n+for (n,m),line in zip(cases,output.splitlines(),strict=True):\n+    assert list(map(int,line.split())) == [v%m for v in formula(n)]\n+start=time.monotonic()\n+output=subprocess.check_output(['./main'],input='1\\n500 999999937\\n',text=True)\n+a=list(map(int,output.split()))\n+c=math.comb(1000,500)//501\n+assert len(a)==501 and a[-1]==c%999999937 and a[-2]==0\n+assert sum(a)%999999937==c*c%999999937\n+print(f'C++ passed varying-modulus checks and n=500 checks ({time.monotonic()-start:.3f}s).')\n\n",
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                  "relativePath": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/main.cpp",
                  "type": "add",
                  "patch": "Index: /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/main.cpp\n===================================================================\n--- /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/main.cpp\n+++ /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/main.cpp\n@@ -0,0 +1,64 @@\n+#include <bits/stdc++.h>\n+using namespace std;\n+\n+using ll = long long;\n+\n+void setIO() {\n+    ios::sync_with_stdio(false);\n+    cin.tie(nullptr);\n+}\n+\n+int main() {\n+    setIO();\n+    int t;\n+    cin >> t;\n+    while (t--) {\n+        int n;\n+        ll mod;\n+        cin >> n >> mod;\n+        auto power = [&](ll a, ll e) {\n+            ll res = 1;\n+            for (; e; e >>= 1, a = a * a % mod)\n+                if (e & 1) res = res * a % mod;\n+            return res;\n+        };\n+        vector<ll> fact(2 * n + 1), invfact(2 * n + 1);\n+        fact[0] = 1;\n+        for (int i = 1; i <= 2 * n; ++i)\n+            fact[i] = fact[i - 1] * i % mod;\n+        invfact[2 * n] = power(fact[2 * n], mod - 2);\n+        for (int i = 2 * n; i > 0; --i)\n+            invfact[i - 1] = invfact[i] * i % mod;\n+        auto choose = [&](int a, int b) {\n+            return fact[a] * invfact[b] % mod * invfact[a - b] % mod;\n+        };\n+\n+        vector<ll> base(n + 1), ans(n + 1);\n+        for (int i = 0; i <= n; ++i) {\n+            ll catalan = fact[2 * i] * invfact[i] % mod * invfact[i + 1] % mod;\n+            base[i] = catalan * catalan % mod;\n+        }\n+        vector<ll> dp = base;\n+        for (int k = 0; k <= n; ++k) {\n+            ll factor = fact[2 * n] * invfact[k + 1] % mod * invfact[2 * n - k] % mod;\n+            ans[k] = factor * dp[n - k] % mod;\n+            if (k == n) break;\n+\n+            int limit = n - k - 1;\n+            vector<ll> next(limit + 1);\n+            for (int d = 0; d <= limit; ++d)\n+                for (int j = 0; j <= d; ++j)\n+                    next[d] = (next[d] + dp[j] * base[d - j]) % mod;\n+            dp.swap(next);\n+        }\n+\n+        for (int k = n; k >= 0; --k) {\n+            for (int j = k + 1; j <= n; ++j) {\n+                ans[k] -= choose(j, k) * ans[j] % mod;\n+                if (ans[k] < 0) ans[k] += mod;\n+            }\n+        }\n+        for (int k = 0; k <= n; ++k)\n+            cout << ans[k] << (k == n ? '\\n' : ' ');\n+    }\n+}\n",
                  "additions": 64,
                  "deletions": 0
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                {
                  "filePath": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/check.py",
                  "relativePath": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/check.py",
                  "type": "add",
                  "patch": "Index: /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/check.py\n===================================================================\n--- /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/check.py\n+++ /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-u4vVKT/check.py\n@@ -0,0 +1,50 @@\n+import math\n+import subprocess\n+import time\n+\n+def formula(n):\n+    base = [(math.comb(2*i, i)//(i+1))**2 for i in range(n+1)]\n+    dp = base[:]\n+    ans = []\n+    for k in range(n+1):\n+        ans.append(math.comb(2*n,k)*dp[n-k]//(k+1))\n+        dp = [sum(dp[j]*base[d-j] for j in range(d+1)) for d in range(n-k)]\n+    for k in range(n,-1,-1):\n+        ans[k] -= sum(math.comb(j,k)*ans[j] for j in range(k+1,n+1))\n+    return ans\n+\n+def pairings(n):\n+    result=[]\n+    def dfs(pos, opens, stack, pairs):\n+        if pos==2*n:\n+            result.append(frozenset(pairs))\n+            return\n+        if opens<n:\n+            dfs(pos+1,opens+1,stack+[pos],pairs)\n+        if stack:\n+            dfs(pos+1,opens,stack[:-1],pairs+[(stack[-1],pos)])\n+    dfs(0,0,[],[])\n+    return result\n+\n+for n in range(1,8):\n+    p = pairings(n)\n+    brute = [0]*(n+1)\n+    for a in p:\n+        for b in p:\n+            brute[len(a & b)] += 1\n+    assert brute == formula(n), (n, brute, formula(n))\n+assert formula(8) == [378790,646560,542960,296800,127400,34944,16016,0,1430]\n+print('Exact formula matches exhaustive enumeration for n=1..7 and all samples.')\n+\n+cases = [(n,m) for m in (100000007,998244353,999999937) for n in range(1,31)]\n+data = str(len(cases))+'\\n'+''.join(f'{n} {m}\\n' for n,m in cases)\n+output=subprocess.check_output(['./main'],input=data,text=True)\n+for (n,m),line in zip(cases,output.splitlines(),strict=True):\n+    assert list(map(int,line.split())) == [v%m for v in formula(n)]\n+start=time.monotonic()\n+output=subprocess.check_output(['./main'],input='1\\n500 999999937\\n',text=True)\n+a=list(map(int,output.split()))\n+c=math.comb(1000,500)//501\n+assert len(a)==501 and a[-1]==c%999999937 and a[-2]==0\n+assert sum(a)%999999937==c*c%999999937\n+print(f'C++ passed varying-modulus checks and n=500 checks ({time.monotonic()-start:.3f}s).')\n",
                  "additions": 50,
                  "deletions": 0
                }
              ],
              "diagnostics": {},
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          "type": "text",
          "text": "The counting formula matches exhaustive enumeration through $n=7$ and all samples. The C++ implementation also passed checks with different prime moduli and $n=500$. The explanation can avoid advanced generating-function machinery: a tree encoding and a cyclic-rotation argument give the needed formula directly.",
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                {
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                  "status": "completed",
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                },
                {
                  "content": "Write five progressive hints, the self-contained editorial, and C++ solution.",
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                }
              ],
              "truncated": false
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            "title": "0 todos",
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              "content": "Counting **exactly** $k$ common pairings forces you to forbid every extra coincidence. That's the annoying part. Instead, count triples $(s,t,S)$ where $S$ is a chosen set of $k$ common pairings, and allow other common pairings to exist."
            },
            {
              "order": 2,
              "content": "Let $a_j$ be the required answer for exactly $j$ common pairings, and let $B_k$ count triples with $k$ chosen common pairings. Then\n$$B_k=\\sum_{j=k}^{n}\\binom{j}{k}a_j.$$\nOnce you know every $B_k$, recover $a_k$ in decreasing order. So the real problem is counting structures with marked common pairings."
            },
            {
              "order": 3,
              "content": "Marked pairings cannot cross, so their nesting forms a tree with $k+1$ regions, including the outside region. In each region, remove its directly nested marked intervals. What remains is an arbitrary ordered pair of balanced sequences of length $2m$, giving $C_m^2$ choices, where $C_m$ is a Catalan number. Its child intervals can be inserted into the $2m+1$ gaps around these remaining positions."
            },
            {
              "order": 4,
              "content": "Let $D[r][q]$ count ordered lists of $r$ region contents containing $q$ unmarked pairs per sequence in total. Then\n$$D[1][q]=C_q^2,\\qquad D[r+1][q]=\\sum_{j=0}^{q}D[r][j]C_{q-j}^2.$$\nFor $k+1$ regions and $q=n-k$, there are $2n-k+1$ gaps altogether. Distributing $k$ child placeholders among them gives $\\binom{2n}{k}$ possibilities. To turn the list into a valid tree, use the fact that exactly one of its $k+1$ cyclic rotations is a valid preorder encoding."
            },
            {
              "order": 5,
              "content": "The marked counts are\n$$B_k=\\frac{1}{k+1}\\binom{2n}{k}D[k+1][n-k].$$\nCompute successive rows of $D$ by convolution, retaining only degrees up to $n-k$. Then recover\n$$a_k=B_k-\\sum_{j=k+1}^{n}\\binom{j}{k}a_j$$\nfrom $k=n$ down to $0$. Factorials and inverse factorials handle all combinatorial factors modulo $M$. This takes $O(n^3)$ time and $O(n)$ memory with rolling rows; division by $k+1$ must be modular division, not integer division of the binomial coefficient."
            }
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          "editorial": "The trick is to **choose some common pairings instead of demanding that they are the only common pairings**. Extra coincidences are allowed for now; we'll subtract them out later.\n\n## 1. Replace exact counts with marked counts\n\nLet $a_k$ be the answer for exactly $k$ common pairings.\n\nDefine $B_k$ as the number of triples $(s,t,S)$, where $s,t$ are balanced sequences of length $2n$ and $S$ is a set of exactly $k$ common pairings. Call the pairings in $S$ **marked**.\n\nIf $(s,t)$ has $j$ common pairings, there are $\\binom{j}{k}$ choices for $S$. Therefore,\n\n$$B_k=\\sum_{j=k}^{n}\\binom{j}{k}a_j.$$\n\nCareful: $B_k$ is not just the number of pairs with at least $k$ common pairings. A pair with more common pairings is counted multiple times, once for each choice of marks. That multiplicity is exactly what makes the counting manageable.\n\nOnce all $B_k$ are known, solve this triangular system backwards:\n\n$$\\boxed{a_k=B_k-\\sum_{j=k+1}^{n}\\binom{j}{k}a_j.}$$\n\nNow we only need to count marked structures.\n\n## 2. Marked pairings split the structure into regions\n\nPairings in a balanced sequence never cross: their intervals are either disjoint or nested. The marked pairings consequently define a rooted, ordered tree:\n\n- The root represents the region outside all marked pairings.\n- Each marked pairing creates a child region consisting of its interior.\n- Its parent is the nearest containing marked pairing, or the outside region if there isn't one.\n\nWith $k$ marked pairings, this tree has $k$ edges and $k+1$ nodes.\n\nConsider one region. Remove every directly nested marked interval, including its two endpoints and everything inside it. In each sequence, the remaining characters form a balanced sequence: removing a complete matched interval preserves balance.\n\nBoth remaining sequences have the same length, say $2m$. Let\n\n$$C_m=\\frac{(2m)!}{m!(m+1)!}$$\n\nbe the $m$-th Catalan number, with $C_0=1$. There are $C_m^2$ choices for this region's ordered pair of remaining sequences.\n\nThere are also $2m+1$ gaps before, between, and after the remaining positions. Each gap can contain an ordered list of child intervals. A child interval has the form `(`, its recursively constructed interior, then `)` in both sequences.\n\nInserting such a balanced interval into the same gap of both sequences preserves balance and creates a common pairing at its endpoints. Conversely, removing the marked intervals recovers this description uniquely.\n\n**Unmarked pairings may still be common.** We are counting a chosen subset of common pairings, so that is intentional.\n\nAcross all regions, the remaining sequences contain $n-k$ unmarked pairs per sequence in total.\n\n## 3. Count lists of region contents\n\nFor the moment, ignore how the regions connect into a tree.\n\nLet $D[r][q]$ count ordered lists of $r$ region contents whose sizes sum to $q$. Equivalently,\n\n$$D[r][q]=\\sum_{m_1+\\cdots+m_r=q}\\prod_{i=1}^{r}C_{m_i}^2.$$\n\nThis is a straightforward convolution DP:\n\n$$D[1][q]=C_q^2,$$\n\n$$\\boxed{D[r+1][q]=\\sum_{j=0}^{q}D[r][j]C_{q-j}^2.}$$\n\nFor $k$ marked pairings, we need $r=k+1$ regions and $q=n-k$.\n\nThe remaining question is: how many ways can these region records be connected into a valid tree?\n\n## 4. Count child positions, then fix the root\n\nTake an ordered list of $k+1$ region contents with sizes summing to $n-k$.\n\nThe total number of gaps in these records is\n\n$$\\sum_{i=1}^{k+1}(2m_i+1)=2(n-k)+(k+1)=2n-k+1.$$\n\nPlace a total of $k$ child placeholders among those gaps. Multiple placeholders may occupy one gap; their order becomes the order of the children.\n\nBy stars and bars, the number of distributions is\n\n$$\\binom{k+(2n-k+1)-1}{k}=\\binom{2n}{k}.$$\n\nThus, before checking whether the records form a valid tree, the total is\n\n$$\\binom{2n}{k}D[k+1][n-k].$$\n\n### Why divide by $k+1$?\n\nEach record now specifies its local sequence pair and its child placeholders. Let $d_i$ be the number of children requested by record $i$, and write $r=k+1$.\n\nWe want to interpret the records as a tree's preorder traversal. Start with one available position for the root. Processing a record consumes one position and creates $d_i$ positions for its children. After $p$ records, the number of available positions is\n\n$$1+\\sum_{i=1}^{p}(d_i-1).$$\n\nA valid traversal must have a position available until the final record, and none afterward. Since $\\sum d_i=k=r-1$, this means\n\n$$\\sum_{i=1}^{p}(d_i-1)\\ge 0\\quad\\text{for }p<r,$$\n\nwhile the full sum is $-1$.\n\nHere is the useful cyclic-rotation fact: **for an integer sequence with total sum $-1$, exactly one cyclic starting point has all proper prefix sums nonnegative**. To see it, form the prefix sums before the final element and start immediately after their first minimum. Prefixes after that minimum cannot go lower; wrapped prefixes stay nonnegative because earlier prefix sums are strictly larger. Those same inequalities also force the starting point to be unique.\n\nApply this to $d_i-1$. Exactly one of the $r=k+1$ rotations of the complete records is a valid tree encoding. Its tree is uniquely reconstructed by filling child placeholders in preorder.\n\nRotating records keeps their contents and gap assignments intact. Also, the rotations are distinct: a shorter repetition would make the total sum $-1$ a multiple of the number of repetitions, which is impossible.\n\nTherefore exactly a $1/(k+1)$ fraction of all record lists are valid. We obtain\n\n$$\\boxed{B_k=\\frac{1}{k+1}\\binom{2n}{k}D[k+1][n-k].}$$\n\nThe division applies to the total count, not to each fixed list of region sizes: rotations can change the order of those sizes.\n\n## 5. Algorithm and modular arithmetic\n\nFor each test case:\n\n1. Precompute factorials and inverse factorials through $2n$ modulo $M$.\n2. Compute $\\text{base}[m]=C_m^2$ for $0\\le m\\le n$.\n3. Initialize the current DP row to $D[1][q]=\\text{base}[q]$.\n4. For $k=0,1,\\ldots,n$:\n   - Calculate $B_k$ from the current row $D[k+1]$.\n   - Convolve the row with `base` to obtain the next row.\n5. Recover $a_k$ in decreasing order using the triangular subtraction formula.\n\nOnly one DP row and the next row are needed. At iteration $k$, we need degrees only through $n-k$. The next row can be truncated to degree $n-k-1$, because later convolutions never decrease degree.\n\nFor implementation, combine the binomial coefficient and division into\n\n$$\\frac{1}{k+1}\\binom{2n}{k}\n=\\frac{(2n)!}{(k+1)!(2n-k)!}.$$\n\nUse inverse factorials for this expression. **Do not use ordinary integer division on $\\binom{2n}{k}$:** that quotient need not be an integer by itself.\n\nSince $M$ is prime and $2n<M$, all factorials we invert are nonzero modulo $M$. One modular exponentiation computes the largest inverse factorial; the rest follow by a backward loop. Recompute these arrays for each test case because the modulus can change.\n\nUse `long long` for products and normalize every modular subtraction.\n\n### Small example\n\nFor $n=2$, the marked counts are\n\n$$B=[4,4,2].$$\n\nBack-substitution gives\n\n$$a_2=2,\\qquad a_1=4-2\\cdot2=0,\\qquad a_0=4-0-2=2,$$\n\nwhich produces $[2,0,2]$.\n\n## Correctness argument\n\nEvery pair of sequences with a chosen set of marked common pairings has a unique region-tree decomposition. Conversely, any such tree with local balanced sequence pairs and child-gap assignments reconstructs a valid marked pair of sequences. Thus this decomposition is a bijection.\n\nThe convolution DP counts all ordered lists of local region contents. Stars and bars counts their child-gap assignments, and the cyclic-rotation argument selects exactly the lists that encode a tree. Hence the formula computes each marked count $B_k$ correctly.\n\nFinally, a pair with exactly $j$ common pairings contributes $\\binom{j}{k}$ to $B_k$. Descending back-substitution removes all contributions from $j>k$, leaving precisely $a_k$. Therefore every output is correct.\n\n## Complexity\n\nThere are $O(n)$ convolution steps, each taking at most $O(n^2)$ time. Back-substitution takes $O(n^2)$ time.\n\n- **Time:** $O(n^3+\\log M)$ per test case.\n- **Extra space:** $O(n)$ using rolling DP rows.\n\nThe bound $\\sum n\\le500$ makes the cubic approach sufficient.",
          "solution": "#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        ll mod;\n        cin >> n >> mod;\n        auto power = [&](ll a, ll e) {\n            ll res = 1;\n            for (; e; e >>= 1, a = a * a % mod)\n                if (e & 1) res = res * a % mod;\n            return res;\n        };\n        vector<ll> fact(2 * n + 1), invfact(2 * n + 1);\n        fact[0] = 1;\n        for (int i = 1; i <= 2 * n; ++i)\n            fact[i] = fact[i - 1] * i % mod;\n        invfact[2 * n] = power(fact[2 * n], mod - 2);\n        for (int i = 2 * n; i > 0; --i)\n            invfact[i - 1] = invfact[i] * i % mod;\n        auto choose = [&](int a, int b) {\n            return fact[a] * invfact[b] % mod * invfact[a - b] % mod;\n        };\n\n        vector<ll> base(n + 1), ans(n + 1);\n        for (int i = 0; i <= n; ++i) {\n            ll catalan = fact[2 * i] * invfact[i] % mod * invfact[i + 1] % mod;\n            base[i] = catalan * catalan % mod;\n        }\n        vector<ll> dp = base;\n        for (int k = 0; k <= n; ++k) {\n            ll factor = fact[2 * n] * invfact[k + 1] % mod * invfact[2 * n - k] % mod;\n            ans[k] = factor * dp[n - k] % mod;\n            if (k == n) break;\n\n            int limit = n - k - 1;\n            vector<ll> next(limit + 1);\n            for (int d = 0; d <= limit; ++d)\n                for (int j = 0; j <= d; ++j)\n                    next[d] = (next[d] + dp[j] * base[d - j]) % mod;\n            dp.swap(next);\n        }\n\n        for (int k = n; k >= 0; --k) {\n            for (int j = k + 1; j <= n; ++j) {\n                ans[k] -= choose(j, k) * ans[j] % mod;\n                if (ans[k] < 0) ans[k] += mod;\n            }\n        }\n        for (int k = 0; k <= n; ++k)\n            cout << ans[k] << (k == n ? '\\n' : ' ');\n    }\n}"
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                  "content": "Counting **exactly** $k$ common pairings forces you to forbid every extra coincidence. That's the annoying part. Instead, count triples $(s,t,S)$ where $S$ is a chosen set of $k$ common pairings, and allow other common pairings to exist."
                },
                {
                  "order": 2,
                  "content": "Let $a_j$ be the required answer for exactly $j$ common pairings, and let $B_k$ count triples with $k$ chosen common pairings. Then\n$$B_k=\\sum_{j=k}^{n}\\binom{j}{k}a_j.$$\nOnce you know every $B_k$, recover $a_k$ in decreasing order. So the real problem is counting structures with marked common pairings."
                },
                {
                  "order": 3,
                  "content": "Marked pairings cannot cross, so their nesting forms a tree with $k+1$ regions, including the outside region. In each region, remove its directly nested marked intervals. What remains is an arbitrary ordered pair of balanced sequences of length $2m$, giving $C_m^2$ choices, where $C_m$ is a Catalan number. Its child intervals can be inserted into the $2m+1$ gaps around these remaining positions."
                },
                {
                  "order": 4,
                  "content": "Let $D[r][q]$ count ordered lists of $r$ region contents containing $q$ unmarked pairs per sequence in total. Then\n$$D[1][q]=C_q^2,\\qquad D[r+1][q]=\\sum_{j=0}^{q}D[r][j]C_{q-j}^2.$$\nFor $k+1$ regions and $q=n-k$, there are $2n-k+1$ gaps altogether. Distributing $k$ child placeholders among them gives $\\binom{2n}{k}$ possibilities. To turn the list into a valid tree, use the fact that exactly one of its $k+1$ cyclic rotations is a valid preorder encoding."
                },
                {
                  "order": 5,
                  "content": "The marked counts are\n$$B_k=\\frac{1}{k+1}\\binom{2n}{k}D[k+1][n-k].$$\nCompute successive rows of $D$ by convolution, retaining only degrees up to $n-k$. Then recover\n$$a_k=B_k-\\sum_{j=k+1}^{n}\\binom{j}{k}a_j$$\nfrom $k=n$ down to $0$. Factorials and inverse factorials handle all combinatorial factors modulo $M$. This takes $O(n^3)$ time and $O(n)$ memory with rolling rows; division by $k+1$ must be modular division, not integer division of the binomial coefficient."
                }
              ],
              "editorial": "The trick is to **choose some common pairings instead of demanding that they are the only common pairings**. Extra coincidences are allowed for now; we'll subtract them out later.\n\n## 1. Replace exact counts with marked counts\n\nLet $a_k$ be the answer for exactly $k$ common pairings.\n\nDefine $B_k$ as the number of triples $(s,t,S)$, where $s,t$ are balanced sequences of length $2n$ and $S$ is a set of exactly $k$ common pairings. Call the pairings in $S$ **marked**.\n\nIf $(s,t)$ has $j$ common pairings, there are $\\binom{j}{k}$ choices for $S$. Therefore,\n\n$$B_k=\\sum_{j=k}^{n}\\binom{j}{k}a_j.$$\n\nCareful: $B_k$ is not just the number of pairs with at least $k$ common pairings. A pair with more common pairings is counted multiple times, once for each choice of marks. That multiplicity is exactly what makes the counting manageable.\n\nOnce all $B_k$ are known, solve this triangular system backwards:\n\n$$\\boxed{a_k=B_k-\\sum_{j=k+1}^{n}\\binom{j}{k}a_j.}$$\n\nNow we only need to count marked structures.\n\n## 2. Marked pairings split the structure into regions\n\nPairings in a balanced sequence never cross: their intervals are either disjoint or nested. The marked pairings consequently define a rooted, ordered tree:\n\n- The root represents the region outside all marked pairings.\n- Each marked pairing creates a child region consisting of its interior.\n- Its parent is the nearest containing marked pairing, or the outside region if there isn't one.\n\nWith $k$ marked pairings, this tree has $k$ edges and $k+1$ nodes.\n\nConsider one region. Remove every directly nested marked interval, including its two endpoints and everything inside it. In each sequence, the remaining characters form a balanced sequence: removing a complete matched interval preserves balance.\n\nBoth remaining sequences have the same length, say $2m$. Let\n\n$$C_m=\\frac{(2m)!}{m!(m+1)!}$$\n\nbe the $m$-th Catalan number, with $C_0=1$. There are $C_m^2$ choices for this region's ordered pair of remaining sequences.\n\nThere are also $2m+1$ gaps before, between, and after the remaining positions. Each gap can contain an ordered list of child intervals. A child interval has the form `(`, its recursively constructed interior, then `)` in both sequences.\n\nInserting such a balanced interval into the same gap of both sequences preserves balance and creates a common pairing at its endpoints. Conversely, removing the marked intervals recovers this description uniquely.\n\n**Unmarked pairings may still be common.** We are counting a chosen subset of common pairings, so that is intentional.\n\nAcross all regions, the remaining sequences contain $n-k$ unmarked pairs per sequence in total.\n\n## 3. Count lists of region contents\n\nFor the moment, ignore how the regions connect into a tree.\n\nLet $D[r][q]$ count ordered lists of $r$ region contents whose sizes sum to $q$. Equivalently,\n\n$$D[r][q]=\\sum_{m_1+\\cdots+m_r=q}\\prod_{i=1}^{r}C_{m_i}^2.$$\n\nThis is a straightforward convolution DP:\n\n$$D[1][q]=C_q^2,$$\n\n$$\\boxed{D[r+1][q]=\\sum_{j=0}^{q}D[r][j]C_{q-j}^2.}$$\n\nFor $k$ marked pairings, we need $r=k+1$ regions and $q=n-k$.\n\nThe remaining question is: how many ways can these region records be connected into a valid tree?\n\n## 4. Count child positions, then fix the root\n\nTake an ordered list of $k+1$ region contents with sizes summing to $n-k$.\n\nThe total number of gaps in these records is\n\n$$\\sum_{i=1}^{k+1}(2m_i+1)=2(n-k)+(k+1)=2n-k+1.$$\n\nPlace a total of $k$ child placeholders among those gaps. Multiple placeholders may occupy one gap; their order becomes the order of the children.\n\nBy stars and bars, the number of distributions is\n\n$$\\binom{k+(2n-k+1)-1}{k}=\\binom{2n}{k}.$$\n\nThus, before checking whether the records form a valid tree, the total is\n\n$$\\binom{2n}{k}D[k+1][n-k].$$\n\n### Why divide by $k+1$?\n\nEach record now specifies its local sequence pair and its child placeholders. Let $d_i$ be the number of children requested by record $i$, and write $r=k+1$.\n\nWe want to interpret the records as a tree's preorder traversal. Start with one available position for the root. Processing a record consumes one position and creates $d_i$ positions for its children. After $p$ records, the number of available positions is\n\n$$1+\\sum_{i=1}^{p}(d_i-1).$$\n\nA valid traversal must have a position available until the final record, and none afterward. Since $\\sum d_i=k=r-1$, this means\n\n$$\\sum_{i=1}^{p}(d_i-1)\\ge 0\\quad\\text{for }p<r,$$\n\nwhile the full sum is $-1$.\n\nHere is the useful cyclic-rotation fact: **for an integer sequence with total sum $-1$, exactly one cyclic starting point has all proper prefix sums nonnegative**. To see it, form the prefix sums before the final element and start immediately after their first minimum. Prefixes after that minimum cannot go lower; wrapped prefixes stay nonnegative because earlier prefix sums are strictly larger. Those same inequalities also force the starting point to be unique.\n\nApply this to $d_i-1$. Exactly one of the $r=k+1$ rotations of the complete records is a valid tree encoding. Its tree is uniquely reconstructed by filling child placeholders in preorder.\n\nRotating records keeps their contents and gap assignments intact. Also, the rotations are distinct: a shorter repetition would make the total sum $-1$ a multiple of the number of repetitions, which is impossible.\n\nTherefore exactly a $1/(k+1)$ fraction of all record lists are valid. We obtain\n\n$$\\boxed{B_k=\\frac{1}{k+1}\\binom{2n}{k}D[k+1][n-k].}$$\n\nThe division applies to the total count, not to each fixed list of region sizes: rotations can change the order of those sizes.\n\n## 5. Algorithm and modular arithmetic\n\nFor each test case:\n\n1. Precompute factorials and inverse factorials through $2n$ modulo $M$.\n2. Compute $\\text{base}[m]=C_m^2$ for $0\\le m\\le n$.\n3. Initialize the current DP row to $D[1][q]=\\text{base}[q]$.\n4. For $k=0,1,\\ldots,n$:\n   - Calculate $B_k$ from the current row $D[k+1]$.\n   - Convolve the row with `base` to obtain the next row.\n5. Recover $a_k$ in decreasing order using the triangular subtraction formula.\n\nOnly one DP row and the next row are needed. At iteration $k$, we need degrees only through $n-k$. The next row can be truncated to degree $n-k-1$, because later convolutions never decrease degree.\n\nFor implementation, combine the binomial coefficient and division into\n\n$$\\frac{1}{k+1}\\binom{2n}{k}\n=\\frac{(2n)!}{(k+1)!(2n-k)!}.$$\n\nUse inverse factorials for this expression. **Do not use ordinary integer division on $\\binom{2n}{k}$:** that quotient need not be an integer by itself.\n\nSince $M$ is prime and $2n<M$, all factorials we invert are nonzero modulo $M$. One modular exponentiation computes the largest inverse factorial; the rest follow by a backward loop. Recompute these arrays for each test case because the modulus can change.\n\nUse `long long` for products and normalize every modular subtraction.\n\n### Small example\n\nFor $n=2$, the marked counts are\n\n$$B=[4,4,2].$$\n\nBack-substitution gives\n\n$$a_2=2,\\qquad a_1=4-2\\cdot2=0,\\qquad a_0=4-0-2=2,$$\n\nwhich produces $[2,0,2]$.\n\n## Correctness argument\n\nEvery pair of sequences with a chosen set of marked common pairings has a unique region-tree decomposition. Conversely, any such tree with local balanced sequence pairs and child-gap assignments reconstructs a valid marked pair of sequences. Thus this decomposition is a bijection.\n\nThe convolution DP counts all ordered lists of local region contents. Stars and bars counts their child-gap assignments, and the cyclic-rotation argument selects exactly the lists that encode a tree. Hence the formula computes each marked count $B_k$ correctly.\n\nFinally, a pair with exactly $j$ common pairings contributes $\\binom{j}{k}$ to $B_k$. Descending back-substitution removes all contributions from $j>k$, leaving precisely $a_k$. Therefore every output is correct.\n\n## Complexity\n\nThere are $O(n)$ convolution steps, each taking at most $O(n^2)$ time. Back-substitution takes $O(n^2)$ time.\n\n- **Time:** $O(n^3+\\log M)$ per test case.\n- **Extra space:** $O(n)$ using rolling DP rows.\n\nThe bound $\\sum n\\le500$ makes the cubic approach sufficient.",
              "solution": "#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        ll mod;\n        cin >> n >> mod;\n        auto power = [&](ll a, ll e) {\n            ll res = 1;\n            for (; e; e >>= 1, a = a * a % mod)\n                if (e & 1) res = res * a % mod;\n            return res;\n        };\n        vector<ll> fact(2 * n + 1), invfact(2 * n + 1);\n        fact[0] = 1;\n        for (int i = 1; i <= 2 * n; ++i)\n            fact[i] = fact[i - 1] * i % mod;\n        invfact[2 * n] = power(fact[2 * n], mod - 2);\n        for (int i = 2 * n; i > 0; --i)\n            invfact[i - 1] = invfact[i] * i % mod;\n        auto choose = [&](int a, int b) {\n            return fact[a] * invfact[b] % mod * invfact[a - b] % mod;\n        };\n\n        vector<ll> base(n + 1), ans(n + 1);\n        for (int i = 0; i <= n; ++i) {\n            ll catalan = fact[2 * i] * invfact[i] % mod * invfact[i + 1] % mod;\n            base[i] = catalan * catalan % mod;\n        }\n        vector<ll> dp = base;\n        for (int k = 0; k <= n; ++k) {\n            ll factor = fact[2 * n] * invfact[k + 1] % mod * invfact[2 * n - k] % mod;\n            ans[k] = factor * dp[n - k] % mod;\n            if (k == n) break;\n\n            int limit = n - k - 1;\n            vector<ll> next(limit + 1);\n            for (int d = 0; d <= limit; ++d)\n                for (int j = 0; j <= d; ++j)\n                    next[d] = (next[d] + dp[j] * base[d - j]) % mod;\n            dp.swap(next);\n        }\n\n        for (int k = n; k >= 0; --k) {\n            for (int j = k + 1; j <= n; ++j) {\n                ans[k] -= choose(j, k) * ans[j] % mod;\n                if (ans[k] < 0) ans[k] += mod;\n            }\n        }\n        for (int k = 0; k <= n; ++k)\n            cout << ans[k] << (k == n ? '\\n' : ' ');\n    }\n}"
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            "output": "Structured output captured successfully.",
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