{
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    "id": "ses_f5341cf67ffemT7NhbK68OoURy",
    "slug": "proud-moon",
    "projectID": "global",
    "directory": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-d1B4yi",
    "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-d1B4yi",
    "title": "Nudge content generation",
    "agent": "nudge-generation",
    "model": {
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      "providerID": "openai",
      "variant": "max"
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      "reasoning": 17348,
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  "messages": [
    {
      "info": {
        "role": "user",
        "time": {
          "created": 1789604540582
        },
        "agent": "nudge-generation",
        "model": {
          "providerID": "openai",
          "modelID": "gpt-5.6-luna",
          "variant": "max"
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        "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",
                    "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",
                          "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",
                    "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",
                    "minLength": 1,
                    "maxLength": 200000
                  },
                  {
                    "type": "null"
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              }
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              "status",
              "reason",
              "hints",
              "editorial",
              "solution"
            ],
            "additionalProperties": false,
            "title": "problem_response",
            "description": "Submit the generated content for the problem, or report if it is unsolvable."
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          "retryCount": 2
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          "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 2262D: \"PLUSworld\".\n\nKnown problem metadata:\n- Contest ID: 2262\n- Index: D\n- Name: PLUSworld\n- Rating: unknown / unrated\n- Tags: constructive algorithms, graphs, greedy\n- Problem URL: https://codeforces.com/contest/2262/problem/D\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\">D. PLUSworld</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 building <span class=\"tex-font-style-it\">PLUSworld</span>, a giant barn complex with $$$n$$$ barns numbered from $$$1$$$ to $$$n$$$. For each $$$1 \\le i \\le n$$$, there is currently a one-way walkway from barn $$$i$$$ to barn $$$a_i$$$.</p><p>Before opening <span class=\"tex-font-style-it\">PLUSworld</span>, Farmer John wants to reconfigure the roads so that, for every $$$1 \\le i \\le n$$$, the road from barn $$$i$$$ points to barn $$$b_i$$$. The target roads have the special property that each road points to its own barn or to a higher-numbered barn $$$(i \\le b_i)$$$.</p><p>To do this, he sends Bessie to one barn of his choice. Let $$$p$$$ be the barn where Bessie is currently located. Bessie may perform the following operations any number of times:</p><ul> <li> Operation $$$1$$$: Increase the destination of the road from the current barn by $$$1$$$. More formally, set $$$a_p := a_p + 1$$$. This operation may only be performed if $$$a_p  \\lt  n$$$. </li><li> Operation $$$2$$$: Follow the road from the current barn. More formally, set $$$p := a_p$$$. </li></ul><p>Operation $$$1$$$ changes the array $$$a$$$, but Bessie stays at the same barn. Operation $$$2$$$ changes Bessie's current barn, but does not change the array.</p><p>Your goal is to make the array $$$a$$$ equal to the array $$$b$$$.</p><p>Determine whether this is possible. If it is possible, output any valid starting barn and any sequence of operations that makes $$$a=b$$$.</p><p>It can be shown that if a solution exists, then there exists one using at most $$$2n^2$$$ operations.</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 $$$t$$$ ($$$1 \\le t \\le 10^4$$$). The description of the test cases follows.</p><p>The first line of each test case contains a single integer $$$n$$$ ($$$1\\le n\\le 1000$$$).</p><p>The second line contains $$$n$$$ integers $$$a_1,a_2,\\ldots,a_n$$$ ($$$1\\le a_i\\le n$$$).</p><p>The third line contains $$$n$$$ integers $$$b_1,b_2,\\ldots,b_n$$$ ($$$\\color{red}{i\\le b_i}\\le n$$$).</p><p>It is guaranteed that the sum of $$$n^2$$$ over all test cases does not exceed $$$1000^2$$$.</p></div><div class=\"output-specification\"><div class=\"section-title\">Output</div><p>For each test case, if it is impossible to make $$$a$$$ equal to $$$b$$$, output a single integer $$$-1$$$.</p><p>Otherwise, output two lines.</p><p>On the first line, output two integers $$$o$$$ and $$$p$$$ — the number of operations you will perform and the initial index you choose ($$$0 \\leq o \\leq 2n^2$$$, $$$1 \\leq p \\leq n$$$). </p><p>On the second line, output $$$o$$$ integers $$$c_1,c_2,\\ldots,c_o$$$ ($$$1\\le c_i\\le 2$$$), where $$$c_i$$$ is the type of the $$$i$$$-th operation.</p><p>If $$$c_i=1$$$, you perform operation $$$1$$$ and set $$$a_p:=a_p+1$$$. This operation may only be used if $$$a_p  \\lt  n$$$ at that moment.</p><p>If $$$c_i=2$$$, you perform operation $$$2$$$ and set $$$p:=a_p$$$. </p><p>After performing all $$$o$$$ operations, the array $$$a$$$ must be equal to $$$b$$$.</p><p>If there are multiple valid answers, you may output any of them.</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\">9</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">3</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">1 2 3</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">2 3 3</div><div class=\"test-example-line test-example-line-even test-example-line-2\">2</div><div class=\"test-example-line test-example-line-even test-example-line-2\">2 2</div><div class=\"test-example-line test-example-line-even test-example-line-2\">1 2</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">5</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">1 1 1 4 5</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">2 3 4 5 5</div><div class=\"test-example-line test-example-line-even test-example-line-4\">3</div><div class=\"test-example-line test-example-line-even test-example-line-4\">1 1 1</div><div class=\"test-example-line test-example-line-even test-example-line-4\">1 2 3</div><div class=\"test-example-line test-example-line-odd test-example-line-5\">3</div><div class=\"test-example-line test-example-line-odd test-example-line-5\">1 1 1</div><div class=\"test-example-line test-example-line-odd test-example-line-5\">3 2 3</div><div class=\"test-example-line test-example-line-even test-example-line-6\">3</div><div class=\"test-example-line test-example-line-even test-example-line-6\">1 1 3</div><div class=\"test-example-line test-example-line-even test-example-line-6\">1 2 3</div><div class=\"test-example-line test-example-line-odd test-example-line-7\">4</div><div class=\"test-example-line test-example-line-odd test-example-line-7\">1 1 1 4</div><div class=\"test-example-line test-example-line-odd test-example-line-7\">3 3 4 4</div><div class=\"test-example-line test-example-line-even test-example-line-8\">4</div><div class=\"test-example-line test-example-line-even test-example-line-8\">1 4 3 4</div><div class=\"test-example-line test-example-line-even test-example-line-8\">1 4 3 4</div><div class=\"test-example-line test-example-line-odd test-example-line-9\">5</div><div class=\"test-example-line test-example-line-odd test-example-line-9\">1 3 4 5 4</div><div class=\"test-example-line test-example-line-odd test-example-line-9\">2 3 4 5 5</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\">5 1</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">1 2 1 2 2 </div><div class=\"test-example-line test-example-line-even test-example-line-2\">-1</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">12 1</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">1 2 1 1 2 1 1 1 2 1 2 2 </div><div class=\"test-example-line test-example-line-even test-example-line-4\">-1</div><div class=\"test-example-line test-example-line-odd test-example-line-5\">-1</div><div class=\"test-example-line test-example-line-even test-example-line-6\">2 2</div><div class=\"test-example-line test-example-line-even test-example-line-6\">1 2 </div><div class=\"test-example-line test-example-line-odd test-example-line-7\">12 1</div><div class=\"test-example-line test-example-line-odd test-example-line-7\">1 1 2 1 2 1 1 2 1 1 2 2 </div><div class=\"test-example-line test-example-line-even test-example-line-8\">0 1</div><div class=\"test-example-line test-example-line-odd test-example-line-9\">7 1</div><div class=\"test-example-line test-example-line-odd test-example-line-9\">1 2 2 2 2 1 2 </div></pre></div></div></div><div class=\"note\"><div class=\"section-title\">Note</div><p>For the first test case, Bessie starts at barn $$$1$$$. She increases $$$a_1$$$ from $$$1$$$ to $$$2$$$, follows the road to barn $$$2$$$, and increases $$$a_2$$$ from $$$2$$$ to $$$3$$$. The resulting array is $$$[2,3,3]$$$, which is equal to $$$b$$$.</p><p>For the second test case, $$$a_1=2 \\gt b_1=1$$$. Since road destinations can only be increased, it is impossible to make $$$a_1$$$ equal to $$$b_1$$$, so the answer is $$$-1$$$.</p><p>For the third test case, Bessie starts at barn $$$1$$$. She changes $$$a_1$$$ from $$$1$$$ to $$$2$$$ and follows the road to barn $$$2$$$. She then changes $$$a_2$$$ from $$$1$$$ to $$$3$$$ and moves to barn $$$3$$$, changes $$$a_3$$$ from $$$1$$$ to $$$4$$$ and moves to barn $$$4$$$, and finally changes $$$a_4$$$ from $$$4$$$ to $$$5$$$. The resulting array is $$$[2,3,4,5,5]$$$, which is equal to $$$b$$$.</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 2262D)\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- 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 JUST the hint text. 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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            },
            "output": "---\nimage: https://codeforces.com/codeforces.org/s/21069/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/21069/images/codeforces-logo-with-telegram-555x130.png \"Codeforces\")](https://codeforces.com/) \n\n[![In English](//codeforces.com/codeforces.org/s/21069/images/flags/24/gb.png \"In English\")](https://codeforces.com/blog/entry/156688?locale=en) [![По-русски](//codeforces.com/codeforces.org/s/21069/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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[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/21069/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; 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Full output saved to: /Users/zayd/.local/share/opencode/tool-output/tool_0acbe4c6a0018n3hSXb9RYZmXX\nUse the Task tool to have explore agent process this file with Grep and Read (with offset/limit). Do NOT read the full file yourself - delegate to save context.",
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              "prompt": "Process the fetched official tutorial output file at /Users/zayd/.local/share/opencode/tool-output/tool_0acbe4c6a0018n3hSXb9RYZmXX. Use Grep and Read (not full file dump) to locate the 2263F/2262D PLUSworld section, then summarize the exact algorithm, impossibility conditions, proof ideas, output construction, and any caveats. Do not edit files. Return a concise but technically complete research report suitable for independently writing hints/editorial/C++ for the supplied 2262D statement. Verify it matches the supplied problem, noting if the official section contains multiple solutions or an incorrect/AI-generated one.",
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            "output": "<task id=\"ses_f53419288ffeY7WsPDtWiuR2bi\" state=\"completed\">\n<task_result>\n## Research report\n\n### Source verification\n\nThe section is at:\n\n`/Users/zayd/.local/share/opencode/tool-output/tool_0acbe4c6a0018n3hSXb9RYZmXX`\n\n- `2263F/2262D - PLUSworld`: lines 548–550.\n- The tutorial text is missing: lines 552–554 say “Tutorial is loading…”.\n- Primary solution code: lines 556–740.\n- A second “Magic Solution”: lines 742–830.\n\nThe current 2262D statement matches the code: arbitrary arrays \\(a,b\\), target guarantee \\(i\\le b_i\\), operation 1 increments \\(a_p\\), operation 2 sets \\(p=a_p\\), and output is an operation sequence plus initial position. The code converts values to zero-based indices.\n\n---\n\n## Primary SCC solution\n\n### 1. Immediate impossibility\n\nIf any \\(a_i>b_i\\), impossible: destinations only increase.\n\nIf \\(a=b\\), output `0 1`.\n\n### 2. Mark useful vertices\n\nLet a vertex be **bad** if \\(a_i\\ne b_i\\). For every bad \\(i\\), repeatedly follow the target mapping \\(i\\mapsto b_i\\), marking every visited vertex.\n\nBecause \\(b_i\\ge i\\), every such path eventually reaches a fixed point \\(e\\) with \\(b_e=e\\).\n\nOnly useful vertices matter: after a bad vertex is finalized, Bessie follows its now-fixed target edge, so its entire \\(b\\)-suffix may be forced.\n\nIf more than one useful fixed point exists, output `-1`. A fixed-point barn is an absorbing location once reached; one walk cannot service two required terminal branches.\n\n### 3. Build the interval graph\n\nFor every vertex \\(i\\), create directed edges\n\n\\[\ni\\to j\\qquad\\text{for every }a_i\\le j\\le b_i.\n\\]\n\nThis represents all destinations Bessie can choose from \\(i\\) by repeatedly applying operation 1 and then operation 2.\n\nCompute SCCs with Tarjan.\n\nFor each SCC \\(C\\), store:\n\n- `sz[C]`: number of bad vertices in it.\n\nFor every useful \\(i\\) whose target \\(b_i\\) lies in a different SCC, set\n\n```text\npred[SCC(b_i)] = SCC(i)\n```\n\nIf `pred[SCC(b_i)]` was already set, reject. The code rejects any second useful cross-SCC edge into the same target SCC, even if the source SCC is identical.\n\n### 4. Extract the required SCC chain\n\nStart at the SCC containing the unique useful fixed point. Repeatedly follow `pred` backwards. This gives the required components in reverse order; reverse the list.\n\nSum `sz` over this chain. If the sum is not the total number of bad vertices, some required bad vertex lies outside the single usable chain; output `-1`.\n\nThese are the primary algorithm’s complete rejection conditions:\n\n1. Some \\(a_i>b_i\\).\n2. More than one useful fixed point.\n3. Multiple useful cross-SCC target edges entering one SCC.\n4. Not all bad vertices lie on the extracted SCC chain.\n\n### 5. Construct operations inside each SCC\n\nFor each SCC in chain order:\n\n1. Pick its smallest-index vertex as the root.\n2. Run DFS on the reverse interval graph restricted to this SCC. For every vertex \\(v\\), store `p2[v]` such that \\(v\\to p2[v]\\) is an internal edge. This forms a tree toward the root.\n3. Process the SCC’s vertices in increasing index order.\n\nFor each target vertex `v`:\n\n```text\nwhile current != v:\n    while a[current] < max(p2[current], v):\n        emit operation 1\n        ++a[current]\n    emit operation 2\n    current = a[current]\n\nwhile a[v] < b[v]:\n    emit operation 1\n    ++a[v]\n\nemit operation 2\ncurrent = a[v]\n```\n\nThe `max(p2[current], v)` is important: it prevents trying to move to a destination below the already-advanced edge while preserving an internal route through the SCC. The sorted order of vertices and the SCC interval structure guarantee that these jumps remain usable.\n\nAfter processing the final vertex of an SCC, its finalized target edge enters the next SCC in the chain.\n\n### Correctness ideas\n\n- Monotonicity gives \\(a_i\\le b_i\\).\n- The interval graph captures every possible one-step destination before an edge is exhausted.\n- Target edges of useful vertices impose a forced order after finalization.\n- A single walk cannot leave distinct SCCs and later return; therefore the useful SCCs must form one chain, not branches.\n- The reverse-DFS tree gives internal routes, while increasing-index processing handles the fact that edges cannot be decreased.\n- Every bad vertex on the chain is eventually set exactly to \\(b_i\\).\n\n### Complexity\n\nThe interval graph has at most \\(n^2\\) edges.\n\n- SCC computation: \\(O(n^2)\\).\n- Construction: \\(O(n^2)\\).\n- Memory: \\(O(n^2)\\).\n- The produced sequence is intended to stay within the promised \\(2n^2\\) operations.\n\n---\n\n## “Magic Solution” alternative\n\nThe second listing is a much simpler greedy simulation:\n\n1. Reject if any \\(a_i>b_i\\).\n2. Let `start` be the first index with \\(a_i<b_i\\).\n3. Let `remaining` be the number of initially bad positions.\n4. At current position `p`:\n\n```text\nwhile a[p] < b[p] and\n      (a[p] >= p or a[a[p]] == b[a[p]]):\n    emit 1\n    ++a[p]\n\nif a[p] == b[p]:\n    --remaining\n\nif remaining == 0 or a[p] == p:\n    stop\n\nemit 2\np = a[p]\n```\n\nIf `remaining>0` after stopping, output `-1`; otherwise output the generated operations and `start`.\n\nInterpretation:\n\n- If the current edge points forward/nondecreasingly, advance it.\n- If it points to an already finalized vertex, advance past it.\n- Otherwise it points to an unfinished lower-index vertex, so follow it and process that vertex first.\n- Reaching a finalized self-loop before all bad vertices are done is fatal.\n\n### Caveat\n\nThis alternative is explicitly marked:\n\n```cpp\n// Note: AI generated solution\n```\n\n(lines 746–750), and the preceding text says its proof will be added later. The fetched artifact contains no proof or operation-bound argument for it. It appears intended as a valid simplification, but it should be treated as unverified unless independently proved. For a dependable editorial, use the SCC solution; present the greedy only as an optional conjectural/alternative solution with a proof supplied separately.\n</task_result>\n</task>",
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            "output": "Title: Problem - 2262D\nURL: https://codeforces.com/\nPublished: N/A\nAuthor: N/A\nHighlights:\n\n\nEnter| Register\n\n- Home\n- Top\n- Catalog\n- Contests\n- Gym\n- Problemset\n- Groups\n- Rating\n- Edu\n- API\n- Calendar\n- Help\n\n→ Pay attention\n\nBefore contest\nCodeforces Round (Div. 3)\n6 days\n\n→ Top rated\n\n| # | User | Rating |\n| --- | --- | --- |\n| 1 | jiangly | 3810 |\n| 2 | Kevin114514 | 3655 |\n| 3 | Benq | 3650 |\n| 4 | maroonrk | 3427 |\n| 5 | strapple | 3413 |\n| 6 | Um_nik | 3387 |\n| 7 | heuristica | 3322 |\n| 8 | tourist | 3301 |\n| 9 | turmax | 3297 |\n| 10 | squareOf105 | 3262 |\n\n| Countries | Cities | Organizations | View all → |\n| --- | --- | --- |\n\n→ Top contributors\n\n| # | User | Contrib. |\n| --- | --- | --- |\n| 1 | Qingyu | 157 |\n| 2 | maspy | 150 |\n| 3 | nik_exists | 149 |\n| 4 | Um_nik | 144 |\n| 5 | Errichto | 139 |\n| 6 | adamant | 136 |\n| 7 | AmShZ | 134 |\n| 8 | maroonrk | 133 |\n| 8 | BledDest | 133 |\n| 10 | DNR | 129 |\n\n| | View all → |\n| | ---------------------------------------------------- |\n\n→ Find user\n\nHandle:\n\n→ Recent actions\n\n- vovuh → Codeforces Round #377 (Div. 2) Editorial \n- QazLon → Why I am so unlucky \n- prabowo → Invitation to TOKI Indonesian NOI Open Contest 2026 \n- sonquanpdp2025 → What I Wish I Knew When I Started Codeforces \n- DuyMinh3005 → Is this acceptable? \n- qwexd → Codeforces Round 1121 (Div. 2) \n- MikeMirzayanov → Telegram Returns as Title Sponsor of Codeforces! \n- farruhbektojimirzayev → I am pupil now \n- qwexd → Codeforces Round 1121 (Div. 2) Editorial \n- 123gjweq2 → Is this a good problem + can you please help me gauge its difficulty? \n- Arbuzik342 → What many are afraid to say about the contribution system \n- mariza_CY → Balkan OI 2026 teams \n- turmax → Are we ready for AGI? \n- kondasujay2 → Codeforces Round 1120 (Div. 1, Div. 2) \n- Grizoo → [GYM] Aleppo Collegiate Programming Contest 2026 \n- atcoder_official → AtCoder Beginner Contest 443 Announcement \n- aryanc403 → ICPC India Regionals 2026 QnA \n- atcoder_official → AtCoder Beginner Contest 475 Announcement \n- kondasujay2 → Codeforces Round 1120 (Div 1, Div 2) Editorial \n- slycelote → Codeforces Beta Round #26 tutorial \n- Eason_He → 2026 Kruskal Cup \n- MikeMirzayanov → Rule about third-party code is changing \n- y_soul090 → I built cses-kit — a CLI workflow for solving CSES problems \n- ICPCNews → ICPC Challenge powered by Huawei \n- TheTitan008 → Free Competitive Programming Lectures (In Russian) \n\n| | Detailed → |\n| | --------------------------------------------------- |\n\n Telegram Returns as Title Sponsor of Codeforces! \n\nBy MikeMirzayanov, 5 days ago, \n\nHi, Codeforces!\n\nI have some great news. But maybe you've already noticed? Seen the updated logo?\n\nI'm glad to announce that Telegram is back as our title sponsor. I'm happy, and not just because this means we can keep developing and maintaining the system and (most importantly) running rounds for a long time to come.\n\nBut there's something else, too. It was Pavel Durov who, way back in 2010, spent several hours asking questions and trying to understand what I had in mind. Then he put his faith in our community — and Codeforces received support for the very first time. That support helped us for many years. So this means a lot to me personally, and I'm proud that now, more than 16 years after that first conversation, that support is coming back.\n\nThis helps with every round we run and gives us a chance to do new things, too. I'm glad to share that thanks to renewed support from Telegram, we've already started moving our infrastructure to a professional data center, making the system faster and more reliable.\n\nI'm sure this won't be the only new thing happening soon. Expect more good news.\n\nMike\n\nFull text and comments »\n\n codeforces, telegram, sponsor, sponsorship\n\n- +1989\n- MikeMirzayanov \n- 5 days ago\n- 101 \n\n Codeforces Round 1121 (Div. 2) \n\nBy qwexd, 3 days ago, \n\n1121. Another Round\n\ntime limit per participant\n\n2 hours\n\nmemory limit per participant\n\n1 brain\n\nThis is an interactive problem a contest with an unusual start time.\n\n$$\nn = 6\n$$\n\n$$\n1 \\\\le i \\\\le n\n$$\n\nThe maximum scores are\n\n---\n\nTitle: Problem - D - Codeforces\nURL: https://mirror.codeforces.com/contest/1909/problem/D\nPublished: N/A\nAuthor: N/A\nHighlights:\nD. Split Plus K\n...\nThere are $$$n$$$ positive integers $$$a_1, a_2, \\dots, a_n$$$ on a blackboard. You are also given a positive integer $$$k$$$. You can perform the following operation some (possibly $$$0$$$) times:\n...\n- choose a number $$$x$$$ on the blackboard;\n- erase one occurrence of $$$x$$$;\n- write two positive integers $$$y$$$, $$$z$$$ such that $$$y+z = x+k$$$ on the blackboard.\n...\nIs it possible to make all the numbers on the blackboard equal? If yes, what is the minimum number of operations you need?\n...\nEach test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 10^4$$$). The description of the test cases follows.\n...\nThe first line of each test case contains two integers $$$n$$$, $$$k$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$, $$$1 \\leq k \\leq 10^{12}$$$) — the number of integers initially on the blackboard and the constant $$$k$$$.\n...\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le 10^{12}$$$) — the initial state of the blackboard.\n...\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \\cdot 10^5$$$.\n...\nFor each test case, output a single line containing an integer: the minimum number of operations you need to make all the numbers on the blackboard equal, or $$$-1$$$ if it is impossible.\n...\nIn the first test case, $$$k = 1$$$. You can make all the numbers on the blackboard equal to $$$2$$$ with the following operations:\n...\n- Erase $$$x = 4$$$ and write $$$(y, z) = (2, 3)$$$. Note that $$$y+z=x+k$$$. The blackboard now contains the multiset $$$\\{3, 2, 3\\}$$$.\n...\n- Erase $$$x = 3$$$ and write $$$(y, z) = (2, 2)$$$. Note that $$$y+z=x+k$$$. The blackboard now contains $$$\\{2, 2, 2, 3\\}$$$.\n...\n- Erase $$$x = 3$$$ and write $$$(y, z) = (2, 2)$$$. Note that $$$y+z=x+k$$$. The blackboard now contains $$$\\{2, 2, 2, 2, 2\\}$$$.\n...\nThis makes all the numbers equal in $$$3$$$ operations. It can be shown that you cannot make all the numbers equal in less than $$$3$$$ operations.\n...\nIn the second test case, $$$k = 3$$$. You can make all the numbers on the blackboard equal to $$$7$$$ with the following operation:\n...\n- Erase $$$x = 11$$$ and write $$$(y, z) = (7, 7)$$$. Note that $$$y+z=x+k$$$. The blackboard now contains $$$\\{7, 7, 7\\}$$$.\n...\nIn the third test case, $$$k = 10$$$. You can make all the numbers on the blackboard equal to $$$40$$$ with the following operations:\n...\n- Erase $$$x = 100$$$ and write $$$(y, z) = (70, 40)$$$. Note that $$$y+z=x+k$$$. The blackboard now contains $$$\\{70, 40, 40, 100\\}$$$.\n...\n- Erase $$$x = 70$$$ and write $$$(y, z) = (40, 40)$$$. Note that $$$y+z=x+k$$$. The blackboard now contains $$$\\{40, 40, 40, 40, 100\\}$$$.\n...\n- Erase $$$x = 100$$$ and write $$$(y, z) = (40, 70)$$$. Note that $$$y+z=x+k$$$. The blackboard now contains $$$\\{40, 40, 40, 40, 40, 70\\}$$$.\n...\n- Erase $$$x = 70$$$ and write $$$(y, z) = (40, 40)$$$. Note that $$$y+z=x+k$$$. The blackboard now contains $$$\\{40, 40, 40, 40, 40, 40, 40\\}$$$.\n...\nIn the fourth and in the fifth test case, you can show that it is impossible to make all the numbers on the blackboard equal.\n\n---\n\nTitle: Plus-Minus Split(Codeforces)\nURL: https://rupeshdangi.vercel.app/codeforces-plus-minus\nPublished: N/A\nAuthor: N/A\nHighlights:\nProblem - B - CodeforcesCodeforces. Programming competitions and contests, programming communityhttps://codeforces.com/contest/1919/problem/B\n...\nYou're given a string consisting of '+' and '-' characters, representing an array. Your task is to split this array into subarrays in such a way that the sum of penalties for all subarrays is minimized. The penalty for a subarray is calculated as the absolute sum of its elements multiplied by its length.\n...\nIn solving the problem, I chose to take a different route by simulating the process of forming subarrays without actually creating them. This simulation was achieved using a deque, which acted like a stack.\n...\n- I noticed that the penalties depend on the arrangement of '+' and '-' elements within the array. This observation guided my approach.\n...\n- Instead of forming subarrays directly, I decided to simulate their formation using a deque This allowed me to understand how the penalties would be influenced by the arrangement of elements.\n...\n- I paid special attention to how I should handle adjacent '+' and '-' characters within the array. By making specific adjustments to the deque based on these pairs, I aimed to minimize the penalties.\n...\n- Goal was to add as many pairs of (+,-) to a subarray because they would cancel out each other thus making penalty 0 of the subarray.\n...\n- My ultimate goal was to find a configuration where the total penalties for all subarrays would be minimized. By using the deque to simulate the formation of subarrays and adjusting it based on the '+' and '-' pairs, I worked towards achieving this goal effectively.\n...\n- If such pair is not possible we just make that individual character a single array and add 1 penaltiy to whole penalty value. For example, for array [+,+,+] it is optimal to make it [+],[+],[+] which has total penalty 3 compared to [+,+,+] which has penaltiy 9.\n...\nI approached the problem with the idea of minimizing penalties by cleverly simulating subarray formations using a deque. By making smart choices about which elements to group together, I aimed to achieve the lowest possible total penalty.\n\n---\n\nTitle: CodeForces 1909D Split Plus K-JZTXT\nURL: https://www.jztxt.com/t/Ummc/\nPublished: N/A\nAuthor: N/A\nHighlights:\n# CodeForces 1909D Split Plus K\n...\n设最后每个数都相等时为 \\(t\\)。那么一次操作变成了合并两个数 \\(x, y\\)，再增加 \\(x + y - k\\)。于是每个 \\(a_i\\) 可以被表示成 \\(b_i t - (b_i - 1)k\\) 的形式，化简得 \\(a_i - k = b_i (t - k)\\)。\n...\n因为 \\(t - k\\) 对于每个 \\(i\\) 都相同，又因为我们的目标是最小化 \\(\\sum\\limits_{i = 1}^n b_i\\)，所以 \\(t - k\\) 取 \\(\\gcd\\limits_{i = 1}^n (a_i - k)\\) 最优。\n...\n注意所有 \\(a_i - k\\) 同号才合法，以及特判掉初始 \\(a_i\\) 全部相同的情况后不能出现 \\(a_i = k\\)。\n...\n```\n// Problem: D. Split Plus K\n// Contest: Codeforces - Pinely Round 3 (Div. 1 + Div. 2)\n// URL: https://codeforces.com/contest/1909/problem/D\n// Memory Limit: 256 MB\n// Time Limit: 1000 ms\n// \n// Powered by CP Editor (https://cpeditor.org)\n...\nvoid solve() {\n\tscanf(\"%lld%lld\", &n, &m);\n\tll s = 0, g = 0;\n\tbool fl = 1;\n\tfor (int i = 1; i <= n; ++i) {\n\t\tscanf(\"%lld\", &a[i]);\n\t\ts += a[i];\n\t\tg = __gcd(g, abs(a[i] - m));\n\t\tfl &= (a[i] == a[1]);\n\t}\n\tif (fl) {\n\t\tputs(\"0\");\n\t\treturn;\n\t}\n\tll ans = 0;\n\tint x = 0;\n\tfor (int i = 1; i <= n; ++i) {\n\t\tif (a[i] == m) {\n\t\t\tputs(\"-1\");\n\t\t\treturn;\n\t\t}\n\t\tif (a[i] > m) {\n\t\t\tx |= 1;\n\t\t} else {\n\t\t\tx |= 2;\n\t\t}\n\t\tans += abs(a[i] - m) / g;\n\t}\n\tprintf(\"%lld\\n\", x == 3 ? -1LL : ans - n);\n}\n\n---\n\nTitle: Round%20477/D.cpp\nURL: https://github.com/blisstoner/Codeforces/blob/master/Round%20477/D.cpp\nPublished: N/A\nAuthor: N/A\nHighlights:\npll c[300005];\nll D1[300005]; // c[i]의 resource로 x1을 커버할 때 몇 개의 server가 자유로운가\nll D2[300005]; // c[i]의 resource로 x2를 커버할 때 몇 개의 server가 자유로운가\n...\nint main(void) {\n scanf(\"%lld %lld %lld\", &n, &x1, &x2);\n for (int i = 0; i < n; i++) {\n scanf(\"%lld\", &c[i].X);\n c[i].Y = i+1;\n }\n sort(c, c + n);\n ll greedy1 = MX;\n ll greedy2 = MX;\n for (int i = n-1; i >= 0; i--) {\n if (greedy1 == MX && (n - i)*c[i].X >= x1)\n greedy1 = i;\n if (greedy2 == MX && (n - i)*c[i].X >= x2)\n greedy2 = i;\n D1[i] = (n - i) - (x1 + c[i].X - 1) / c[i].X;\n D2[i] = (n - i) - (x2 + c[i].X - 1) / c[i].X;\n }\n if (greedy1 != MX) {\n for (ll i = 0; i < greedy1; i++) {\n if (D2[i] >= n - greedy1) {\n printf(\"Yes\\n\");\n printf(\"%lld %lld\\n\", n - greedy1, greedy1 - i);\n for (ll j = greedy1; j < n; j++)\n printf(\"%lld \", c[j].Y);\n printf(\"\\n\");\n for (ll j = i; j < greedy1; j++)\n printf(\"%lld \", c[j].Y);\n return 0;\n }\n }\n }\n if (greedy2 != MX) {\n for (ll i = 0; i < greedy2; i++) {\n if (D1[i] >= n - greedy2) {\n printf(\"Yes\\n\");\n printf(\"%lld %lld\\n\", greedy2 - i,n - greedy2);\n for (ll j = i; j < greedy2; j++)\n printf(\"%lld \", c[j].Y);\n printf(\"\\n\");\n for (ll j = greedy2; j < n; j++)\n printf(\"%lld \", c[j].Y); \n return 0;\n }\n }\n }\n printf(\"No\");\n}\n\n---\n\nTitle: Problem - D - Codeforces\nURL: https://codeforces.com/contest/1987/problem/D\nPublished: N/A\nAuthor: N/A\nHighlights:\nProblem - D - Codeforces\n...\nD. World is Mine\n...\nAlice and Bob are playing a game. Initially, there are n cakes, with the i-th cake having a tastiness value of ai.\n...\nAlice and Bob take turns eating them, with Alice starting first:\n...\n- In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.\n- In his turn, Bob chooses any remaining cake and eats it.\n...\nThe game ends when the current player can't eat a suitable cake. Let x be the number of cakes that Alice ate. Then, Alice wants to maximize x, while Bob wants to minimize x.\n...\nFind out how many cakes Alice will eat if both players play optimally.\n...\nEach test contains multiple test cases. The first line of input contains a single integer t (1≤t≤500) — the number of test cases. The description of the test cases follows.\n...\nThe first line of each test case contains a single integer n (1≤n≤5000) — the number of cakes.\n...\nThe second line of each test case contains n integers a1,a2,…,an (1≤ai≤n) — the tastiness values of the cakes.\n...\nIt is guaranteed that the sum of n over all test cases does not exceed 5000.\n...\nFor each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.\n...\nIn the first test case, one possible sequence of turns is:\n...\n1. Alice eats a cake with a tastiness value of 1. The remaining cakes are [4,2,3].\n2. Bob eats a cake with a tastiness value of 2. The remaining cakes are [4,3].\n3. Alice eats a cake with a tastiness of 3. The remaining cakes are [4].\n4. Bob eats a cake with a tastiness value of 4. The remaining cakes are [].\n5. Since there are no more cakes left, the game ends.\n...\nIn the second test case, one possible sequence of turns is:\n...\n1. Alice eats a cake with a tastiness value of 1. The remaining cakes are [1,1].\n2. Bob eats a cake with a tastiness value of 1. The remaining cakes are [1].\n3. Since Alice has already eaten a cake with a tastiness value of 1, she cannot make a turn, so the game ends.\n\n---\n\nTitle: Problem - D - Codeforces\nURL: https://codeforces.com/contest/2126/problem/D\nPublished: N/A\nAuthor: N/A\nHighlights:\nProblem - D - Codeforces\n...\nYou are given n casinos, numbered from 1 to n. Each casino is described by three integers: li, ri, and reali (li≤reali≤ri). You initially have k coins.\n...\nYou can play at casino i only if the current number of coins x satisfies li≤x≤ri. After playing, your number of coins becomes reali.\n...\nYou can visit the casinos in any order and are not required to visit all of them. Each casino can be visited no more than once.\n...\nYour task is to find the maximum final number of coins you can obtain.\n...\nThe first line contains a single integer t (1≤t≤\n...\nThe first line of each test case contains two integers n and k (1≤n≤105, 0≤k≤109) — the number of casinos and the initial number of coins.\n...\nThis is followed by n lines. In the i-th line, there are three integers li, ri, reali (0≤li≤reali≤ri≤109) — the parameters of the i-th casino.\n...\nIt is guaranteed that the sum of all n across all test cases does not exceed 105.\n...\nFor each test case, output a single integer — the maximum number of coins you can obtain after optimally choosing the order of visiting the casinos.\n...\nIn the first test case, you can first play at the 2-nd casino. After that, you will have 2 coins. Then you can play at the 1-st casino, and the number of coins will increase to 3. Finally, after playing at the 3-rd casino, you will have 10 coins — this is the maximum possible amount.\n...\nIn the second test case, you have no money, so you cannot earn more.\n...\nIn the fourth test case, it is beneficial to play at the 2-nd casino right away and earn 4 coins.\n\n---\n\nTitle: Codeforces Round 1092 (Div. 1, Div. 2, Based on THUPC 2026 — Finals) Editorial - Codeforces\nURL: https://codeforces.com/blog/entry/152930?mobile=true\nPublished: N/A\nAuthor: N/A\nHighlights:\n, EXPloitation, and Gain Some EXPerience!\n\nSolution\n\nWe can identify a key strategy: if player A leads player B by a sufficiently large margin, A can make three consecutive moves of distance $$$1$$$ to occupy four consecutive positions. This effectively blocks B from reaching any position beyond A's current location.\n\nSpecifically, suppose it is the leading player A's turn, with A at position $$$u$$$ and B at position $$$v$$$, where $$$u - v \\geqslant 6$$$. In this case, A can force a block on B using a state sequence like the following (taking $$$v = u-6$$$ as an example): $$$(u-6,u) \\to (u-6,u+1) \\to (u-2,u+1) \\to (u-2,u+2) \\to (u-1,u+2) \\to (u-1,u+3)$$$.\n\nThis constitutes the optimal strategy for both sides. Player A can guarantee that the final score difference is at most the value achieved in this scenario, because once B is blocked, A is guaranteed to claim all remaining positions in the suffix. Conversely, player B can guarantee that the score difference is at least this value, since regardless of how A plays, B can secure all positions in this intermediate segment.\n\nConsequently, we can solve the problem using a backward DP. The state needs to track the position of the leading player, the distance gap $$$\\in [0,5]$$$ between them, and a $$$2^4$$$ bitmask to record the occupancy status of the positions between the two players. This yields an $$$O(n)$$$ time complexity, and space complexity is easily manageable using rolling array optimization.\n\nDuring implementation, note that after making a move, the leading player might temporarily extend their lead to a distance of $$$\\geqslant 6$$$. However, on the subsequent turn, they will either move back to reduce the gap, or the situation will naturally fall into the blocking case described above.\n\nCode(C++)\n\n```\n#include<bits/stdc++.h>\nusing namespace std;\nconst int maxn=200005,maxs=(1<<4)+5;\nint T,n;\nint a[maxn];\nlong long sum[maxn];\nlong long f[2][5][maxn][maxs];\nbool g[2][5][maxn][maxs];\nlong long dfs(int dir,int fr,int bk,int S){\n    int hav=0;\n    if(fr<n)\n        hav=1;\n    for(int i=bk+1;i<=n&&i<=bk+4;i++)\n        if(((S>>(i-bk-1))&1)==0&&i!=fr)\n            hav=1;\n    if(hav==0)\n        return 0;\n    if(fr-bk>=6&&dir==0){\n        long long res=sum[n]-sum[fr];\n        for(int i=bk+1;i<=fr-1;i++)\n            if(((S>>(i-bk-1))&1)==0)\n                res-=a[i];\n        return res;\n    }\n    if(fr-bk<6&&g[dir][fr-bk-1][bk][S])\n        return f[dir][fr-bk-1][bk][S];\n    long long res=-1e18;\n    if(dir==0){\n        int flg=0;\n        for(int i=fr+1;i<=n&&i<=fr+4;i++)\n            flg=1,res=max(res,a[i]-dfs(1,i,bk,S|(1<<(fr-bk-1))));\n        if(flg==0)\n            res=-dfs(1,fr,bk,S);\n    }\n    else{\n        int flg=0;\n        for(int i=bk+1;i<=n&&i<=bk+4;i++)\n            if(((S>>(i-bk-1))&1)==0&&i!=fr){\n                flg=1;\n                if(i<fr)\n                    res=max(res,a[i]-dfs(0,fr,i,S>>(i-bk)));\n                if(i>fr)\n                    res=max(res,a[i]-dfs(1,i,fr,0));\n            }\n        if(flg==0)\n            res=-dfs(0,fr,bk,S);\n    }\n    if(fr-bk<6)\n        f[dir][fr-bk-1][bk][S]=res,g[dir][fr-bk-1][bk][S]=1;\n    return res;\n}\nint main(){\n    memset(g,0,sizeof(g));\n    scanf(\"%d\",&T);\n    while(T--){\n        scanf(\"%d\",&n);\n        for(int i=3;i<=n;i++)\n            scanf(\"%d\",&a[i]),sum[i]=sum[i-1]+a[i];\n        for(int i=n;i>=1;i--)\n            for(int j=i-1;j>=1&&j>=i-5;j--)\n                for(int k=0;k<=1;k++)\n                    for(int s=0;s<(1<<(i-j-1));s++)\n                        dfs(k,i,j,s);\n        printf(\"%lld\\n\",dfs(1,2,1,0));\n        for(int i=0;i<=1;i++)\n            for(int j=0;j<=4;j++)\n                for(int k=0;k<=n;k++)\n                    for(int s=0;s<maxs;s++)\n                        g[i][j][k][s]=0;\n    }\n    return 0;\n}\n\n```\n\n2215E - Star Map\n\nSolution\n\nConsider the planar graph corresponding to the triangulation. Let the number of points on its outermost face be $$$x$$$, and the size of the triangulation be $$$t$$$. By Euler's formula for planar graphs, we have $$$V+F-E=1$$$. Together with $$$F\\ge t$$$ and $$$E= \\dfrac{3F+x}2$$$, we obtain $$$t\\le F=2(n-1)-x$$$. The problem thus reduces to minimizing this $$$x$$$ and achieving equality simultaneously.\n\nWe now identify which points must lie on the outermost face. Intuitively, if, when taking a point as the origin, there exists a quadrant containing no other points, then this point must be on the convex hull (outer face). We next give a construction under which all equalities hold.\n\nSort all points by their $$$y$$$-coordinate in increasing order and add them one by one, while maintaining two monotonic stacks $$$q_1,q_2$$$ satisfying: $$$x_{q_{1,i}} \\gt x_{q_{1,i-1}},\\ y_{q_{1,i}} \\gt y_{q_{1,i-1}}$$$; $$$x_{q_{2,i}} \\lt x_{q_{2,i-1}},\\ y_{q_{2,i}} \\gt y_{q_{2,i-1}}$$$, and the top elements of the two stacks are identical.\n\nWhen inserting a point $$$(x_i,y_i)$$$, update the two stacks as follows:\n\nIf $$$q_1$$$ is nonempty, take its top element $$$j$$$. If $$$x_i \\lt x_j$$$, pop $$$j$$$ from the stack; if the stack is still nonempty, let $$$k$$$ be its new top, and add $$$(i,j,k)$$$ to $$$T$$$. Repeat until $$$x_i \\gt x_j$$$.\n\nIf $$$q_2$$$ is nonempty, take its top element $$$j$$$. If $$$x_i \\gt x_j$$$, pop $$$j$$$ from the stack; if the stack is still nonempty, let $$$k$$$ be its new top, and add $$$(i,j,k)$$$ to $$$T$$$. Repeat until $$$x_i \\lt x_j$$$.\n\nIt can be easily verified that this construction achieves equality.\n\nCode(C++)\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\nconst int N = 2e5 + 5;\nint T, n, ql[N], qr[N], tpl, tpr, cnt;\nstruct point {\n    int x, y, id;\n    bool operator<(const point &other) const {\n        return y < other.y;\n    }\n} p[N];\nstruct triangle { int a, b, c; } ans[N << 1];\nint main() {\n    scanf(\"%d\", &T);\n    while (T--) {\n        scanf(\"%d\", &n);\n        for (int i = 1; i <= n; i++) \n            scanf(\"%d%d\", &p[i].x, &p[i].y), p[i].id = i;\n        sort(p + 1, p + n + 1);\n        cnt = 0;\n        ql[1] = qr[1] = 1;\n        tpl = tpr = 1;\n        for (int i = 2; i <= n; i++) {\n            while (tpl && p[i].x < p[ql[tpl]].x) {\n                if (tpl >= 2) ans[++cnt] = triangle{p[i].id, p[ql[tpl]].id, p[ql[tpl - 1]].id};\n                tpl--;\n            }\n            ql[++tpl] = i;\n            while (tpr && p[i].x > p[qr[tpr]].x) {\n                if (tpr >= 2) ans[++cnt] = triangle{p[i].id, p[qr[tpr]].id, p[qr[tpr - 1]].id};\n                tpr--;\n            }\n            qr[++tpr] = i;\n        }\n        printf(\"%d\\n\", cnt);\n        for (int i = 1; i <= cnt; i++) \n            printf(\"%d %d %d\\n\", ans[i].a, ans[i].b, ans[i].c);\n    }\n    return 0;\n}\n\n```\n\nSolution\n\nThe game cannot go on indefinitely. When $$$n \\le k$$$, Alice must end the game immediately; otherwise, Bob will remove the green card on the next turn, making the score $$$0$$$. Moreover, Bob can continuously remove cards to make the deck thinner. Therefore, Bob has the ability to prevent Alice from continuing the game indefinitely.\n\nFirst, consider the case where $$$s=1$$$. In this situation, Alice must decide whether to take the green card or put it back. Once she puts it back, who gets the green card next only depends on its position. It can be observed that only when $$$n \\equiv k \\pmod{2k}$$$, regardless of how Alice puts it back, she cannot get the green card next time. In this case, it is clearly optimal to remove the green card directly; otherwise, Alice can always find a way to ensure she gets the green card next time by placing it either before or after all the other removed cards. Thus, putting it back is definitely better than removing it directly.\n\nStarting from Alice putting the green card back, both players will perform $$$x = \\left\\lfloor \\frac{n+k}{2k} \\right\\rfloor$$$ operations, after which it will again become a situation where Alice can take the green card. The key question is how much the size of the pile will change.\n\nLet $$$f_k(n)$$$ be the final score when the pile size is",
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            "output": "Title: Codeforces Round 1120 (Div 1, Div 2) Editorial\nURL: https://codeforces.com/\nPublished: N/A\nAuthor: N/A\nHighlights:\nexcited to invite you to participate in Codeforces Round 1120 (Div. 1) and Codeforces Round 1120 (Div. 2) on Sep/12/2026 17:35 (Moscow time).\n\nThe round will be rated for participants from both divisions. You will be given 6 problems and 3 hours to solve them. At least one of the problems will be divided into subtasks.\n\nThe problems were authored and prepared by me, CutSandstone, and sukon.\n\nWe would like to thank:\n\n- abc864197532 for being the coordinator.\n- Um_nik for preliminary reviewing the round.\n- Alexdat2000 for translating the statements to Russian.\n- baluteshih, dlu, kevinyu, MattTheNub, n685, nadya271, yunz_qiao, _istil, alvingogo, avnithv, awoo, BurnedChicken, Fysty, Misuki, passwordisa, pbj2006, snowythecat, WiwiHo, and YuukiS for testing the round and providing valuable feedback.\n- KAN and MikeMirzayanov for the great Codeforces and Polygon platforms.\n- And finally, You, for participating!\n\nThe scoring distribution is:\n\n1. Div. 1: $$ — $$ — $$ — $$ — $$ — $$\n2. Div. 2: $$ — $$ — $$ — $$ — $$ — $$\n\nWe hope you enjoy the problems.\n\nGood luck and have fun!\n\nUPD:\n\nCongratulations to the winners (subject to change):\n\nDiv. 1:\n\n1. Kevin114514\n2. XorGhost\n3. strapple\n4. PiGoat\n5. zjy2008\n\nDiv. 2 (trusted participants):\n\n1. gausset\n2. Dream_Sky\n3. Guddu_06\n4. Saransh_Gupta_\n5. zhudaxia\n\nUPD 2:\n\nSorry for the late tutorial, its posted now!\n\nFull text and comments »\n\n Announcement of Codeforces Round 1120 (Div. 1)\n\n Announcement of Codeforces Round 1120 (Div. 2)\n\n- +278\n- kondasujay2 \n- 5 days ago\n- 123 \n\n Educational Codeforces Round 194 [rated for Div. 2] \n\nBy BledDest, 8 days ago, translation, \n\nHello Codeforces!\n\nThe series of Educational Rounds continues thanks to the support of the Neapolis University Pafos.\n\nEducational Codeforces Round 194 (Rated for Div. 2) will start on Sep/08/2026 17:35 (Moscow time).\n\nThis round will be rated for the participants with rating lower than 2100. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest, you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.\n\nYou will be given 7 problems and 2 hours to solve them. I hope everyone participating finds interesting problems for them to solve.\n\nI would like to express my gratitude to Mike MikeMirzayanov Mirzayanov for creating Codeforces and Polygon, two platforms which made conducting programming contests and preparing problems for them much more convenient. Also big thanks to testers: awoo, FelixArg, Alenochka.\n\nOur friends at Neapolis University Pafos have a message for you:\n\nPlanning to apply for the Computer Science and Artificial Intelligence (CSAI) program at Neapolis University Pafos?\n\nJetBrains Youth Clubs can help you prepare.\n\n🤖 AI Club is a chance to explore one of the core areas you’ll study in the CSAI program. You’ll work with AI topics, strengthen your understanding, and get more experience before university.\n\n➗ Math Club helps you sharpen the mathematical problem-solving skills you’ll need for the program. There’s also a direct application benefit: participation in the Math Club can give you 30% of your entrance test mark for the Computer Science and Artificial Intelligence BSc supported by JetBrains Foundation.\n\nBoth clubs are designed for high school students with a strong interest in STEM who want to challenge themselves and prepare for the next step.\n\n👉 Join JetBrains Youth Clubs: AI and Math\n\nGood luck in solving the problems! See you at the contest.\n\nUPD: The editorial has been published.\n\nFull text and comments »\n\n Announcement of Educational Codeforces Round 194 (Rated for Div. 2)\n\n- +214\n- BledDest \n- 8 days ago\n- 86 \n\n Codeforces Round 1119 (Div. 3) \n\nBy nik_exists, 2 weeks ago, \n\n`#include \"codeforces.h\"\n#include\n#include\n\nusing namespace std;\nusing namespace std::chrono_literals;\nusing namespace cf;\n\nint main() {\nBlog announcement(\"Codeforces Round 1119 (Div. 3)\");\n\n```\nannouncement << \"Hello Codeforces!\\n\";\nannouncement << \"I am delighted to invite you to participate in \" << `[Codeforces Round 1119 (Div. 3)](https://codeforces.com/contest/2259 \"Codeforces Round 1119 (Div. 3)\")` << \"!\\n\";\nannouncement << \"The problems were written and prepared by \" << `[nik_exists](https://codeforces.com/profile/nik%5Fexists \"Expert nik_exists\")` << \".\\n\\n\";\n\nannouncement << ContestInfo {\n    .start_time = `[Sep/05/2026 17:45 (Moscow time)](https://www.timeanddate.com/worldclock/fixedtime.html?day=5&month=9&year=2026&hour=17&min=45&sec=0&p1=166)`,\n    .duration = 2h + 15min,\n    .number_of_problems = 7,\n    .subtasks = true // possibly not adjacent\n};\n\nannouncement << RatingRules {\n    .rated_for_rating_below = 1600,\n    .unrated_registration_allowed = true,\n    .trusted_max_rating_below = 1900,\n    .trusted_min_rated_rounds = 5\n};\n\nannouncement << JudgingRules {\n    .scoring = codeforces::ICPC,\n    .wrong_answer_penalty = 10min,\n    .open_hacking_phase = 12h,\n    .final_tests = after_open_hacks\n};\n\nauto is_cheater = [](User user) {\n    // https://codeforces.com/blog/entry/133941\n    if (user.used_ai_in_contest()) return true;\n    if (user.discussed_problems_online_before_contest_end()) return true;\n    if (user.violated_other_rules_im_too_lazy_to_list()) return true;\n    return false;\n};\n\nfor (auto user : get_users(2259)) {\n    if (is_cheater(user)) user.send_to_cry_basement();\n}\n\nannouncement << Acknowledgments {\n    .coordinator = `[cry](https://codeforces.com/profile/cry \"International Master cry\")`,\n    .russian_translation = `[Vladosiya](https://codeforces.com/profile/Vladosiya \"Expert Vladosiya\")`,\n    .red_testers = {`[__baozii__](https://codeforces.com/profile/%5F%5Fbaozii%5F%5F \"International Grandmaster __baozii__\")`, `[AksLolCoding](https://codeforces.com/profile/AksLolCoding \"Grandmaster AksLolCoding\")`, `[awesomeguy856](https://codeforces.com/profile/awesomeguy856 \"Grandmaster awesomeguy856\")`, `[nifeshe](https://codeforces.com/profile/nifeshe \"Grandmaster nifeshe\")`},\n    .golden_testers = {`[Arpa](https://codeforces.com/profile/Arpa \"Master Arpa\")`, `[Intellegent](https://codeforces.com/profile/Intellegent \"International Master Intellegent\")`, `[Proof_by_QED](https://codeforces.com/profile/Proof%5Fby%5FQED \"Master Proof_by_QED\")`, `[temporary1](https://codeforces.com/profile/temporary1 \"Master temporary1\")`},\n    .purple_testers = {`[Argentum47](https://codeforces.com/profile/Argentum47 \"Candidate Master Argentum47\")`, `[Euclid73](https://codeforces.com/profile/Euclid73 \"Candidate Master Euclid73\")`, `[Jteh](https://codeforces.com/profile/Jteh \"Candidate Master Jteh\")`, `[Lilypad](https://codeforces.com/profile/Lilypad \"Candidate Master Lilypad\")`, `[SpyrosAliv](https://codeforces.com/profile/SpyrosAliv \"Candidate Master SpyrosAliv\")`, `[nimoxide](https://codeforces.com/profile/nimoxide \"Candidate Master nimoxide\")`, `[omsincoconut](https://codeforces.com/profile/omsincoconut \"Candidate Master omsincoconut\")`, `[wakanda-forever](https://codeforces.com/profile/wakanda-forever \"Candidate Master wakanda-forever\")`, `[yse](https://codeforces.com/profile/yse \"Candidate Master yse\")`},\n    .blue_testers = {`[CatsAreCool](https://codeforces.com/profile/CatsAreCool \"Expert CatsAreCool\")`, `[Vladosiya](https://codeforces.com/profile/Vladosiya \"Expert Vladosiya\")`, `[chromate00](https://codeforces.com/profile/chromate00 \"Expert chromate00\")`, `[fatespeaker](https://codeforces.com/profile/fatespeaker \"Expert fatespeaker\")`, `[linearspace](https://codeforces.com/profile/linearspace \"Expert linearspace\")`, `[reirugan](https://codeforces.com/profile/reirugan \"Expert reirugan\")`, `[simplelife](https://codeforces.com/profile/simplelife \"Expert simplelife\")`},\n    .green_testers = {`[hannah12345](https://codeforces.com/profile/hannah12345 \"Pupil hannah12345\")`},\n    .codeforces_and_polygon = {`[KAN](https://codeforces.com/profile/KAN\n\n---\n\nTitle: A-B Rockwell  20-750-2262D-2R PowerFlex 750 115V AC IO Option Module | Werner\nURL: https://www.onewerner.com/product/Rockwell-20-750-2262D-2R\nPublished: N/A\nAuthor: N/A\nHighlights:\nA-B Rockwell 20-750-2262D-2R PowerFlex 750 115V AC IO Option Module | Werner\n...\nRockwell Automation ALLEN BRADLEY ROCKWELL AU 20-750-2262D-2R POWERFLEX 750 115V AC IO OPTION MODULE\n...\nManufacturer Rockwell Automation\n\n---\n\nTitle: 115V AC 22-Series I/O module with 2 analog In, 2 analog out, 6 digital in and 2 relay outputs | Allen-Bradley PowerFlex | InstroDirect\nURL: https://instrodirect.com.au/product/115v-ac-22-series-i-o-module-with-2-analog-in-2-analog-out-6-digital-in-and-2-relay-outputs/\nPublished: N/A\nAuthor: N/A\nHighlights:\nAllen-Bradley 20-750-2262D-2R PowerFlex 750 22-Series I/O Option Module — 115V AC 22-Series I/O module with 2 analog In, 2 analog out, 6 digital in and 2 relay outputs.\n...\nThe Allen-Bradley 20-750-2262D-2R is a 22-Series I/O option module for PowerFlex 750 series drives (753 and 755). Function: 115V AC 22-Series I/O module with 2 analog In, 2 analog out, 6 digital in and 2 relay outputs. Installs into an option slot (slot 1–3 for 753, slot 1–5 for 755) to expand the drive’s I/O capability with analogue and digital signals. Enables interfacing of process transmitters, PTC thermistors, and discrete control signals directly at the drive. InstroDirect supplies genuine 20-750-2262D-2R at parallel import pricing.\n\n---\n\nTitle: 20-750-2262D-2R Allen-Bradley PowerFlex 750 I/O Module | New\nURL: https://www.5gplc.com/products/allen-bradley-20-750-2262d-2r-powerflex-750-series-i-o-option-module\nPublished: 2026-04-03T00:00:00.000Z\nAuthor: N/A\nHighlights:\n20-750-2262D-2R Allen-Bradley PowerFlex 750 I/O Module | New\n...\n# Allen-Bradley 20-750-2262D-2R PowerFlex 750-Series I/O Option Module\n\n Only 10 left - Selling fast \n\n PRODUCT TYPE : Operator Terminal\n\n PRODUCT VENDOR : Allen-Bradley\n...\nThe Allen-Bradley 20-750-2262D-2R functions as a high-versatility I/O expansion module specifically engineered for the PowerFlex 750-Series (PowerFlex 753 and 755) variable frequency drives. This option module significantly expands the drive's control capabilities by providing a dedicated mix of digital and analog interfaces. We supply this unit as 100% Brand New original factory stock, ensuring seamless integration into your drive’s option slots to facilitate complex automation sequences and feedback loops without external PLC intervention.\n...\n| Feature | Specification |\n| --- | --- |\n| Manufacturer | Allen-Bradley / Rockwell Automation |\n| Product Line | PowerFlex 750-Series Options |\n| Part Number | 20-750-2262D-2R |\n| Digital Inputs | Two (2) 24V DC Inputs |\n| Relay Outputs | Two (2) Form C Relay Outputs |\n| Analog Inputs | Two (2) Differential/Bipolar Inputs |\n| Compatibility | PowerFlex 753 and PowerFlex 755 Drives |\n| Mounting | Internal Option Slot (1 through 5) |\n| I/O Terminal Type | Multi-point pluggable terminal block |\n| Weight | 0.35 lbs (0.16 kg) |\n...\n0-7\n...\nmanage interlocks, brake\n...\nors, or\n...\nThe 20-750-2262D-2R fits into any of the five available option slots (Slots 1–5) on the PowerFlex 755 control pod. Always ensure you configure the drive parameters to recognize the specific slot location.\n...\nAbsolutely. We ship only 100% Brand New 20-750-2262D-2R modules in original Allen-Bradley static-shielded packaging. This ensures the firmware and hardware components meet current Rockwell Automation safety and performance standards.\n\n---\n\nTitle: 20-750-2262D-2R | Option Module,115V AC IO,6 Digital Input | PLC Exchange\nURL: https://plcexchange.net/product/allen-bradley-20-750-2262d-2r-option-module115v-ac-io6-digital-inputs2-analog-inputs2/\nPublished: 2026-04-20T00:00:00.000Z\nAuthor: N/A\nHighlights:\n20-750-2262D-2R | Option Module,115V AC IO,6 Digital Input | PLC Exchange\n\n 980-202-0882 (9802020882) support@plcexchange.net \n\n# Allen-Bradley 20-750-2262D-2R Option Module,115V AC IO,6 Digital Inputs,2 Analog Inputs,2\n...\nManufacturer: Allen-Bradley MPN: 20-750-2262D-2R UPC/GTIN: 884951062542\n...\nAllen-Bradley 20-750-2262D-2R — PowerFlex 750-Series AC Drives\n...\n| Catalog Number | 20-750-2262D-2R |\n| --- | --- |\n| Manufacturer | Allen-Bradley / Rockwell Automation |\n| Power Consumption | 4.4 W |\n| Weight | 91 g |\n| Mounting | and ventilation provisions along with the control pod and its correspondin |\n| Certifications | and Specifications 51 |\n...\n### Allen-Bradley 20-750-2262D-2R\n...\nPowerFlex 750-Series AC Drives\n...\n| Module Overview | |\n| --- | --- |\n| Catalog Number | 20-750-2262D-2R |\n| Manufacturer | Allen-Bradley / Rockwell Automation |\n...\n## Technical Notes for 20-750-2262D-2R\n...\n- Mounts directly on the Compact 5000 I/O bus alongside other 5069-series modules.\n- Removable terminal blocks (RTBs) allow pre-wiring before module installation.\n- Module configuration is performed in Studio 5000 Logix Designer.\n- All 20-750-2262D-2R units sold by PLC Exchange are tested and ship in original factory packaging.\n- For application-specific questions or system compatibility verification, contact PLC Exchange technical support at [email protected] or 980-202-0882.\n...\nThe 20-750-2262D-2R by Allen-Bradley may also be found under the following alternative names, catalog numbers, or common search variations:\n...\n- 207502262D2R\n- Allen Bradley 20-750-2262D-2R\n- Allen-Bradley 20-750-2262D-2R\n- 20 750 2262D 2R\n...\n| Weight | 0.2 lbs |\n| --- | --- |\n| warranty | 2.5 Year PLC Exchange Warranty, 30 Days worry-free return from delivery |\n| upc | 884951062542 |\n| mpn | 20-750-2262D-2R |\n| manufacturer | Allen-Bradley |\n| condition | Original Packaging |\n...\nThe official manufacturer datasheet is available for download on the 20-750-2262D-2R product page at PLC Exchange. Look for the Tech Data PDF link in the product description.\n...\nYes. Every 20-750-2262D-2R sold by PLC Exchange includes a 2.5-year warranty (30 months, full replacement). Details: https://plcexchange.net/warranty-information/\n...\nWho manufactures the\n...\n750-2262D-2R?\n...\nThe 20-750-2262D-2R is manufactured by Allen-Bradley. PLC Exchange is an independent reseller specializing in Allen-Bradley industrial automation products.\n\n 6816 April Lane Charlotte, NC 28215 \n\n +1 (980) 202-0882 (+19802020882) support@plcexchange.net \n\n Mon–Sat 9am–5pm Sun Closed\n\n---\n\nTitle: A-B Rockwell  20-750-2262D-2R PowerFlex 750 115V AC IO Option Module | Werner\nURL: https://www.onewerner.com/catalog/Rockwell-Automation/Drives/Rockwell-20-750-2262D-2R\nPublished: N/A\nAuthor: N/A\nHighlights:\nA-B Rockwell 20-750-2262D-2R PowerFlex 750 115V AC IO Option Module | Werner\n...\nA-B Rockwell 20-750-2262D-2R PowerFlex 750 115V AC IO Option Module\n...\nRockwell Automation ALLEN BRADLEY ROCKWELL AU 20-750-2262D-2R POWERFLEX 750 115V AC IO OPTION MODULE\n...\nManufacturer Rockwell Automation\n\n---\n\nTitle: Rockwell Automation 20-750-2262D-2R PF750-115V I/O Module-2AI,2AO,6DI,  – Voomi Supply\nURL: https://voomisupply.com/products/rockwell-automation-20-750-2262d-2r-pf750-115v-i-o-module-2ai-2ao-6di-2ro\nPublished: N/A\nAuthor: N/A\nHighlights:\nRockwell Automation 20-750-2262D-2R PF750-115V I/O Module-2AI,2AO,6DI, – Voomi Supply --\n...\n# Rockwell Automation 20-750-2262D-2R PF750-115V I/O Module-2AI,2AO,6DI,2RO\n...\nBrand: ROCKWELL AUTOMATION\n...\nSKU: 20-750-2262D-2R\n...\nThe Rockwell Automation 20-750-2262D-2R is designed as a PF750-115V I/O module featuring two analog inputs, two analog outputs, six digital inputs, and two relay outputs. This module supports various automation tasks by facilitating communication between controllers and field devices.\n...\n- Part Number: 20-750-2262D-2R AB\n- Manufacturer: Rockwell Automation\n...\n- Category: PowerFlex AC Drive User Installed Options\n- Availability: In Stock\n- Min Order Quantity: 1\n- UPC: 884951062542\n...\n#### Rockwell Automation 20-750-2262D-2R PF750-115V I/O Module-2AI,2AO,6DI,2RO\n...\n#### Rockwell Automation 20-750-2262D-2R PF750-115V I/O Module-2AI,2AO,6DI,2RO\n\n---\n\nTitle: PLC 20-750-2262D-2R POWERFLEX POWER OPTION MODULE\nURL: https://www.xiatongautomation.com/doc/41405051/plc-20-750-2262d-2r-powerflex-power-option-module.pdf\nPublished: N/A\nAuthor: N/A\nHighlights:\n## PLC 20-750-2262D-2R POWERFLEX POWER OPTION MODULE\n...\n20-750-2262D-2R\n...\n20-750-2262D-2R\n...\n### Wuhan Sean Automation Equipment Co.,Ltd xiatongautomation.com\n...\nPLC 20-750-2262D-2R POWERFLEX POWER OPTION MODULE\n...\nModel 20-750-2262D-2R\n...\nBrand PLC Description POWERFLEX 750-SERIES Datasheet/Catalogue PDF File Weight 1.5KG Condition Normal\n...\nOur PowerFlex® 755 AC Drives provide ease of use, application flexibility, and high performance. They include multiple control and hardware options, as well as optional embedded safety features. These drives are well-suited for a wide variety of motor control applications including pumps, fans, and conveyors. With Integrated Motion on EtherNet/IP™, you can place these drives on the same network as Kinetix® drives. A Dual-port EtherNet/IP™ Option Module supports ring, linear, and star topologies, as well as Device Level Ring (DLR) functionality for fault tolerant connectivity and high drive availability.",
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              "prompt": "Independently analyze the supplied PLUSworld problem and the simple greedy algorithm from the official page's 'Magic Solution': reject a_i>b_i; start at first bad index; at current p repeatedly increment while a[p]<b[p] and (a[p]>=p or a[a[p]]==b[a[p]]), then decrement remaining if finalized, stop if all done or a[p]==p, otherwise follow. Determine whether this algorithm is actually correct and respects <=2n^2 operations. Use exhaustive state-space reasoning or write/run temporary brute-force checks for small n if useful. Return either a rigorous proof (including why -1 is sound) or a concrete counterexample and recommend whether to use it. Do not edit repository files.",
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              "prompt": "The full statement is: n barns 1..n, current one-way edge i->a_i. Target is b with i<=b_i. Start Bessie at any p. Operations: type 1 increments a_p by 1 if a_p<n and stays at p; type 2 sets p=a_p. Need reach a=b, output <=2n^2 ops. All arrays 1-based. Re-evaluate the exact greedy from my previous message under this statement. Please perform an exhaustive brute-force comparison against reachability for small n if possible, and return a rigorous proof or counterexample. Do not edit files.",
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            "output": "<task id=\"ses_f5330a1d8ffeGbTravJQJPfyTl\" state=\"completed\">\n<task_result>\nThe greedy is correct under one necessary bookkeeping condition: `remaining` must count unfinished coordinates, and each coordinate may decrement it only once.\n\n**Exhaustive Check**\n\nI compared it with independent reverse BFS over states `(a, p)`.\n\n- `n=1`: 1 case, 0 mismatches\n- `n=2`: 8 cases, 0 mismatches\n- `n=3`: 162 cases, 0 mismatches\n- `n=4`: 6,144 cases, 0 mismatches\n- `n=5`: 375,000 cases, 0 mismatches\n- `n=6`: all 7,484,400 feasible `a <= b` cases, 0 mismatches\n- Additional random `n=7` targets: 12,990,390 states, 0 mismatches\n\nThe reverse BFS used inverse operations:\n\n- Decrease `a_p` by one, reversing type 1.\n- Move from `(a,p)` backward to `(a,r)` for every `r` with `a_r=p`, reversing type 2.\n\n**Correctness**\n\nCall barn `i` finished when `a_i=b_i`. Since values only increase, a finished barn stays finished.\n\nUse this normal-form lemma:\n\n> If a state is solvable, it has a solvable execution obeying:\n>\n> - If `a_p<p` and barn `a_p` is unfinished, move to `a_p`.\n> - Otherwise, if `a_p<b_p`, increment `a_p`.\n> - If `a_p=b_p`, move along the edge.\n\nWhy this normalization is valid:\n\n- Incrementing `a_p` from `x<p` skips the unfinished barn `x`. Instead, move to `x` first. Any recursive dependency goes to a strictly smaller index, so this process terminates.\n- Once a barn is finished, its edge is `b_x>=x`; it can never create a new edge downward.\n- If `a_p>=p`, or the destination is already finished, no unfinished lower-index dependency is being skipped, so the increment can be moved earlier without harming feasibility.\n- The smallest unfinished barn has no unfinished lower prerequisite, so it may be chosen as the starting barn.\n\nThe greedy performs exactly this normal-form execution.\n\nIf it stops with `a_p=p` while unfinished barns remain, then necessarily `b_p=p` too. Barn `p` is a finished self-loop:\n\n- Type 1 would overshoot the target.\n- Type 2 leaves Bessie at `p`.\n\nTherefore no remaining coordinate can ever be changed. Returning `-1` is sound.\n\n**Operation Bound**\n\nType 1 operations:\n\n\\[\n\\sum_i (b_i-a_i)\\le n(n-1).\n\\]\n\nFor type 2, fix a source barn `p`. Between two departures from `p`, some previously unfinished barn with index `<p` must become finished. Otherwise the greedy walk either strictly descends forever or cannot return to `p`. Each such barn finishes only once, so `p` is departed at most `p` times.\n\nThus:\n\n\\[\n\\text{type 2 operations}\\le \\sum_{p=1}^n p=\\frac{n(n+1)}2.\n\\]\n\nTotal:\n\n\\[\nn(n-1)+\\frac{n(n+1)}2\n=\\frac{3n^2-n}{2}\n<2n^2.\n\\]\n\n**Important Bug**\n\nIf the earlier pseudocode decrements `remaining` every time it sees `a_p=b_p`, without checking whether `p` was already finished, it is incorrect.\n\nCounterexample:\n\n```text\nn = 3\na = [1, 1, 3]\nb = [3, 2, 3]\n```\n\nThe algorithm fixes barn 1, moves to barn 3, observes that barn 3 is already finished, and may incorrectly claim success while `a_2=1 != 2`.\n\nUse a `done[]` array, or decrement only when `a_p` changes from unfinished to finished.\n</task_result>\n</task>",
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          "type": "tool",
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          "state": {
            "status": "completed",
            "input": {
              "query": "\"a[position] >= position\" \"a[a[position]]\" PLUSworld",
              "numResults": 10,
              "livecrawl": "preferred",
              "type": "deep",
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            "output": "Title: Problem - A - Codeforces\nURL: https://mirror.codeforces.com/contest/2072/problem/A\nPublished: N/A\nAuthor: N/A\nHighlights:\nNatsume Akito has just woken up in a new world and immediately received his first quest! The system provided him with an array $$$a$$$ of $$$n$$$ zeros, an integer $$$k$$$, and an integer $$$p$$$.\n...\nIn one operation, Akito chooses two integers $$$i$$$ and $$$x$$$ such that $$$1 \\le i \\le n$$$ and $$$-p \\le x \\le p$$$, and performs the assignment $$$a_i = x$$$.\n...\nAkito is still not fully accustomed to controlling his new body, so help him calculate the minimum number of operations required to make the sum of all elements in the array equal to $$$k$$$, or tell him that it is impossible.\n...\nThe only line of each test case contains three integers $$$n$$$, $$$k$$$, $$$p$$$ ($$$1 \\le n \\le 50$$$, $$$-2500 \\le k \\le 2500$$$, $$$1 \\le p \\le 50$$$) — the length of the array, the required sum, and the boundary of the segment from which numbers can be replaced.\n...\nFor each test case, output the minimum number of operations to achieve the final sum $$$k$$$ in the array, or $$$-1$$$ if it is impossible to achieve the sum $$$k$$$.\n...\nIn the fifth example, the sum of the array is initially zero, so no operations are needed.\n...\nIn the sixth example, the maximum sum in the array that we can achieve is $$$9$$$ (by assigning the number $$$9$$$ to the single element), so the sum $$$10$$$ cannot be obtained by any operations.\n...\nIn the seventh example, only one operation $$$a_3 = -7$$$ is needed.\n\n---\n\nTitle: Plusworld Realtors\nURL: https://linkedin.com/company/plusworld-realtors\nPublished: N/A\nAuthor: N/A\nHighlights:\n# Plusworld Realtors\n...\nPlusworld Realtors is a Real Estate company. Plusworld Realtors is Nigeria's leading real estate company that helps individuals find, buy, sell, and rent properties across Nigeria. Plusworld Realtors employs 7 people (+16.7% YoY, +1 person). Headquartered in Nigeria, with an office in Lagos.\n...\nWe are a full-service real estate and investment company headquartered in Lagos, operating across Nigeria's most prime residential and commercial addresses — Lekki, Ikoyi, Victoria Island, Banana Island, and beyond. Our work covers the full spectrum of real estate: sales, leasing, facility management, investment advisory, project development, and property consultancy for individuals, corporates, and diaspora clients.\n...\n📍 Ojomu Royal Plaza, Km 23, Lekki-Epe Expressway, Lagos\n...\n## Company Details\n...\n- Industry: Real Estate\n- Type: Privately held\n- Homepage: plusworldrealtors.com.ng\n- Aliases: plusworld realtors\n- LinkedIn: linkedin.com/company/plusworld-realtors\n- Linkedin Followers: 222\n- Emails: info@plusworldrealtors.com.ng\n- Phone: +234 812 670 8162\n...\n- Employees: 7\n- Company Size: 11-50 employees\n  - Yearly Growth: +16.7%\n- Key Executives:\n  - John Igbaifua: Chief Executive Officer\n- Breakdown:\n  - By Country: Nigeria: 4 (57%)\n  - By Department: General Management: 1 (14%), Hr: 1 (14%), Marketing: 1 (14%), Sales: 1 (14%)\n  - By Seniority: Specialist: 3 (43%), Clevel: 1 (14%)\n- Salary ranges:\n  - Facility Manager: median 1,442,520 NGN per monthly\n\n---\n\nTitle: PLUSWORLD, INC. - Ohio Company\nURL: https://ohio-corp.com/co/plusworld-inc\nPublished: N/A\nAuthor: N/A\nHighlights:\nPLUSWORLD, INC. - Ohio Company\n\n##### Company Info\n\n###### Entity Number:\n\n1185191\n\n###### Business Name:\n\n###### Original Filing Date:\n\n09/21/2000\n\n###### Business Status:\n\nActive\n\n###### Business Type:\n\nCORPORATION FOR PROFIT\n\n###### Exp. Date:\n\n-\n\n###### Location:\n\nCINCINNATI, HAMILTON, -\n\nAdded: 2018-12-20 10:19 Updated: 2021-10-29 20:27\n\n##### Filings\n\n###### No Data\n\n##### Registered Agent Info\n\n###### Registered Agent Name\n\nSCOTT WALLACE\n\n###### Status\n\nActive\n\n###### Address:\n\n5985 SOVEREIGN DRIVE, CINCINNATI OH 45241\n\n« GREATER TOLEDO AREA CHAPTER OF THE BRAIN INJURY ASSOCIATION OF OHIO MICKELI'S PUB, INC. »\n\n##### Comments\n\nThis is not the official website of this company. Don't seek support service here please.\n\nNo comment\n\nAdd Comment\n\nSubmit\n\n##### Registered On 09/21/2000\n\nBROWN CARTOON LIKE PILLAR W/ BLUE MARBLIZED BALL\n\nOPTIMAL EMPLOYMENT & CONSULTING W/ LETTER 'O' IN BROWN MARBLIZED SQUARE\n\nS Z DEVELOPMENT , LLC\n\nCURRENCY BROKERS (USA) INC.\n\nTHE BEDFORD TWENTY FIRST CENTURY DEVELOPMENT CORPORATION\n\nFIRST VERSE RECORDS, LTD.\n\nQUALITY PLUS HOME HEALTH CARE SERVICES, INC.\n\n##### Similar Names\n\n---\n\nTitle: Plusworld Coating\nURL: https://linkedin.com/company/plusworld-coating\nPublished: N/A\nAuthor: N/A\nHighlights:\n# Plusworld Coating\n...\nPlusworld Coating is a Manufacturing company. Plusworld Coating employs 4 people.\n...\n- Industry: Manufacturing\n...\n- Type: Public company\n...\n- Homepage: plusworldcoating.\n...\nAliases:\n...\ncom/company/\n...\n- Employees: 4\n- Company Size: 11-50 employees\n- Breakdown:\n  - By Country: Nigeria: 4 (100%)\n  - By Department: Hr: 1 (25%), Operations: 1 (25%), Other Department: 1 (25%), Project Management: 1 (25%), Sales: 1 (25%)\n  - By Seniority: Manager: 2 (50%), Specialist: 2 (50%), Other Management: 1 (25%)\n\n---\n\nTitle: 会社概要｜株式会社プラスワールド｜愛知県春日井市\nURL: https://www.plusworld.co.jp/company\nPublished: 2019-05-27T00:00:00.000Z\nAuthor: N/A\nHighlights:\n会社概要｜株式会社プラスワールド｜愛知県春日井市\n...\n# 会社\n...\n| 社名 | 株式会社プラスワールド |\n| --- | --- |\n| 代表 | 満園 昌吾 |\n| 所在地 | 〒486-0901 愛知県春日井市牛山町2780番地1 |\n| TEL | 0568-34-0261 |\n| FAX | 0568-34-3777 |\n| メール | info@plusworld.co.jp |\n| URL | https://www.plusworld.co.jp |\n| 営業品目 | 1.製缶溶接 2.板金溶接 3.肉盛溶接 4.機械加工 5.塗装 |\n| 資本金 | 1,000万円 |\n| 設立年月日 | 平成10年3月20日 |\n| 従業員 | 35名（グループ会社含む） |\n\n---\n\nTitle: PLUSWORLD INVESTMENT PTE. LTD. (202325979R) - Holding companies of firms engaged in non-financial and insurance activities  - Singapore Company\nURL: https://www.companies.sg/business/202325979R/PLUSWORLD-INVESTMENT-PTE-LTD-\nPublished: N/A\nAuthor: N/A\nHighlights:\n# PLUSWORLD INVESTMENT PTE. LTD.\n...\n- Entity name: PLUSWORLD INVESTMENT PTE. LTD.\n- UEN: 202325979R\n- Profile URL: https://www.companies.sg/business/202325979R/PLUSWORLD-INVESTMENT-PTE-LTD-\n- Markdown URL: https://www.companies.sg/business/202325979R/PLUSWORLD-INVESTMENT-PTE-LTD-.md\n- Company type: Exempt Private Company Limited by Shares\n- Registration date: 2023-07-03\n- UEN issue date: 2023-07-03\n- Current status: Live Company\n- Registered address: 7 TEMASEK BOULEVARD, #12-07, SUNTEC TOWER ONE, Singapore 38987\n- Postal code: 038987\n- Primary SSIC code: 64202\n- Primary SSIC description: Holding companies of firms engaged in non-financial and insurance activities\n- Paid-up capital: SGD 5,000\n- Issuance agency: ACRA\n- Data source note: Public Singapore company information organized by companies.sg.\n...\nPLUSWORLD INVESTMENT PTE. LTD. is a Singapore Exempt Private Company Limited by Shares. The company was incorporated on 03 Jul 2023, which is 2.9 years ago. The address of the Business's registered office is 7 TEMASEK BOULEVARD, #12-07, SUNTEC TOWER ONE, Singapore 38987. The Business current operating status is Live Company. The Business's principal activity is Holding companies of firms engaged in non-financial and insurance activities . The company's paid-up capital is SGD 5,000. The company UEN is 202325979R, registered with ACRA on 2023-07-03.\n\n---\n\nTitle: TvoozMagnificent/BrainFuckPlusInterpreter\nURL: https://github.com/TvoozMagnificent/BrainFuckPlusInterpreter\nPublished: 2022-01-28T12:14:12.000Z\nAuthor: N/A\nHighlights:\na : move 1 cell to the left; if already at position 0 raise an error|Move|Not really, same as < in bf\n...\n|Loop|Yes, as goes all the loop features except \\[CODEl\\]\n...\n\\,CODEl\\; : repeatly execute code inside as long as current cell remains >= 0 (position of 'current cell' aforementioned might change due to CODE inside) (code doesn't include character 'l', the letter in jk'l'mn)|Loop|Yes, as goes all the loop features except \\[CODEl\\]\n\n---\n\nTitle: BrainFuck+ - Esolang\nURL: https://esolangs.org/wiki/BrainFuck%2B\nPublished: N/A\nAuthor: N/A\nHighlights:\n| a | move 1 cell to the left; if already at position 0 raise an error | Move | Not really, same as < in bf | |\n| d | move 1 cell to the right | Move | Not really, same as > in bf | |\n...\n| ,CODE; | execute code inside if current cell is >= 0 | Loop | Yes, as goes all the loop features except [CODEl] | |\n...\n| ,CODEl; | repeatly execute code inside as long as current cell remains >= 0 (position of 'current cell' aforementioned might change due to CODE inside) (code doesn't include character 'l', the letter in jk'l'mn) | Loop | Yes, as goes all the loop features except [CODEl] | |\n\n---\n\nTitle: Career Plusworld\nURL: https://linkedin.com/company/career-plusworld\nPublished: N/A\nAuthor: N/A\nHighlights:\n# Career Plusworld\n...\nCareer Plusworld is a Higher Education company. Career Plusworld employs 1 person. Headquartered in New Delhi, Delhi, India.\n...\nWe are the preferred counselors of different top management colleges across the breath of the states. We at career plus, encourage students to chose best suitable place of study through.\n...\n- Industry: Higher Education\n- Headquarters: New Delhi, India\n- Homepage: careerplusworld.com\n- Aliases: career plusworld\n- LinkedIn: linkedin.com/company/career-plusworld\n...\n- Employees: 1\n\n---\n\nTitle: Looking for a house in Lekki… but your budget is closer to Ikorodu? | Plusworld Realtors\nURL: https://www.linkedin.com/posts/plusworld-realtors_lagosrealestate-propertyinvestment-lekkirealestate-activity-7434894475555946496-yj8h\nPublished: 2026-03-04T00:00:00.000Z\nAuthor: Plusworld Realtors\nHighlights:\n**Plusworld Realtors**: Plusworld Realtors is Nigeria's leading real estate company that helps individuals find, buy, sell, and rent properties across Nigeria. Plusworld Realtors has 1-10 employees. Headquartered in Nigeria, with an office in Lagos.",
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              "command": "g++ -std=c++26 -O2 -pipe -x c++ -o \"/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/plusworld\" - <<'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    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        vector<int> a(n + 1), b(n + 1);\n        for (int i = 1; i <= n; ++i) cin >> a[i];\n        for (int i = 1; i <= n; ++i) cin >> b[i];\n\n        bool possible = true;\n        int need = 0, start = 1;\n        for (int i = 1; i <= n; ++i) {\n            if (a[i] > b[i]) possible = false;\n            if (a[i] < b[i]) {\n                if (need == 0) start = i;\n                ++need;\n            }\n        }\n\n        if (!possible) {\n            cout << -1 << '\\n';\n            continue;\n        }\n        if (need == 0) {\n            cout << \"0 1\\n\\n\";\n            continue;\n        }\n\n        vector<int> ops;\n        int p = start;\n        while (need > 0) {\n            if (a[p] < b[p]) {\n                while (a[p] < b[p] &&\n                       (a[p] >= p || a[a[p]] == b[a[p]])) {\n                    ++a[p];\n                    ops.push_back(1);\n                }\n                if (a[p] == b[p]) --need;\n            }\n\n            if (need == 0 || a[p] == p) break;\n            ops.push_back(2);\n            p = a[p];\n        }\n\n        if (need > 0) {\n            cout << -1 << '\\n';\n            continue;\n        }\n\n        cout << ops.size() << ' ' << start << '\\n';\n        for (int x : ops) cout << x << ' ';\n        cout << '\\n';\n    }\n}\nCPP\n/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/plusworld <<'EOF'\n9\n3\n1 2 3\n2 3 3\n2\n2 2\n1 2\n5\n1 1 1 4 5\n2 3 4 5 5\n3\n1 1 1\n1 2 3\n3\n1 1 1\n3 2 3\n3\n1 1 3\n1 2 3\n4\n1 1 1 4\n3 3 4 4\n4\n1 4 3 4\n1 4 3 4\n5\n1 3 4 5 4\n2 3 4 5 5\nEOF",
              "workdir": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-d1B4yi",
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            "title": "g++ -std=c++26 -O2 -pipe -x c++ -o \"/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/plusworld\" - <<'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    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        vector<int> a(n + 1), b(n + 1);\n        for (int i = 1; i <= n; ++i) cin >> a[i];\n        for (int i = 1; i <= n; ++i) cin >> b[i];\n\n        bool possible = true;\n        int need = 0, start = 1;\n        for (int i = 1; i <= n; ++i) {\n            if (a[i] > b[i]) possible = false;\n            if (a[i] < b[i]) {\n                if (need == 0) start = i;\n                ++need;\n            }\n        }\n\n        if (!possible) {\n            cout << -1 << '\\n';\n            continue;\n        }\n        if (need == 0) {\n            cout << \"0 1\\n\\n\";\n            continue;\n        }\n\n        vector<int> ops;\n        int p = start;\n        while (need > 0) {\n            if (a[p] < b[p]) {\n                while (a[p] < b[p] &&\n                       (a[p] >= p || a[a[p]] == b[a[p]])) {\n                    ++a[p];\n                    ops.push_back(1);\n                }\n                if (a[p] == b[p]) --need;\n            }\n\n            if (need == 0 || a[p] == p) break;\n            ops.push_back(2);\n            p = a[p];\n        }\n\n        if (need > 0) {\n            cout << -1 << '\\n';\n            continue;\n        }\n\n        cout << ops.size() << ' ' << start << '\\n';\n        for (int x : ops) cout << x << ' ';\n        cout << '\\n';\n    }\n}\nCPP\n/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/plusworld <<'EOF'\n9\n3\n1 2 3\n2 3 3\n2\n2 2\n1 2\n5\n1 1 1 4 5\n2 3 4 5 5\n3\n1 1 1\n1 2 3\n3\n1 1 1\n3 2 3\n3\n1 1 3\n1 2 3\n4\n1 1 1 4\n3 3 4 4\n4\n1 4 3 4\n1 4 3 4\n5\n1 3 4 5 4\n2 3 4 5 5\nEOF",
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              "content": "Look at the current barn $p$ and its destination $q=a_p$. If $q<p$ and barn $q$ is still unfinished, do not jump over it: follow the edge to $q$ first. In every other case, increasing $a_p$ is the useful move."
            },
            {
              "order": 3,
              "content": "Implement that rule literally. While $a_p<b_p$, perform type 1 while either $a_p\\ge p$ or the destination barn $a_p$ is already finished. If the loop stops before reaching $b_p$, operation 2 must move to the unfinished lower-index destination."
            },
            {
              "order": 4,
              "content": "Start at the first unfinished barn. Keep a count of unfinished barns. If the count reaches zero, you are done. If the walk stops at $a_p=p$ while work remains, then $p$ is already a target self-loop, so neither operation can ever get Bessie out."
            },
            {
              "order": 5,
              "content": "The construction is small enough for the limit: type 1 is used exactly $\\sum_i(b_i-a_i)\\le n(n-1)$ times. A fixed barn $p$ is left at most $p$ times, so all type 2 operations are at most $\\sum_{p=1}^n p$."
            }
          ],
          "editorial": "Treat barn $i$ as **unfinished** while $a_i<b_i$. We never want to touch a finished barn again.\n\n## The local rule\n\nAssume first that every $a_i\\le b_i$; otherwise the answer is obviously impossible because destinations only increase.\n\nSuppose Bessie is at barn $p$ and currently its edge is $p\\to q$, where $q=a_p$.\n\n- If $q\\ge p$, increase $a_p$. The edge is already pointing forward, so there is no unfinished lower-index dependency to deal with first.\n- If $q<p$ but barn $q$ is finished, increase $a_p$ as well. Passing over a finished barn is harmless.\n- If $q<p$ and barn $q$ is unfinished, follow the edge to $q$ first. This is the only dangerous case: increasing $a_p$ would skip an unfinished lower-index barn.\n\nTherefore, while barn $p$ is unfinished, we repeatedly perform type 1 exactly while\n\n$$\na_p<b_p\\quad\\text{and}\\quad (a_p\\ge p\\ \\text{or}\\ a_{a_p}=b_{a_p}).\n$$\n\nIf this stops because $a_p=b_p$, barn $p$ is finished. If it stops earlier, then $a_p<p$ and its destination is unfinished, so we perform type 2 and continue from that destination.\n\nThe walk is basically a DFS with one bizarre feature: its edges can be increased but never decreased. Charming.\n\n## Why the greedy choice is safe\n\nThe target edges satisfy $b_i\\ge i$, so every finished edge points right or stays at its own vertex.\n\nConsider a valid solution and look at an unfinished lower-index destination $q=a_p<p$. Before the solution raises the edge of $p$ past $q$, we may move the work that finishes $q$ earlier. Barn $p$ stays at $q$ during this rearrangement, and finishing $q$ cannot make any required increment illegal. If handling $q$ encounters another unfinished lower destination, apply the same argument recursively. Indices strictly decrease during these dependencies, so the process terminates.\n\nAfter $q$ is finished, crossing it while increasing $a_p$ is safe. The same exchange argument lets us perform every immediately safe increment before moving. Thus, whenever a solution exists, there is also a solution following the local rule above.\n\nThe first unfinished index is a valid starting point: every smaller barn is already finished, so there is no unfinished work below it that could be skipped.\n\n## Algorithm\n\n1. If some $a_i>b_i$, print `-1`.\n2. Find the first unfinished barn `start` and count all unfinished barns as `need`.\n3. Start Bessie at `start`.\n4. At the current barn `p`:\n   - If it is unfinished, apply type 1 while\n     $$a_p<b_p\\quad\\text{and}\\quad(a_p\\ge p\\ \\text{or}\\ a_{a_p}=b_{a_p}).$$\n   - If it just reached $b_p$, decrement `need`.\n   - If all barns are finished, stop successfully.\n   - If $a_p=p$ while work remains, stop and report `-1`.\n   - Otherwise apply type 2 and set `p=a_p`.\n\nThe decrement of `need` happens only inside the `a_p<b_p` branch, so an already finished barn is never counted twice.\n\n## Why stopping means impossible\n\nIf the algorithm stops with work remaining, it has $a_p=p$. If barn $p$ were unfinished, then $a_p\\ge p$ would satisfy the inner loop condition, and the algorithm would increase it. Hence $p$ is finished and $b_p=p$.\n\nBessie is now trapped at a target self-loop: type 2 keeps her at $p$, while type 1 would make $a_p>b_p$. By the safe exchange argument, any solvable instance has a solution following our canonical choices, so reaching this trap proves that the original instance is impossible.\n\nIf `need` reaches zero, every coordinate has been increased only up to its target, so the final array is exactly $b$.\n\n## Complexity and operation bound\n\nEach barn receives exactly $b_i-a_i$ type 1 operations, hence\n\n$$\n\\#\\text{type 1}=\\sum_i(b_i-a_i)\\le n(n-1).\n$$\n\nFor type 2, fix a barn $p$. After its first departure, every later departure from $p$ requires some previously unfinished barn with index smaller than $p$ to become finished before Bessie can return and leave again. There are only $p-1$ such barns, so barn $p$ is left at most $p$ times. Therefore\n\n$$\n\\#\\text{type 2}\\le\\sum_{p=1}^n p=\\frac{n(n+1)}2.\n$$\n\nThe total is less than $2n^2$, as required. The simulation takes $O(n^2)$ time and $O(n)$ memory.",
          "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\n    int t;\n    cin >> t;\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) cin >> a[i];\n        for (int i = 1; i <= n; ++i) cin >> b[i];\n\n        bool possible = true;\n        int need = 0;\n        int start = 1;\n\n        for (int i = 1; i <= n; ++i) {\n            if (a[i] > b[i]) possible = false;\n            if (a[i] < b[i]) {\n                if (need == 0) start = i;\n                ++need;\n            }\n        }\n\n        if (!possible) {\n            cout << -1 << '\\n';\n            continue;\n        }\n\n        if (need == 0) {\n            cout << \"0 1\\n\\n\";\n            continue;\n        }\n\n        vector<int> operations;\n        int p = start;\n\n        while (need > 0) {\n            if (a[p] < b[p]) {\n                while (a[p] < b[p] &&\n                       (a[p] >= p || a[a[p]] == b[a[p]])) {\n                    ++a[p];\n                    operations.push_back(1);\n                }\n\n                if (a[p] == b[p]) --need;\n            }\n\n            if (need == 0 || a[p] == p) break;\n\n            operations.push_back(2);\n            p = a[p];\n        }\n\n        if (need > 0) {\n            cout << -1 << '\\n';\n            continue;\n        }\n\n        cout << operations.size() << ' ' << start << '\\n';\n        for (int op : operations) cout << op << ' ';\n        cout << '\\n';\n    }\n\n    return 0;\n}\n"
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      "parts": [
        {
          "type": "step-start",
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          "sessionID": "ses_f5341cf67ffemT7NhbK68OoURy",
          "messageID": "msg_0acf2c97d001eQ3n44JsjS5ZmO"
        },
        {
          "type": "reasoning",
          "text": "**Finalizing structured output format**",
          "time": {
            "start": 1789607989228,
            "end": 1789607996318
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          "sessionID": "ses_f5341cf67ffemT7NhbK68OoURy",
          "messageID": "msg_0acf2c97d001eQ3n44JsjS5ZmO"
        },
        {
          "type": "reasoning",
          "text": "**Analyzing canonical lemma for walk correctness**",
          "time": {
            "start": 1789607996325,
            "end": 1789607998691
          },
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            "openai": {
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          "messageID": "msg_0acf2c97d001eQ3n44JsjS5ZmO"
        },
        {
          "type": "reasoning",
          "text": "**Clarifying canonical start justification**",
          "time": {
            "start": 1789607998696,
            "end": 1789607998700
          },
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          "type": "reasoning",
          "text": "**Planning inductive proof for operation ordering**",
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        {
          "type": "reasoning",
          "text": "**Finalizing proof explanation and style notes**",
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          "tool": "StructuredOutput",
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              "hints": [
                {
                  "order": 1,
                  "content": "Operation 1 can never decrease an edge, so any $a_i>b_i$ is an immediate `-1`. Otherwise call barn $i$ unfinished while $a_i<b_i$; once it reaches $b_i$, it never needs to be touched again."
                },
                {
                  "order": 2,
                  "content": "Look at the current barn $p$ and its destination $q=a_p$. If $q<p$ and barn $q$ is still unfinished, do not jump over it: follow the edge to $q$ first. In every other case, increasing $a_p$ is the useful move."
                },
                {
                  "order": 3,
                  "content": "Implement that rule literally. While $a_p<b_p$, perform type 1 while either $a_p\\ge p$ or the destination barn $a_p$ is already finished. If the loop stops before reaching $b_p$, operation 2 must move to the unfinished lower-index destination."
                },
                {
                  "order": 4,
                  "content": "Start at the first unfinished barn. Keep a count of unfinished barns. If the count reaches zero, you are done. If the walk stops at $a_p=p$ while work remains, then $p$ is already a target self-loop, so neither operation can ever get Bessie out."
                },
                {
                  "order": 5,
                  "content": "The construction is small enough for the limit: type 1 is used exactly $\\sum_i(b_i-a_i)\\le n(n-1)$ times. A fixed barn $p$ is left at most $p$ times, so all type 2 operations are at most $\\sum_{p=1}^n p$."
                }
              ],
              "editorial": "Treat barn $i$ as **unfinished** while $a_i<b_i$. We never want to touch a finished barn again.\n\n## The local rule\n\nAssume first that every $a_i\\le b_i$; otherwise the answer is obviously impossible because destinations only increase.\n\nSuppose Bessie is at barn $p$ and currently its edge is $p\\to q$, where $q=a_p$.\n\n- If $q\\ge p$, increase $a_p$. The edge is already pointing forward, so there is no unfinished lower-index dependency to deal with first.\n- If $q<p$ but barn $q$ is finished, increase $a_p$ as well. Passing over a finished barn is harmless.\n- If $q<p$ and barn $q$ is unfinished, follow the edge to $q$ first. This is the only dangerous case: increasing $a_p$ would skip an unfinished lower-index barn.\n\nTherefore, while barn $p$ is unfinished, we repeatedly perform type 1 exactly while\n\n$$\na_p<b_p\\quad\\text{and}\\quad (a_p\\ge p\\ \\text{or}\\ a_{a_p}=b_{a_p}).\n$$\n\nIf this stops because $a_p=b_p$, barn $p$ is finished. If it stops earlier, then $a_p<p$ and its destination is unfinished, so we perform type 2 and continue from that destination.\n\nThe walk is basically a DFS with one bizarre feature: its edges can be increased but never decreased. Charming.\n\n## Why the greedy choice is safe\n\nThe target edges satisfy $b_i\\ge i$, so every finished edge points right or stays at its own vertex.\n\nConsider a valid solution and look at an unfinished lower-index destination $q=a_p<p$. Before the solution raises the edge of $p$ past $q$, we may move the work that finishes $q$ earlier. Barn $p$ stays at $q$ during this rearrangement, and finishing $q$ cannot make any required increment illegal. If handling $q$ encounters another unfinished lower destination, apply the same argument recursively. Indices strictly decrease during these dependencies, so the process terminates.\n\nAfter $q$ is finished, crossing it while increasing $a_p$ is safe. The same exchange argument lets us perform every immediately safe increment before moving. Thus, whenever a solution exists, there is also a solution following the local rule above.\n\nThe first unfinished index is a valid starting point: every smaller barn is already finished, so there is no unfinished work below it that could be skipped.\n\n## Algorithm\n\n1. If some $a_i>b_i$, print `-1`.\n2. Find the first unfinished barn `start` and count all unfinished barns as `need`.\n3. Start Bessie at `start`.\n4. At the current barn `p`:\n   - If it is unfinished, apply type 1 while\n     $$a_p<b_p\\quad\\text{and}\\quad(a_p\\ge p\\ \\text{or}\\ a_{a_p}=b_{a_p}).$$\n   - If it just reached $b_p$, decrement `need`.\n   - If all barns are finished, stop successfully.\n   - If $a_p=p$ while work remains, stop and report `-1`.\n   - Otherwise apply type 2 and set `p=a_p`.\n\nThe decrement of `need` happens only inside the `a_p<b_p` branch, so an already finished barn is never counted twice.\n\n## Why stopping means impossible\n\nIf the algorithm stops with work remaining, it has $a_p=p$. If barn $p$ were unfinished, then $a_p\\ge p$ would satisfy the inner loop condition, and the algorithm would increase it. Hence $p$ is finished and $b_p=p$.\n\nBessie is now trapped at a target self-loop: type 2 keeps her at $p$, while type 1 would make $a_p>b_p$. By the safe exchange argument, any solvable instance has a solution following our canonical choices, so reaching this trap proves that the original instance is impossible.\n\nIf `need` reaches zero, every coordinate has been increased only up to its target, so the final array is exactly $b$.\n\n## Complexity and operation bound\n\nEach barn receives exactly $b_i-a_i$ type 1 operations, hence\n\n$$\n\\#\\text{type 1}=\\sum_i(b_i-a_i)\\le n(n-1).\n$$\n\nFor type 2, fix a barn $p$. After its first departure, every later departure from $p$ requires some previously unfinished barn with index smaller than $p$ to become finished before Bessie can return and leave again. There are only $p-1$ such barns, so barn $p$ is left at most $p$ times. Therefore\n\n$$\n\\#\\text{type 2}\\le\\sum_{p=1}^n p=\\frac{n(n+1)}2.\n$$\n\nThe total is less than $2n^2$, as required. The simulation takes $O(n^2)$ time and $O(n)$ memory.",
              "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\n    int t;\n    cin >> t;\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) cin >> a[i];\n        for (int i = 1; i <= n; ++i) cin >> b[i];\n\n        bool possible = true;\n        int need = 0;\n        int start = 1;\n\n        for (int i = 1; i <= n; ++i) {\n            if (a[i] > b[i]) possible = false;\n            if (a[i] < b[i]) {\n                if (need == 0) start = i;\n                ++need;\n            }\n        }\n\n        if (!possible) {\n            cout << -1 << '\\n';\n            continue;\n        }\n\n        if (need == 0) {\n            cout << \"0 1\\n\\n\";\n            continue;\n        }\n\n        vector<int> operations;\n        int p = start;\n\n        while (need > 0) {\n            if (a[p] < b[p]) {\n                while (a[p] < b[p] &&\n                       (a[p] >= p || a[a[p]] == b[a[p]])) {\n                    ++a[p];\n                    operations.push_back(1);\n                }\n\n                if (a[p] == b[p]) --need;\n            }\n\n            if (need == 0 || a[p] == p) break;\n\n            operations.push_back(2);\n            p = a[p];\n        }\n\n        if (need > 0) {\n            cout << -1 << '\\n';\n            continue;\n        }\n\n        cout << operations.size() << ' ' << start << '\\n';\n        for (int op : operations) cout << op << ' ';\n        cout << '\\n';\n    }\n\n    return 0;\n}\n"
            },
            "output": "Structured output captured successfully.",
            "metadata": {
              "valid": true
            },
            "title": "Structured Output",
            "time": {
              "start": 1789608044306,
              "end": 1789608044310
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