{
  "info": {
    "id": "ses_f53419cd5ffe6m5ZjKx96hm4tQ",
    "slug": "calm-otter",
    "projectID": "global",
    "directory": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Z0OhDi",
    "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Z0OhDi",
    "title": "Nudge content generation",
    "agent": "nudge-generation",
    "model": {
      "id": "gpt-5.6-luna",
      "providerID": "openai",
      "variant": "max"
    },
    "version": "1.18.30",
    "summary": {
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    "tokens": {
      "input": 31740,
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      "reasoning": 4854,
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  "messages": [
    {
      "info": {
        "role": "user",
        "time": {
          "created": 1789604553529
        },
        "agent": "nudge-generation",
        "model": {
          "providerID": "openai",
          "modelID": "gpt-5.6-luna",
          "variant": "max"
        },
        "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": [
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                  "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"
                  }
                ],
                "description": "If status is 'success', a complete C++ solution. Otherwise null."
              }
            },
            "required": [
              "status",
              "reason",
              "hints",
              "editorial",
              "solution"
            ],
            "additionalProperties": false,
            "title": "problem_response",
            "description": "Submit the generated content for the problem, or report if it is unsolvable."
          },
          "retryCount": 2
        },
        "id": "msg_0acbe63390014Bf2sgGEtnLlte",
        "sessionID": "ses_f53419cd5ffe6m5ZjKx96hm4tQ"
      },
      "parts": [
        {
          "type": "text",
          "text": "Generate Codeforces learning content: five progressive hints, an original editorial, and AC-quality C++.\n\nFollow the editorial-first research workflow in the system instructions. Use the supplied statement to verify that the source solves this exact problem.\n\nWrite clean Markdown with LaTeX as needed. Use quick and clever humor when appropriate. Tell it like it is (don't sugar-coat responses), and use very casual language. You are fully allowed to swear, just don't overdo it like a sailor (be natural). Deconstruct any false assumptions.\n\nGenerate content for Codeforces problem 2262A1: \"Floor of MEX (Easy Version)\".\n\nKnown problem metadata:\n- Contest ID: 2262\n- Index: A1\n- Name: Floor of MEX (Easy Version)\n- Rating: unknown / unrated\n- Tags: constructive algorithms, greedy\n- Problem URL: https://codeforces.com/contest/2262/problem/A1\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\">A1. Floor of MEX (Easy Version)</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><span class=\"tex-font-style-bf\">This is the easy version of the problem. The difference between the versions is that in this version, you only need to find one subset that satisfies the constraints. You can hack only if you solved all versions of this problem.</span></p><p>Let $$$$$$ f(S,x)=\\text{mex}\\left(\\left\\{\\left\\lfloor \\frac{y}{x}\\right\\rfloor : y\\in S\\right\\}\\right), $$$$$$ where $$$S$$$ is a set of non-negative integers and $$$x$$$ is a positive integer. $$$^{\\text{∗}}$$$</p><p>Farmer John chooses a (possibly empty) subset $$$A\\subseteq \\{0,1,\\ldots,n-1\\}$$$. He then constructs an array $$$a$$$ of length $$$n$$$, where $$$a_k = f(A,k)$$$ for every $$$1\\le k\\le n$$$.</p><p>Bessie, being a mischievous cow, hides the set $$$A$$$. Farmer John is left only with the array $$$a_1,a_2,\\ldots,a_n$$$.</p><p>Your task is to construct any (possibly empty) subset $$$B\\subseteq \\{0,1,\\ldots,n-1\\}$$$ such that $$$f(B,k)=a_k$$$ for every $$$1 \\le k \\le n$$$. <span class=\"tex-font-style-bf\">Farmer John will only give you arrays where such a $$$B$$$ exists</span>.</p><div class=\"statement-footnote\"><p>$$$^{\\text{∗}}$$$$$$\\operatorname{mex}(c)$$$ denotes the minimum excluded (MEX)$$$^{\\text{∗}}$$$ of the collection $$$c$$$. (in this case, the minimum non-negative integer not present in the set)</p></div></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 $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).</p><p>The second line of each test case contains $$$n$$$ integers $$$a_1,a_2,\\ldots,a_n$$$ ($$$0 \\leq a_i \\leq n$$$).</p><p>You will only be given arrays where there exists $$$B \\subseteq \\{0,1,\\ldots,n-1\\}$$$ such that $$$f(B,k)=a_k$$$ for every $$$1 \\le k \\le n$$$.</p><p>It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.</p></div><div class=\"output-specification\"><div class=\"section-title\">Output</div><p>For each test case, output two lines.</p><p>On the first line, output an integer $$$m$$$ — the size of the set $$$B$$$.</p><p>On the second line, output $$$m$$$ <span class=\"tex-font-style-bf\">distinct</span> integers $$$b_1,b_2,\\ldots,b_m$$$ — the elements of $$$B$$$. If $$$m=0$$$, the second line should be empty.</p><p>The elements of $$$B$$$ may be printed in any order.</p><p>If there are multiple solutions, 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\">3</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">6</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">0 3 2 2 2 1</div><div class=\"test-example-line test-example-line-even test-example-line-2\">5</div><div class=\"test-example-line test-example-line-even test-example-line-2\">2 1 1 1 1</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">6</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">1 2 1 1 1 1</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</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">1 2 3 4 5 </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\">0 1 </div><div class=\"test-example-line test-example-line-odd test-example-line-3\">2</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">0 2 </div></pre></div></div></div><div class=\"note\"><div class=\"section-title\">Note</div><p>For the first test case, one possible answer is $$$$$$ B=\\{1,2,3,4,5\\}. $$$$$$ Then $$$$$$ f(B,1)=\\text{mex}(\\{1,2,3,4,5\\})=0, $$$$$$ $$$$$$ f(B,2)=\\text{mex}(\\{0,1,2\\})=3, $$$$$$ and $$$$$$ f(B,3)=f(B,4)=f(B,5)=2. $$$$$$ Finally, $$$$$$ f(B,6)=\\text{mex}(\\{0\\})=1. $$$$$$ Thus the array is $$$[0,3,2,2,2,1]$$$.</p><p>For the second test case, one possible answer is $$$$$$ B=\\{0,1\\}. $$$$$$ Then $$$f(B,1)=2$$$, and for every $$$2\\le k\\le 5$$$, we have $$$$$$ \\left\\{\\left\\lfloor y/k\\right\\rfloor : y\\in B\\right\\}=\\{0\\}, $$$$$$ so $$$f(B,k)=1$$$. Therefore, the array is $$$[2,1,1,1,1]$$$.</p></div>\n</problem-statement>\n\n\nSource lookup status:\n<source-lookup-status>\n- Tutorial (en): https://codeforces.com/blog/entry/156688 loaded (200 OK; title: \"Codeforces Round 1120 (Div 1, Div 2) Editorial - Codeforces\"; mentions 2262A1)\n</source-lookup-status>\n\nGenerate:\n1. Five progressive hints, from a gentle nudge to the key insight.\n2. A self-contained editorial that teaches the solution clearly, including the key insight, algorithm, correctness argument, complexity, and necessary implementation details. Explain non-obvious steps without padding or a discovery-process transcript.\n3. A complete C++26 solution that gets AC on Codeforces.\n\nFormatting & Style Rules:\n- Hints and editorial: valid Markdown with LaTeX as needed (e.g., $dp[i]$, $$\\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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        "variant": "max",
        "path": {
          "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Z0OhDi",
          "root": "/"
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        "modelID": "gpt-5.6-luna",
        "providerID": "openai",
        "time": {
          "created": 1789604553659,
          "completed": 1789604561922
        },
        "finish": "tool-calls",
        "id": "msg_0acbe63bb001ROEUMWPuLnKztx",
        "sessionID": "ses_f53419cd5ffe6m5ZjKx96hm4tQ"
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      "parts": [
        {
          "type": "step-start",
          "id": "prt_0acbe6ab3001XDbO9ZTJzw4Qp4",
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        {
          "type": "reasoning",
          "text": "**Planning parallel tutorial fetch and web search**",
          "time": {
            "start": 1789604555445,
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          "sessionID": "ses_f53419cd5ffe6m5ZjKx96hm4tQ",
          "messageID": "msg_0acbe63bb001ROEUMWPuLnKztx"
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        {
          "type": "tool",
          "tool": "webfetch",
          "callID": "call_MfJqgQIMX7VLt5V1B5urUoNx",
          "state": {
            "status": "completed",
            "input": {
              "url": "https://codeforces.com/blog/entry/156688",
              "format": "markdown",
              "timeout": 30
            },
            "output": "---\nimage: https://codeforces.com/codeforces.org/s/30864/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/30864/images/codeforces-logo-with-telegram-555x130.png \"Codeforces\")](https://codeforces.com/) \n\n[![In English](//codeforces.com/codeforces.org/s/30864/images/flags/24/gb.png \"In English\")](https://codeforces.com/blog/entry/156688?locale=en) [![По-русски](//codeforces.com/codeforces.org/s/30864/images/flags/24/ru.png \"По-русски\")](https://codeforces.com/blog/entry/156688?locale=ru) \n\n[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/30864/images/flags/24/gb.png \"In English\") \n\nThank you to everyone who participated! \n\nWe are sorry for the poor testcases on C1, along with C1 and C2's solutions being so similar to the recent Div3 E. \n\n[2263A - Min Max Game](https://codeforces.com/contest/2263/problem/A \"Codeforces Round 1120 (Div. 2)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nvoid tc() {\n    int n; cin >> n;\n    int cnt = 0;\n    vector<int> a(n);\n    for(int i = 0; i < n; i++) {\n        cin >> a[i];\n        cnt += a[i];\n    }\n    if(cnt >= n - cnt) {\n        cout << \"Bessie\\n\";\n    } else {\n        cout << \"Elsie\\n\";\n    }\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263B - Min Matrices](https://codeforces.com/contest/2263/problem/B \"Codeforces Round 1120 (Div. 2)\")\n\nAuthor: [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include<bits/stdc++.h>\n \nusing namespace std;\n \nvoid solve() {\n    int n, k; cin >> n >> k;\n    int x = k-n+1;\n    if(!(1 <= x && x <= n)) { cout << -1; return; }\n    vector<vector<int>> ans(n, vector<int>(n));\n    ans[0][0] = 1;\n    int on = 2;\n    for(int i = 1; i < x; i++) {\n        ans[0][i] = on; on++;\n        ans[i][0] = on; on++;\n    }\n    for(int i = x; i < n; i++) {\n        ans[i][i] = on; on++;\n    }\n\n    for(int i = 0; i < n; i++) {\n        for(int j = 0; j < n; j++) {\n            if(!ans[i][j]) {\n                ans[i][j] = on; on++;\n            }\n        }\n    }\n\n    for(int j = 0; j < n; j++) {\n        if(j) cout << \"\\n\";\n        for(auto &i : ans[j]) cout << i << \" \";\n    }\n}\n\nint main() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) {\n        solve();\n        if(t)   cout << \"\\n\";\n    }\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263C1 - Floor of MEX (Easy Version)](https://codeforces.com/contest/2263/problem/C1 \"Codeforces Round 1120 (Div. 2)\") / [2262A1 - Floor of MEX (Easy Version)](https://codeforces.com/contest/2262/problem/A1 \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [CutSandstone](https://codeforces.com/profile/CutSandstone \"Candidate Master CutSandstone\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nvoid tc() {\n    int n; cin >> n;\n    vector<int> b(n + 1);\n    for(int i = 1; i <= n; i++) cin >> b[i];\n    vector<pair<int, int>> goodivs;\n    vector<pair<int, int>> badivs;\n    for(int i = 1; i <= n; i++) {\n        badivs.push_back({b[i] * i, (b[i] + 1) * i - 1});\n        for(int j = 0; j < b[i]; j++) {\n            goodivs.push_back({j * i, (j + 1) * i - 1});\n        }\n    }\n    vector<int> d(n + 1);\n    for(auto [l, r] : badivs) {\n        d[min(n, l)]++;\n        d[min(n, r + 1)]--;\n    }\n    int csm = 0;\n    vector<int> gnums;\n    for(int i = 0; i < n; i++) {\n        csm += d[i];\n        if(csm == 0) {\n            gnums.push_back(i);\n        }\n    }\n    cout << gnums.size() << endl;\n    for(int i : gnums) cout << i << \" \";\n    cout << endl;\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263C2 - Floor of MEX (Hard Version)](https://codeforces.com/contest/2263/problem/C2 \"Codeforces Round 1120 (Div. 2)\") / [2262A2 - Floor of MEX (Hard Version)](https://codeforces.com/contest/2262/problem/A2 \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nconst int MOD = 1e9 + 7;\n\nvoid tc() {\n    int n; cin >> n;\n    vector<int> b(n + 1);\n    for(int i = 1; i <= n; i++) cin >> b[i];\n    vector<pair<int, int>> goodivs;\n    vector<pair<int, int>> badivs;\n    for(int i = 1; i <= n; i++) {\n        if(b[i] > (n - 1) / i + 1) {\n            cout << 0 << endl;\n            return;\n        }\n        badivs.push_back({b[i] * i, (b[i] + 1) * i - 1});\n        for(int j = 0; j < b[i]; j++) {\n            goodivs.push_back({j * i, (j + 1) * i - 1});\n        }\n    }\n    vector<int> d(n + 1);\n    for(auto [l, r] : badivs) {\n        d[min(n, l)]++;\n        d[min(n, r + 1)]--;\n    }\n    int csm = 0;\n    vector<int> gnums;\n    for(int i = 0; i < n; i++) {\n        csm += d[i];\n        if(csm == 0) {\n            gnums.push_back(i);\n        }\n    }\n    set<pair<int, int>> st;\n    sort(goodivs.begin(), goodivs.end(), [&] (pair<int, int> iv, pair<int, int> iv2) {\n        return iv.second - iv.first < iv2.second - iv2.first;\n    });\n    for(auto [l, r] : goodivs) {\n        auto it = st.lower_bound({l, 0});\n        if(it != st.end() && it->second <= r) {\n            continue;\n        }\n        st.insert({l, r});\n    }\n\n    vector<pair<int, int>> impivs(st.begin(), st.end());\n    int l = 0, r = 0;\n    vector<int> dp(impivs.size() + 1);\n    dp[0] = 1;\n    int poss = 1;\n    for(int i : gnums) {\n        while(l < impivs.size() && impivs[l].second < i) {\n            poss += MOD - dp[l];\n            poss %= MOD;\n            l++;\n        }\n        while(r < impivs.size() && impivs[r].first <= i) r++;\n        dp[r] += poss;\n        dp[r] %= MOD;\n        poss += poss;\n        poss %= MOD;\n    }\n    cout << dp[impivs.size()] << endl;\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263D - Culling Game](https://codeforces.com/contest/2263/problem/D \"Codeforces Round 1120 (Div. 2)\") / [2262B - Culling Game](https://codeforces.com/contest/2262/problem/B \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nstruct BIT {\n    vector<ll> tr;\n    BIT(int n) : tr(n + 1) {}\n    void add(int k, ll x) {\n        while(k < tr.size()) {\n            tr[k] += x;\n            k += k&-k;\n        }\n    }\n    ll sum(int k) {\n        ll res = 0;\n        while(k > 0) {\n            res += tr[k];\n            k -= k&-k;\n        }\n        return res;\n    }\n};\n\nvoid tc() {\n    int n; cin >> n;\n    vector<int> a(n), p(n);\n    for(int i = 0; i < n; i++) {\n        cin >> a[i];\n    }\n    for(int i = 0; i < n; i++) {\n        cin >> p[i]; p[i]--;\n    }\n    reverse(p.begin(), p.end());\n    BIT bit(n);\n    set<int> imp; \n    vector<int> ans(n);\n    for(int i = 0; i < n; i++) {\n        int k = p[i];\n        auto it = imp.upper_bound(k);\n        int pimp;\n        if(it == imp.begin() || bit.sum(k) - bit.sum(*prev(it)) < a[k]) {\n            imp.insert(k);\n            pimp = k;\n        } else {\n            pimp = *prev(it);\n        }\n        bit.add(k + 1, a[k]);\n        while(imp.upper_bound(k) != imp.end()) {\n            int j = *imp.upper_bound(k);\n            if(bit.sum(j) - bit.sum(pimp) >= a[j]) {\n                imp.erase(j);\n            } else {\n                break;\n            }\n        }\n        ans[i] = imp.size() - 1;\n    }\n    reverse(ans.begin(), ans.end());\n    for(int i : ans) cout << i << \" \";\n    cout << endl;\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263E - Traveling the World](https://codeforces.com/contest/2263/problem/E \"Codeforces Round 1120 (Div. 2)\") / [2262C - Traveling the World](https://codeforces.com/contest/2262/problem/C \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nconst int MOD = 1e9 + 7;\n\nvoid tc() {\n    int n; cin >> n;\n    vector<ll> a(n);\n    for(int i = 0; i < n; i++) cin >> a[i];\n    int k = n / 2;\n\n    int val = 1;\n    for(int i = 1; i <= k; i++) {\n        val = (ll)i * val % MOD;\n    }\n    for(int i = 1; i <= k - (n % 2 == 0 ? 2 : 1); i++) {\n        val = (ll)i * val % MOD;\n    }\n\n    set<ll> xposs;\n    xposs.insert(a[n - 1] - a[n - 2]);\n    xposs.insert(a[n - 1] - a[n - 3]);\n\n    int ans = 0;\n    for(ll x : xposs) {\n        vector<ll> b;\n        for(int i = 0; i < k; i++) {\n            b.push_back(i * x);\n        }\n        for(int i = 0; i < n - k; i++) {\n            b.push_back(a[n - 1] - i * x);\n        }\n        sort(b.begin(), b.end());\n        if(a == b) {\n            ans += val;\n            if(ans >= MOD) ans -= MOD;\n        }\n    }\n    cout << ans << endl;\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263F - PLUSworld](https://codeforces.com/contest/2263/problem/F \"Codeforces Round 1120 (Div. 2)\") / [2262D - PLUSworld](https://codeforces.com/contest/2262/problem/D \"Codeforces Round 1120 (Div. 1)\")\n\nAuthors: [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") & [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nvoid tc() {\n    int n; cin >> n;\n\n    vector<int> a(n), b(n);\n    for(int i = 0; i < n; i++) cin >> a[i], a[i]--;\n    for(int i = 0; i < n; i++) cin >> b[i], b[i]--;\n\n    int badcnt = 0;\n    for(int i = 0; i < n; i++) {\n        badcnt += a[i] != b[i];\n    }\n    if(badcnt == 0) {\n        cout << \"0 1\\n\";\n        return;\n    }\n\n    // mark everything which will ever be reached\n    vector<bool> useful(n);\n    for(int i = 0; i < n; i++) {\n        if(a[i] == b[i]) continue;\n        int x = i;\n        while(!useful[x]) {\n            useful[x] = true;\n            x = b[x];\n        }\n    }\n\n    // anything with b[i] == i once we reach we can't reach anything else, \n    // at most one of them can exit\n    vector<int> ends;\n    for(int i = 0; i < n; i++) {\n        if(useful[i] && b[i] == i) ends.push_back(i);\n    }\n\n    if(ends.size() > 1) {\n        cout << \"-1\\n\";\n        return;\n    }\n\n    vector<vector<int>> adj(n), radj(n);\n    for(int i = 0; i < n; i++) {\n        if(a[i] > b[i]) {\n            cout << \"-1\\n\";\n            return;\n        }\n        for(int j = a[i]; j <= b[i]; j++) {\n            adj[i].push_back(j);\n            radj[j].push_back(i);\n        }\n    }\n\n    // tarjan algorithm, from cp-algorithms\n    vector<int> st;    // - stack holding the unclaimed vertices\n    vector<int> roots(n, -1); // - keeps track of the SCC roots of the vertices\n    int timer = 0;         // - dfs timestamp counter\n    vector<int> t_in(n, -1);  // - keeps track of the dfs timestamp of the vertices\n    vector<int> t_low(n, -1); // - keeps track of the lowest t_in of unclaimed vertices\n    // reachable in the subtree\n\n    // implements the tarjan algorithm for strongly connected components\n    auto dfs = [&](auto&& self, int s) -> void {\n        t_low[s] = t_in[s] = timer++;\n        st.push_back(s);\n\n        for (auto u : adj[s]) {\n            if (t_in[u] == -1) { // tree-edge\n                self(self, u);\n                t_low[s] = min(t_low[s], t_low[u]);\n            } else if (roots[u] == -1) { // back-edge, cross-edge or forward-edge to an unclaimed vertex\n                t_low[s] = min(t_low[s], t_in[u]);\n            }\n        }\n\n        if (t_low[s] == t_in[s]) { // vertex is a root\n            while (true) {\n                int u = st.back();\n                st.pop_back();\n                roots[u] = s; // claims the vertex\n                if (u == s)\n                    break;\n            }\n        }\n    };\n\n    // applies the tarjan algorithm to all the vertices\n    for (int v = 0; v < n; v++) {\n        if (t_in[v] == -1) {\n            dfs(dfs, v);\n        }\n    }\n    vector<vector<int>> comps(n);\n    for(int v = 0; v < n; v++) {\n        comps[roots[v]].push_back(v);\n    }\n\n    vector<int> p(n, -1);\n    vector<int> sz(n);\n    for(int i = 0; i < n; i++) {\n        if(a[i] != b[i]) {\n            sz[roots[i]]++;\n        }\n        if(useful[i] && roots[i] != roots[b[i]]) {\n            if(p[roots[b[i]]] != -1) {\n                cout << \"-1\\n\";\n                return;\n            }\n            p[roots[b[i]]] = roots[i];\n        }\n    }\n    vector<int> chain;\n    int x = roots[ends[0]];\n    int csz = 0;\n    while(x != -1) {\n        chain.push_back(x);\n        csz += sz[x];\n        x = p[x];\n    }\n    if(csz != badcnt) {\n        cout << \"-1\\n\";\n        return;\n    }\n    reverse(chain.begin(), chain.end());\n\n    {\n        x = comps[chain[0]][0];\n        vector<int> ops;\n        auto op1 = [&] () {\n            ops.push_back(1);\n            a[x]++;\n        };\n        auto op2 = [&] () {\n            ops.push_back(2);\n            x = a[x];\n        };\n        auto solve_scc = [&] () {\n            int rt = roots[x];\n            vector<int> comp = comps[rt];\n\n            vector<int> p2(n, -1);\n            // only one thing in comp case\n            auto rdfs = [&] (auto&& self, int s) -> void {\n                for(int u : radj[s]) {\n                    if(roots[u] != rt) continue;\n                    if(p2[u] == -1) {\n                        p2[u] = s;\n                        self(self, u);\n                    }\n                }\n            };\n\n            // root a tree\n            rdfs(rdfs, comp[0]);\n\n            for(int i = 0; i < comp.size(); i++) {\n                while(x != comp[i]) {\n                    while(a[x] < max(p2[x], comp[i])) op1();\n                    op2();\n                }\n                while(a[x] < b[x]) op1();\n                op2();\n            }\n        };\n        for(int i = 0; i < chain.size(); i++) solve_scc();\n\n        cout << ops.size() << \" \" << comps[chain[0]][0] + 1 << endl;\n        for(int i : ops) cout << i << \" \";\n        cout << endl;\n\n        for(int i = 0; i < n; i++) assert(a[i] == b[i]);\n    }\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Magic**\n\nWith this, you can prove a significantly simpler solution. We will add this proof later.\n\n**Magic Solution**\n\n```\n// Note: AI generated solution\n#include <bits/stdc++.h>\nusing namespace std;\n\nint main() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n\n    int t;\n    cin >> t;\n\n    while (t--) {\n        int n;\n        cin >> n;\n\n        vector<int> a(n + 1), b(n + 1);\n        for (int i = 1; i <= n; i++) {\n            cin >> a[i];\n        }\n        for (int i = 1; i <= n; i++) {\n            cin >> b[i];\n        }\n\n        int start = 1, remaining = 0;\n        bool possible = true;\n\n        for (int i = 1; i <= n; i++) {\n            if (a[i] > b[i]) {\n                possible = false;\n            }\n            if (a[i] < b[i]) {\n                if (remaining == 0) {\n                    start = i;\n                }\n                remaining++;\n            }\n        }\n\n        if (!possible) {\n            cout << -1 << '\\n';\n            continue;\n        }\n\n        vector<int> operations;\n        int position = start;\n\n        while (remaining > 0) {\n            if (a[position] < b[position]) {\n                while (a[position] < b[position] &&\n                       (a[position] >= position ||\n                        a[a[position]] == b[a[position]])) {\n                    a[position]++;\n                    operations.push_back(1);\n                }\n\n                if (a[position] == b[position]) {\n                    remaining--;\n                }\n            }\n\n            if (remaining == 0 || a[position] == position) {\n                break;\n            }\n\n            operations.push_back(2);\n            position = a[position];\n        }\n\n        if (remaining > 0) {\n            cout << -1 << '\\n';\n            continue;\n        }\n\n        cout << operations.size() << ' ' << start << '\\n';\n        for (int operation : operations) {\n            cout << operation << ' ';\n        }\n        cout << '\\n';\n    }\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2262E - Paired Bracket Sequences](https://codeforces.com/contest/2262/problem/E \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nint MOD;\n\nvoid madd(int& a, int b) {\n    a += b;\n    if(a >= MOD) a -= MOD;\n}\n\nvoid msub(int& a, int b) {\n    a -= b;\n    if(a < 0) a += MOD;\n}\n\nint mmul(int a, int b) {\n    return 1LL * a * b % MOD;\n}\n\nint bpow(int x, int y) {\n    return y == 0 ? 1 : mmul((y % 2) ? x : 1, bpow(mmul(x, x), y / 2));\n}\n\nvoid tc() {\n    int n; cin >> n >> MOD;\n    vector<int> f(2 * n + 1), invf(2 * n + 1);\n    f[0] = 1;\n    for(int i = 1; i <= 2 * n; i++) {\n        f[i] = mmul(f[i - 1], i);\n    }\n    invf[2 * n] = bpow(f[2 * n], MOD - 2);\n    for(int i = 2 * n; i >= 1; i--) {\n        invf[i - 1] = mmul(invf[i], i);\n    }\n    auto F = [&] (int n, int k) -> int {\n        return mmul(f[n], mmul(invf[k + 1], invf[n - k]));\n    };\n    auto choose = [&] (int n, int k) -> int {\n        return mmul(f[n], mmul(invf[k], invf[n - k]));\n    };\n    vector<vector<int>> p(n + 1, vector<int>(n + 1));\n    for(int i = 0; i <= n; i++) {\n        p[0][i] = mmul(F(2 * i, i), F(2 * i, i));\n    }\n    for(int i = 1; i <= n; i++) {\n        for(int j = 0; j <= n; j++) {\n            for(int k = 0; j + k <= n; k++) {\n                madd(p[i][j + k], mmul(p[i - 1][j], p[0][k]));\n            }\n        }\n    }\n    vector<int> ans(n + 1);\n    for(int i = 0; i <= n; i++) {\n        ans[i] = mmul(F(2 * n, i), p[i][n - i]);\n    }\n    for(int i = n - 1; i >= 0; i--) {\n        for(int j = i + 1; j <= n; j++) {\n            msub(ans[i], mmul(ans[j], choose(j, i)));\n        }\n    }\n    for(int i = 0; i <= n; i++) {\n        cout << ans[i] << \" \";\n    }\n    cout << endl;\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t; \n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2262F - Rank Removal](https://codeforces.com/contest/2262/problem/F \"Codeforces Round 1120 (Div. 1)\")\n\nAuthors: [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") & [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nconst int MXN = 400;\n\ntemplate<size_t SZ> int compute_inv(const vector<bitset<SZ>>& mat, vector<bitset<SZ>>& inv) {\n    int n = mat.size();\n    vector<bool> empty(n, true);\n    vector<bitset<SZ>> basis(n);\n\n    for(int i = 0; i < n; i++) {\n        bitset<SZ> cur = mat[i];\n        bitset<SZ> curinv;\n        curinv[i] = 1;\n        for(int j = 0; j < n; j++) {\n            if(cur[j]) {\n                if(empty[j]) {\n                    basis[j] = cur;\n                    inv[j] = curinv;\n                    empty[j] = false;\n                    break;\n                } else {\n                    cur ^= basis[j];\n                    curinv ^= inv[j];\n                }\n            }\n        }\n    }\n    for(int i = 0; i < n; i++) {\n        if(empty[i]) return 0;\n    }\n\n    for(int i = n - 1; i >= 1; i--) {\n        for(int j = 0; j < i; j++) {\n            if(basis[j][i]) {\n                basis[j] ^= basis[i];\n                inv[j] ^= inv[i];\n            }\n        }\n    }\n    return 1;\n}\n\nbool try_removal(vector<pair<int, int>> cells, vector<bitset<MXN>>& mat, vector<bitset<MXN>>& inv, vector<int>& row_cnt) {\n    int n = mat.size(), k = cells.size();\n    for(auto [x, y] : cells) {\n        if(mat[x][y] == 0) {\n            return false;\n        }\n    }\n    assert(k < 4);\n    vector<bitset<3>> K(k); \n    for(int i = 0; i < k; i++) {\n        for(int j = 0; j < k; j++) {\n            K[i][j] = (i == j) ^ inv[cells[i].second][cells[j].first];\n        }\n    }\n    vector<bitset<3>> invK(cells.size());\n    if(!compute_inv(K, invK)) {\n        return false;\n    }\n    for(auto [x, y] : cells) {\n        mat[x][y] = 0;\n        row_cnt[x]--;\n    }\n\n    auto old = inv;\n    for(int i = 0; i < n; i++) {\n        bitset<3> coef;\n        for(int alpha = 0; alpha < k; alpha++) {\n            int r = cells[alpha].first;\n\n            if(old[i][r]) {\n                coef ^= invK[alpha];\n            }\n        }\n\n        // delta = coef * (V^T * old)\n        for(int beta = 0; beta < k; beta++) {\n            if(coef[beta]) {\n                int c = cells[beta].second;\n                inv[i] ^= old[c];\n            }\n        }\n    }\n    return true;\n}\n\nvector<int> compute_matching(const vector<bitset<MXN>>& mat) {\n    int n = mat.size();\n    vector<vector<int>> adj(n);\n    for(int i = 0; i < n; i++) {\n        for(int j = 0; j < n; j++) {\n            if(mat[i][j]) adj[i].push_back(j);\n        }\n    }\n\n    vector<int> match(n, -1);\n    vector<bool> done(n);\n\n    function<bool(int)> dfs = [&] (int s) {\n        for(int u : adj[s]) {\n            if(!done[u]) {\n                done[u] = true;\n\n                if(match[u] == -1 || dfs(match[u])) {\n                    match[u] = s;\n                    return true;\n                }\n            }\n        }\n        return false;\n    };\n\n    for(int i = 0; i < n; i++) {\n        fill(done.begin(), done.end(), false);\n        dfs(i);\n    }\n\n    for(int i = 0; i < n; i++) {\n        assert(match[i] != -1);\n    }\n    return match;\n}\n\nvoid tc() {\n    int n, m; cin >> n >> m;\n\n    vector<bitset<MXN>> mat(n);\n    vector<int> row_cnt(n);\n    for(int i = 0; i < m; i++) {\n        int x, y; cin >> x >> y;\n        x--, y--;\n        mat[x][y] = 1;\n        row_cnt[x]++;\n    }\n\n    vector<bitset<MXN>> inv(n);\n    compute_inv(mat, inv);\n\n    int sm = m;\n    int left = (sm + n - 1) / n;\n    cout << left << '\\n';\n\n    while(sm > 0) {\n        vector<bitset<MXN>> omat = mat;\n        if(sm <= n) {\n            for(int i = 0; i < n; i++) {\n                for(int j = 0; j < n; j++) {\n                    mat[i][j] = 0;\n                }\n            }\n            row_cnt.assign(n, 0);\n        } else if(sm <= 2 * n) {\n            vector<int> matching = compute_matching(mat);\n            for(int i = 0; i < n; i++) {\n                for(int j = 0; j < n; j++) {\n                    mat[i][j] = 0;\n                }\n            }\n            row_cnt.assign(n, 0);\n            for(int i = 0; i < sm - n; i++) {\n                mat[matching[i]][i] = 1;\n                row_cnt[matching[i]]++;\n            }\n        } else if(sm <= 3 * n) {\n            vector<int> matching = compute_matching(mat);\n            int above = 0;\n            for(int i = 0; i < n; i++) {\n                for(int j = matching[i] + 1; j < n; j++) {\n                    above += mat[j][i];\n                }\n            }\n\n            vector<bitset<MXN>> nmat(n);\n            vector<int> nrow_cnt(n);\n            int ones_left = sm - n;\n            for(int i = 0; i < n; i++) {\n                nmat[matching[i]][i] = 1;\n                nrow_cnt[matching[i]]++;\n                ones_left--;\n            }\n            // make matrix either upper or lower triangular, whichever one\n            // removes a sufficient number of ones\n            if(above >= ones_left) {\n                for(int i = 0; i < n; i++) {\n                    for(int j = matching[i] + 1; j < n; j++) {\n                        if(mat[j][i]) {\n                            ones_left--;\n                            nmat[j][i] = 1;\n                            nrow_cnt[j]++;\n                            if(ones_left == 0) break;\n                        }\n                    }\n                    if(ones_left == 0) break;\n                }\n            } else {\n                for(int i = 0; i < n; i++) {\n                    for(int j = 0; j < matching[i]; j++) {\n                        if(mat[j][i]) {\n                            ones_left--;\n                            nmat[j][i] = 1;\n                            nrow_cnt[j]++;\n                            if(ones_left == 0) break;\n                        }\n                    }\n                    if(ones_left == 0) break;\n                }\n            }\n            mat = nmat;\n            row_cnt = nrow_cnt;\n        } else {\n            int csm = sm;\n            if(n % 2) {\n                // odd case (remove either 1 or 3)\n                int dup_row = -1;\n                for(int i = 0; i < n; i++) {\n                    if(row_cnt[i] >= 2) {\n                        dup_row = i;\n                        break;\n                    }\n                }\n                assert(dup_row != -1);\n\n                vector<int> cols;\n                for(int i = 0; i < n; i++) {\n                    if(mat[dup_row][i]) cols.push_back(i);\n                }\n\n                assert(cols.size() >= 2);\n\n                for(int i = 0; i < n; i++) {\n                    if(try_removal({{i, cols[0]}}, mat, inv, row_cnt)) {\n                        csm--;\n                        break;\n                    }\n                    if(try_removal({{i, cols[1]}}, mat, inv, row_cnt)) {\n                        csm--;\n                        break;\n                    }\n                }\n\n                if(csm == sm) {\n                    for(int i = 0; i < n; i++) {\n                        if(i != dup_row) {\n                            // remove L shape\n                            if(try_removal(\n                                    {{i, cols[0]}, \n                                     {dup_row, cols[0]}, \n                                     {dup_row, cols[1]}},\n                                mat, inv, row_cnt)) {\n                                csm -= 3;\n                                break;\n                            }\n                        }\n                    }\n                    assert(sm != csm);\n                }\n            }\n\n            while(csm > sm - n) {\n                // remove 2 at a time\n                int dup_row = -1;\n                for(int i = 0; i < n; i++) {\n                    if(row_cnt[i] >= 3) {\n                        dup_row = i;\n                        break;\n                    }\n                }\n\n                vector<int> cols;\n                for(int i = 0; i < n; i++) {\n                    if(mat[dup_row][i]) cols.push_back(i);\n                }\n\n                assert(cols.size() >= 3);\n\n                if(!try_removal({{dup_row, cols[0]}, {dup_row, cols[1]}}, mat, inv, row_cnt)) {\n                    if(!try_removal({{dup_row, cols[1]}, {dup_row, cols[2]}}, mat, inv, row_cnt)) {\n                        if(!try_removal({{dup_row, cols[0]}, {dup_row, cols[2]}}, mat, inv, row_cnt)) {\n                            assert(false);\n                        }\n                    }\n                }\n\n                csm -= 2;\n            }\n            assert(csm == sm - n);\n        }\n\n        vector<pair<int, int>> move;\n        for(int i = 0; i < n; i++) {\n            for(int j = 0; j < n; j++) {\n                if(omat[i][j] == 1 && mat[i][j] == 0) {\n                    move.push_back({i, j});\n                }\n            }\n        }\n        cout << move.size() << \"\\n\";\n        for(auto [x, y] : move) cout << x + 1 << \" \" << y + 1 << \"\\n\";\n        sm -= n;\n\n        int csm = 0;\n        for(int i = 0; i < n; i++) {\n            csm += row_cnt[i];\n        }\n        assert(csm == max(0, sm));\n    }\n}\n\nint main() {\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Tutorial 2**\n\nCredit to [BurnedChicken](https://codeforces.com/profile/BurnedChicken \"International Grandmaster BurnedChicken\") for finding this in testing!\n\nWe will first prove that for any row/column of a rank $$$n$$$ matrix, we can zero out everything in that row/column except one, and maintain rank $$$n$$$. This is because rank $$$n$$$ means the determinant is $$$1$$$, so there are exactly an odd number of perfect matchings in total (read Solution 1 for more understanding). Therefore at least $$$1$$$ of the ones in the row/column must be in an odd number of perfect matchings (since if they were all even then the total number would be even). This means if we delete all other ones but that one, then we would keep rank $$$n$$$.\n\nThis actually all you need to prove to solve the problem. Note that after we clear out one row except one number, we can clear out that column except that one number (even partially) and the determinant will not change (and the same works when clearing out a column with the leftover row). \n\nFor every turn, we want to try to find $$$n$$$ things to remove without changing the rank. Therefore, we begin by finding the one in the biggest row/column we can keep while removing everything else. Then we remove everything in the column/row of that one, and the everything in the row/column into the queue. If the queue + the current removals is big enough to reach $$$n$$$ removals, then we can just remove the ones in the queue. If the queue is too small, then we repeat one the next biggest row/column. Because the queue contains the row/column with the most ones, when the current removals will never exceed $$$n$$$ if the the queue + current removals is less than $$$n$$$. Once we run out of row/columns to clear, we can start removing the ones we excluded removing, since we have to reduce the rank at this point.\n\nTo find the one we must keep, you have to utilize the inverse, since we can clear out the row/column on $$$M\\_{i, j}$$$ if $$$M^{-1}\\_{j, i} = 1$$$ (proof is by Cramer's Rule). We can keep track of the inverse after this clear out operation in $$$O(n^2 / w)$$$ if done properly (using something similar to Woodbury Matrix Identity, look at Solution 1 for details). If implemented properly, this results in a $$$O(n^3 / w)$$$ solution.\n\n**Solution 2**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n \nconst int SZ = 400;\n \n \ntemplate<size_t SZ> int compute_inv(const vector<bitset<SZ>>& mat, vector<bitset<SZ>>& inv) {\n    int n = mat.size();\n    vector<bool> empty(n, true);\n    vector<bitset<SZ>> basis(n);\n \n    for(int i = 0; i < n; i++) {\n        bitset<SZ> cur = mat[i];\n        bitset<SZ> curinv;\n        curinv[i] = 1;\n        for(int j = 0; j < n; j++) {\n            if(cur[j]) {\n                if(empty[j]) {\n                    basis[j] = cur;\n                    inv[j] = curinv;\n                    empty[j] = false;\n                    break;\n                } else {\n                    cur ^= basis[j];\n                    curinv ^= inv[j];\n                }\n            }\n        }\n    }\n    for(int i = 0; i < n; i++) {\n        if(empty[i]) return 0;\n    }\n \n    for(int i = n - 1; i >= 1; i--) {\n        for(int j = 0; j < i; j++) {\n            if(basis[j][i]) {\n                basis[j] ^= basis[i];\n                inv[j] ^= inv[i];\n            }\n        }\n    }\n    return 1;\n}\n \nvoid tc() {\n    int n, m; cin >> n >> m;\n    vector<bitset<SZ>> mat(n);\n    for(int i = 0; i < m; i++) {\n        int x, y; cin >> x >> y;\n        x--, y--;\n        mat[x][y] = 1;\n    }\n \n    vector<bitset<SZ>> inv(n);\n    compute_inv(mat, inv);\n    vector<int> cnts(2 * n);\n    for(int i = 0; i < n; i++) {\n        for(int j = 0; j < n; j++) {\n            cnts[i] += mat[i][j];\n            cnts[n + j] += mat[i][j];\n        }\n    }\n    queue<pair<int, int>> mustq, freeq;\n    vector<pair<int, int>> centers;\n    vector<vector<pair<int, int>>> moves;\n    while(true) {\n        while(mustq.size() + freeq.size() < n) {\n            int best_rc = 0;\n            for(int rc = 0; rc < 2 * n; rc++) {\n                if(cnts[rc] >= cnts[best_rc]) best_rc = rc;\n            }\n            if(cnts[best_rc] == 0) {\n                break;\n            }\n \n            int x = -1, y = -1;\n            if(best_rc < n) {\n                x = best_rc;\n                for(int j = 0; j < n; j++) {\n                    if(mat[x][j] && inv[j][x]) {\n                        y = j;\n                        break;\n                    }\n                }\n            } else {\n                y = best_rc - n;\n                for(int j = 0; j < n; j++) {\n                    if(mat[j][y] && inv[y][j]) {\n                        x = j;\n                        break;\n                    }\n                }\n            }\n            assert(x != -1 && y != -1);\n \n            centers.push_back({x, y});\n            vector<pair<int, int>> must, free;\n            auto erase = [&] (int a, int b) {\n                mat[a][b] = 0;\n                cnts[a]--;\n                cnts[b + n]--;\n            };\n            for(int j = 0; j < n; j++) {\n                if(mat[x][j] && j != y) {\n                    must.push_back({x, j});\n                    erase(x, j);\n                }\n            }\n            for(int j = 0; j < n; j++) {\n                if(mat[j][y] && j != x) {\n                    free.push_back({j, y});\n                    erase(j, y);\n                }\n            }\n            erase(x, y);\n            assert(cnts[x] == 0 && cnts[n + y] == 0);\n \n            if(best_rc < n) swap(must, free);\n            for(auto u : must) mustq.push(u);\n            for(auto u : free) freeq.push(u);\n \n            bitset<SZ> row = inv[y];\n            for(int j = 0; j < n; j++) {\n                if(cnts[j + n] > 0 && inv[j][x]) {\n                    inv[j] ^= row;\n                }\n            }\n        }\n        int left = n;\n        vector<pair<int, int>> move;\n        while(left > 0 && mustq.size() > 0) {\n            left--;\n            move.push_back(mustq.front());\n            mustq.pop();\n        }\n        while(left > 0 && freeq.size() > 0) {\n            left--;\n            move.push_back(freeq.front());\n            freeq.pop();\n        }\n        while(left > 0 && centers.size() > 0) {\n            left--;\n            move.push_back(centers.back());\n            centers.pop_back();\n        }\n        moves.push_back(move);\n        if(centers.size() == 0) break;\n    }\n    cout << moves.size() << \"\\n\";\n    for(auto move : moves) {\n        cout << move.size() << \"\\n\";\n        for(auto [u, v] : move) cout << u + 1 << \" \" << v + 1 << \"\\n\";\n    }\n}\n \nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n\n...108135 bytes truncated...\n\nThe tool call succeeded but the output was truncated. Full output saved to: /Users/zayd/.local/share/opencode/tool-output/tool_0acbe83fa001KnTRsS4K4QbMbk\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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            "output": "Title: Codeforces Round 1120 (Div 1, Div 2) Editorial\nURL: https://codeforces.com/\nPublished: N/A\nAuthor: N/A\nHighlights:\nIt has been shown that every problem has a solution. The proofs are now available in the editorial.\n...\nUPD2 Editorial\n\n---\n\nTitle: CF2262A1 Floor of MEX (Easy Version)\nURL: https://www.luogu.com.cn/\nPublished: N/A\nAuthor: N/A\nHighlights:\nRemoteJudge 服务中断情况公告 洛谷 In\n...\n站务版 @ 04-19 15:12 5578回复 置顶 \n\n 奖项认证规则更新 chen_zhe Aya In 站务版 @ 09-15 16:31 130回复 置顶 \n\n 洛谷算法竞赛秋令营 - 基础组/提… kkksc03 吉祥物 In 站务版 @ 09-08 16:40 154回复 置顶 \n\n 自动封禁 AI 已正式上线 kkksc04 管理员 In 站务版 @ 07-28 16:53 1673回复 置顶 \n\n 新版难度体系更新下的临时难度定… chen_zhe Aya In 站务版 @ 07-02 18:29 724回复 置顶 \n\n 玄关求容斥好题 watermouthhang In 学术版 @ 09-15 19:01 2回复 \n\n 我有个疑问 zly20150126 In P17455 [GESP20260… @ 09-15 19:00 1回复 \n\n 求助 CSP 初赛。 clx201022 In 学术版 @ 09-15 19:00 8回复 \n\n 求问 Can_I_do_OI In 学术版 @ 09-15 18:59 1回复 \n\n 建议降黄 cxr2023 In P14467 [COCI 2025… @ 09-15 18:58 0回复 \n\n 关于初赛 Kyousuk In 学术版 @ 09-15 18:57 13回复 \n\n 2025CSP初赛试题在哪 dH_Hb In 学术版 @ 09-15 18:55 2回复 \n\n 求助 CSP 初赛 诗久桜月 In 学术版 @ 09-15 18:54 2回复 \n\n 关于初赛 chenxumin1017 In 学术版 @ 09-15 18:53 13回复 \n\n 问 JQH_jiqihang In 学术版 @ 09-15 18:52 3回复 \n\n 求助csp初赛 PumpPump In 学术版 @ 09-15 18:51 8回复 \n\n 求比赛 JQH_jiqihang In 学术版 @ 09-15 18:50 6回复 \n\n 求问CCF少爷机 formu1 In 学术版 @ 09-15 18:50 1回复\n\n---\n\nTitle: Codeforces #1116 : Div1. A2. Floor of MEX (Hard Version) - kmjp's blog\nURL: https://kmjp.hatenablog.jp/entry/2026/09/14/1030\nPublished: 2026-09-14T01:30:00.000Z\nAuthor: kmjp\nHighlights:\nCodeforces #1116 : Div1. A2. Floor of MEX (Hard Version) - kmjp's blog\n\n 2026- 09- 14 \n\n# Codeforces #1116 : Div1. A2. Floor of MEX (Hard Version) \n\nうーん、前半から手間取った。 https://codeforces.com/contest/2262/problem/A2\n...\nf(S,x)とは、Sの各要素をxで割った集合に対するmex値とする。\n...\n整数Nと、N要素の整数列Aが与えられる。 0～(N-1)の部分集合Bのうち、f(B,x) = A[x]となるようなBは何通りか。\n...\n### 解法\n...\nA[x]=dの場合、B中にx*d～x*(d+1)-1の範囲の値は含まれないことがわかる。 そこで、まずBに現れない値を確定させよう。\n...\n次に小さい順にBに何の値を入れるかを考える。 dp(n,m) := Bにnまで入れるかどうか考えたとき、Aを満たす範囲で最大値がmである\n...\n次にn+1を加えるか加えないかを考える。 まずBに現れないこと確定の場合は、そもそも加えない確定。 そうでない場合、n+1を加えないとAに違反するようなmに対してはdp(n+1,m)=0とする。 具体的にはn+1が、dで割ると(d-1)余る場合、すなわちn+1=kd-1の形の場合、(k-1)d～(kd-1)のうち１個以上値が無いと、A[d]はk+1より大きくできない。 このような場合、m＜(k-1)dとなるmに対しdp(n+1,m)=0にする。\n...\nvoid solve() {\n\tint i,j,k,l,r,x,y; string s;\n\t\n\tfor(i=1;i<=100002;i++) {\n\t\tfor(j=i;j<=100002;j+=i) D[j-1].push_back(i);\n\t}\n\t\n\tcin>>T;\n\twhile(T--) {\n\t\tcin>>N;\n\t\tFOR(i,N+10) S[i]=0;\n\t\tll ma=-1;\n\t\tfor(i=1;i<=N;i++) {\n\t\t\tcin>>A[i];\n\t\t\tS[(A[i])*i]++;\n\t\t\tS[(A[i]+1)*i]--;\n\t\t\tma=max(ma,1LL*(A[i]-1)*i);\n\t\t}\n\t\t\n\t\tmap<int,ll> dp;\n\t\tFOR(i,N) S[i+1]+=S[i];\n\t\t\n\t\tfor(i=1;i<=N;i++) if(A[i]!=0) break;\n\t\t\n\t\tif(i==N+1) {\n\t\t\tcout<<1<<endl;\n\t\t\tcontinue;\n\t\t}\n\t\tdp[i-1]=1;\n\t\tll sum=1;\n\t\tfor(;i<=N;i++) {\n\t\t\tif(i<N&&S[i]==0) {\n\t\t\t\tdp[i]=sum;\n\t\t\t\tsum=sum*2%mo;\n\t\t\t}\n\t\t\tint mi=-1;\n\t\t\tFORR(d,D[i]) {\n\t\t\t\tif(1LL*A[d]*d>i) {\n\t\t\t\t\tmi=max(mi,i-d);\n\t\t\t\t}\n\t\t\t}\n\t\t\t\n\t\t\twhile(dp.size()&&dp.begin()->first<=mi) {\n\t\t\t\t(sum+=mo-dp.begin()->second)%=mo;\n\t\t\t\tdp.erase(dp.begin());\n\t\t\t}\n\t\t}\n\t\t\n\t\tll ret=0;\n\t\tFORR2(a,b,dp) if(a>=ma) ret+=b;\n\t\t\n\t\t\n\t\tcout<<ret%mo<<endl;\n\t\t\n\t\t\n\t}\n}\n\n---\n\nTitle: Codeforces Round 1054 (Div. 3) Editorial - Codeforces\nURL: https://codeforces.com/blog/entry/146793\nPublished: N/A\nAuthor: N/A\nHighlights:\nRound 105\n...\n(Div. 3) Editorial\n...\n2149C - MEX rose\n...\nLet $$$Z$$$ be the number of integers in $$$[0, k-1]$$$ that are missing from the array, and let $$$C$$$ be the number of elements equal to $$$k$$$.\n...\nTo make $$$\\text{MEX}(a) = k$$$, every number from $$$0$$$ to $$$k-1$$$ must appear, which costs $$$Z$$$ operations. Also, $$$k$$$ must not appear in the array, so we need $$$C$$$ operations to remove them. While removing copies of $$$k$$$, we can directly replace them with missing numbers, so the total number of operations is the maximum of these two values.\n...\nThus, the answer is: $$$\\text{ans} = \\max(Z, C)$$$.\n...\nThe time complexity is $$$O(n)$$$ per test case.\n\n---\n\nTitle: Hello 2026 Editorial - Codeforces\nURL: https://codeforces.com/blog/entry/149944\nPublished: N/A\nAuthor: N/A\nHighlights:\n2183B — Yet Another MEX Problem\n\nIdea & Preparation: wangmarui\n\nRate The Problem!\n\n- Good Problem\n- Okay Problem\n- Bad Problem\n- Didn't Solve\n\nHint 1\n\nHint 2\n\nSolution\n\nSince only $$$k - 1$$$ numbers will remain in the end, it is relatively easy to notice that for any interval of length $$$k$$$, numbers within this interval that are greater than or equal to $$$k-1$$$ or have duplicate values are invalid. One of these two types of numbers can be removed. Since there are only $$$k - 1$$$ numbers from $$$0$$$ to $$$k-2$$$, in any interval of length $$$k$$$, there must be at least one number that can be deleted. Therefore, the answer is $$$\\min(\\text{mex}(a), k-1)$$$.\n\nTime complexity: $$$O(\\sum n)$$$.\n\nImplementation\n\n```\n#include<bits/stdc++.h>\nusing namespace std;\n#define re register\n#define ll long long\n#define forl(i,a,b) for(re ll (i)=(a);i<=(b);(i)++)\n#define forr(i,a,b) for(re ll (i)=(a);i>=(b);(i)--)\n#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);\n#define endl '\\n'\nll _t_;\nll n,m;\nll maxn;\nll a[1000010];\nll b[1000010];\nvoid _clear(){}\nvoid solve()\n{\n    _clear();\n    cin>>n>>m;\n    forl(i,1,n)\n        cin>>a[i],a[i]=min(a[i],n+1);\n    forl(i,0,n+5)\n    \tb[i]=0;\n    ll mex=0;\n\tforl(i,1,n)\n\t\tb[a[i]]++;\n\twhile(b[mex])\n\t\tmex++;\n\tcout<<min(mex,m-1)<<endl; \n}\nint main()\n{\n    IOS;\n    _t_=1;\n    cin>>_t_;\n    while(_t_--)\n        solve();\n    return 0;\n}\n\n```\n\n## 2183C — War Strategy\n\nIdea & Preparation: Network_Error\n\nRate The Problem!\n\n- Good Problem\n- Okay Problem\n- Bad Problem\n- Didn't Solve\n\nHint 1\n\nHint 2\n\nHint 3\n\nClaim 1\n\nProof\n\nIf the finally occupied bases do not form an interval containing $$$k$$$, then there must exist some $$$x, y$$$ such that $$$1 \\le x \\lt y \\lt k$$$ or $$$k \\lt y \\lt x \\le n$$$, with $$$x$$$ occupied and $$$y$$$ unoccupied. Consider a soldier at $$$y$$$ starting from $$$k$$$ and moving to $$$x$$$; it must pass through base $$$x$$$ along the way. Consider making this soldier stop exactly at $$$x$$$. Then soldier $$$y$$$ ends up at $$$x$$$. After a finite number of such adjustments, we will reach a state where no further adjustment is possible, which implies the finally occupied bases must be an interval containing $$$k$$$.\n\nClaim 2\n\nSuppose this interval contains $$$a$$$ positions to the left of $$$k$$$ and $$$b$$$ positions to the right of $$$k$$$ (excluding $$$k$$$ itself) ($$$0 \\le a \\le k-1$$$, $$$0 \\le b \\le n-k$$$). It is possible to occupy this interval if and only if $$$a + b + \\max(a, b) - 1 \\le m$$$.\n\nSufficiency:\n\nWe can imagine a simple strategy. Initially, we do nothing until the capital contains $$$a$$$ soldiers. Then we move $$$a$$$ soldiers from $$$k$$$ to $$$k-1$$$, then move $$$a-1$$$ soldiers from $$$k-1$$$ to $$$k-2$$$, and so on. Thus $$$k-1, k-2, \\ldots, k-a$$$ are all filled with soldiers. Then we wait (if necessary) until the capital contains $$$b$$$ soldiers and move them in the same way to $$$k+1, \\ldots, k+b$$$. It can be observed that this strategy requires exactly $$$a + b + \\max(a, b) - 1$$$ days.\n\nNecessity:\n\nWe prove that for any strategy, the required number of days $$$m$$$ is at least $$$a + b + \\max(a, b) - 1$$$.\n\nFirst, we need at least $$$a+b$$$ days to obtain the extra $$$a+b$$$ soldiers. Second, we cannot move new soldiers to new bases every moment. In fact, we can prove that there are at least $$$\\max(a, b) - 1$$$ idle days. The proof is as follows.\n\nAssume without loss of generality that $$$a \\ge b$$$.\n\nDefine the potential function $$$h(S)$$$ of a state $$$S$$$ as the number of soldiers minus the number of occupied bases.\n\nSuppose the first move to the left is moving $$$t$$$ soldiers to the leftmost consecutive $$$t$$$ positions $$$k-a, \\ldots, k-a+t-1$$$.\n\n(If the move is not to the leftmost positions but to some intermediate positions, we can show through adjustment that such a move is suboptimal. Therefore, an optimal solution will not contain such moves.)\n\nLet $$$S_0$$$ be the state before the move, $$$S_1$$$ be the state when moving $$$t$$$ soldiers to $$$k-a+t$$$, $$$S_2$$$ be the state after completing the move of all $$$t$$$ soldiers, and $$$S_3$$$ be the final state.\n\nWe have $$$h(S_2) = h(S_1)$$$, $$$h(S_3) \\ge h(S_2)$$$.\n\n$$$h(S_1) - h(S_0) \\ge a - t$$$ (during the intermediate $$$a-t$$$ steps, the potential increases each time).\n\n$$$h(S_0) \\ge t - 1$$$ (at least $$$t$$$ movable soldiers are needed).\n\nSumming up, we get $$$h(S_3) \\ge a - 1$$$. Therefore, the number of days needed is at least $$$2a + b - 1$$$.\n\nIn summary, at least $$$a + b + \\max(a, b) - 1$$$ days are required.\n\nWe consider a greedy approach. First, if $$$k-1 \\lt n-k$$$, we replace $$$k$$$ with $$$n+1-k$$$.\n\nWe define two variables $$$a$$$ and $$$b$$$, representing how many cells have been extended to the left and right, respectively. Initially, $$$a = 0$$$, $$$b = 0$$$. Then we repeatedly try to increase $$$a$$$ and $$$b$$$ by $$$1$$$ in turns. We first try to increase $$$b$$$ by $$$1$$$ (if $$$b \\lt n-k$$$ and $$$a + (b+1) + \\max(a, b+1) - 1 \\le m$$$), then set $$$b$$$ to $$$b+1$$$. Next, we try to increase $$$a$$$ by $$$1$$$ (if $$$a \\lt k-1$$$ and $$$(a+1) + b + \\max(a+1, b) - 1 \\le m$$$). If we cannot increase $$$a$$$, we break the loop. Finally, $$$a + b + 1$$$ (don't forget the home base) is the answer. The time complexity is $$$O(n)$$$.\n\nImplementation\n\n```\n#include<bits/stdc++.h>\nusing namespace std;\nint n,k,m;\nsigned main(){\n    int T;\n    cin>>T;\n    while(T--){\n        cin>>n>>m>>k;\n        if(k-1<n-k)k=n+1-k;\n        int a=0,b=0;\n        while(1){\n            if(b<n-k&&a+(b+1)+max(a,b+1)-1<=m)++b;\n            if(a<k-1&&(a+1)+b+max(a+1,b)-1<=m)++a;\n            else break;\n        }\n        cout<<a+b+1<<endl;\n    }\n}\n\n```\n\n## 2183D1 — Tree Coloring (Easy Version)\n\nIdea & Preparation: jiazhichen844\n\n- Good Problem\n- Okay Problem\n- Bad Problem\n- Didn't Solve\n\nSuppose the number of points at depth $$$i$$$ is $$$t_i$$$. First, points at the same depth cannot be colored simultaneously, so the lower bound for the answer is $$$\\max t_i$$$.\n\nHowever, note that if points in layer $$$x$$$ share a common parent, they and their parent also cannot be colored simultaneously. Thus, the answer must also be compared with $$$\\max(t_x + 1)$$$.\n\nWe claim this is the answer. Proof follows from the constructive proof in D2.\n\n## 2183D2 — Tree Coloring (Hard Version)\n\nIdea & Preparation: jiazhichen844\n\n- Good Problem\n- Okay Problem\n- Bad Problem\n- Didn't Solve\n\nSolution 1\n\nWe claim the maximum number of operations is $$$\\max t_i + 1$$$. Consider the following constructive proof.\n\nLet $$$S_i$$$ denote the set of points for the $$$i$$$ th operation.\n\nWithout loss of generality, let $$$T=\\max t_i$$$. Find the largest $$$x$$$ such that $$$t_x=T$$$. Clearly, points with $$$d_i=x$$$ cannot be operated on simultaneously; we can place them into $$$S_1, S_2, \\cdots, S_T$$$ respectively.\n\nThen we only need to consider placing points of depth $$$\\neq x$$$ into $$$S$$$. First, consider points of depth $$$ \\gt x$$$. Suppose we have already placed all points of depth $$$k-1$$$. Now, when placing points of depth $$$k$$$, it is clear that each $$$S_i$$$ can contain at most one point of depth $$$k-1$$$.\n\nNow, $$$t_{k-1}$$$ sets contain points of depth $$$k-1$$$, which impose constraints on the depth $$$k$$$ points to be placed; The remaining $$$T-t_{k-1}$$$ sets contain no points of depth $$$k-1$$$ and can freely add any point of depth $$$k$$$ (see the diagram below, where the upper red dots represent sets containing points of depth $$$k-1$$$, and the blue dots represent sets without such points).\n\n \n\nFor restricted sets (where children cannot be placed), we can retain only one child and place the others into unrestricted sets. Since the number of points in the lower layer is strictly less than $$$T$$$, at least one unrestricted set will remain. At this point, we can place each retained child into the set to its right.\n\nNext, we construct the point of depth $$$ \\lt x$$$. Suppose we have already constructed the",
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          "text": "The core construction is unusually clean: mark every value that would make one of the observed MEX values present, then keep every unmarked value. I’m checking the exact proof against the supplied existence guarantee and the easy-version constraints before writing the learner-facing content.",
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              "description": "Extract editorial proof",
              "prompt": "Inspect the fetched Codeforces editorial output saved at /Users/zayd/.local/share/opencode/tool-output/tool_0acbe83fa001KnTRsS4K4QbMbk. Extract only the relevant explanation/code for 2262A1 or its equivalent 2263C1 Floor of MEX Easy Version. Summarize the construction, the proof that all un-forbidden values work given a solution exists, and any implementation caveats. Do not edit files and do not solve unrelated problems. Return a concise research summary.",
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            "output": "<task id=\"ses_f534124b8ffe0ojX16NCY60QGk\" state=\"completed\">\n<task_result>\n## Research summary\n\nSource: `/Users/zayd/.local/share/opencode/tool-output/tool_0acbe83fa001KnTRsS4K4QbMbk` (C1 section, lines 235–285). The fetched editorial has no prose—the tutorial is still “loading”—so the proof is reconstructed from the code and comments.\n\n### Construction\n\nFor each `k`, a candidate `y` produces `floor(y/k) = a[k]` exactly when:\n\n\\[\na[k]k \\le y \\le (a[k]+1)k-1.\n\\]\n\nSince `a[k]` is the MEX, this value must be absent. Mark that interval forbidden for every `k`.\n\nThen choose:\n\n\\[\nB=\\{0,\\ldots,n-1\\}\\setminus\\bigcup_k [a[k]k,(a[k]+1)k-1].\n\\]\n\nThe code marks the union with a difference array and outputs every uncovered value.\n\n### Why taking all un-forbidden values works\n\nLet `A` be the promised existing solution.\n\n- `A` cannot contain any forbidden value, so `A ⊆ B`.\n- For every `q < a[k]`, the fact that `a[k]` is the MEX means some `y ∈ A` satisfies `floor(y/k)=q`.\n- That `y` is not forbidden, hence `y ∈ B`.\n\nTherefore `B` contains every floor value `0,1,…,a[k]-1`, while it contains no value producing `a[k]`. Values larger than `a[k]` do not matter, so `f(B,k)=a[k]`.\n\n### Implementation caveats\n\n- Clamp intervals to candidates `[0,n-1]`; `n` is the exclusive difference-array endpoint.\n- Use `long long` for `a[k]*k` and `(a[k]+1)*k-1`. The pasted code uses `int`, which is safe only because the problem guarantees a solution.\n- The pasted C1 code builds `goodivs` but never uses it. Remove that loop: the clean solution is `O(n)` time and space. As written, it costs `O(n + Σa[k])`; existence bounds this by `O(n log n)`.\n- No impossibility check is needed under the problem’s guarantee. Empty `B` is valid; still print the required blank second line.\n</task_result>\n</task>",
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            "title": "Extract editorial proof",
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            {
              "order": 1,
              "content": "For a fixed divisor $k$, the MEX condition says that the quotient value $a_k$ itself must be absent. Work backward: which values of $y$ satisfy $\\left\\lfloor y/k\\right\\rfloor=a_k$?"
            },
            {
              "order": 2,
              "content": "Those $y$ form one interval: $a_k k\\le y<(a_k+1)k$. Every value in this interval is forbidden, because including even one of them would make the MEX at $k$ different."
            },
            {
              "order": 3,
              "content": "Take $B$ to contain every number from $0$ to $n-1$ that is not forbidden by any $k$. This automatically keeps every observed $a_k$ absent. The remaining question is why adding all these survivors does not lose any smaller quotient."
            },
            {
              "order": 4,
              "content": "Use the promise that some valid hidden set exists; call it $A$. It cannot contain a forbidden value, so $A\\subseteq B$. Thus, for every $q<a_k$, the witness value from $A$ with $\\lfloor y/k\\rfloor=q$ is still in $B$, while $a_k$ remains absent."
            },
            {
              "order": 5,
              "content": "There are $n$ forbidden intervals, so mark their union with a difference array. Treat $[a_k k,(a_k+1)k)$ as half-open, clip it to $[0,n)$, and use `long long` for the products."
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          "editorial": "Do not try to reconstruct the hidden set exactly. The promise that a solution exists lets us build a canonical, possibly larger set.\n\n**Key observation**\n\nFor a fixed $k$, a value $y$ produces the quotient $a_k$ exactly when\n$$\n\\left\\lfloor\\frac{y}{k}\\right\\rfloor=a_k\n\\quad\\Longleftrightarrow\\quad\n a_k k\\le y<(a_k+1)k.\n$$\nTherefore every integer in the interval\n$$\n[a_k k,(a_k+1)k)\n$$\nmust be absent from any valid set. Otherwise $a_k$ would appear among the quotients, so its MEX could not be $a_k$.\n\nCall the union of these intervals forbidden, and construct\n$$\nB=\\{0,1,\\ldots,n-1\\}\\setminus\\text{(forbidden values)}.\n$$\n\n**Why this works**\n\nLet $A$ be the hidden valid set whose existence is guaranteed.\n\nFirst, $A$ contains no forbidden value. If $y\\in A$ belonged to the interval for some $k$, then $\\lfloor y/k\\rfloor=a_k$ would occur, contradicting $f(A,k)=a_k$. Hence $A\\subseteq B$.\n\nFix any $k$.\n\n- No element of $B$ produces quotient $a_k$, because all such elements were marked forbidden.\n- Since $f(A,k)=a_k$, every value $q=0,1,\\ldots,a_k-1$ occurs as $\\lfloor y/k\\rfloor$ for some $y\\in A$. Because $A\\subseteq B$, all those quotient values also occur for $B$.\n\nThus $B$ contains every quotient below $a_k$ and does not contain quotient $a_k$. Its MEX is exactly $a_k$.\n\nThe important trick is that adding survivors cannot hurt: the only dangerous quotient, $a_k$, has been explicitly excluded. Trying to guess the original set is unnecessary busywork.\n\n**Implementation**\n\nFor each $k$, use the half-open forbidden interval $[L,R)$ with\n$$\nL=a_k k,\\qquad R=(a_k+1)k.\n$$\nOnly values in $[0,n-1]$ matter. If $L<n$, add $+1$ at `L` and $-1$ at `min(n, R)` in a difference array. A prefix sum tells whether each value is covered by at least one forbidden interval.\n\nUse 64-bit arithmetic for $L$ and $R$. The input guarantee makes the construction valid, so no impossibility case needs to be handled. If every value is forbidden, output an empty set followed by an empty second line.\n\n**Correctness**\n\nThe interval characterization shows that every value excluded from $B$ is precisely a possible creator of some required MEX value, so $a_k$ is absent from the quotient set for every $k$. The hidden witness $A$ contains no excluded value, hence $A\\subseteq B$; therefore all quotient values smaller than each $a_k$ that occur for $A$ still occur for $B$. These two facts are exactly the definition of MEX, so $f(B,k)=a_k$ for every $k$.\n\n**Complexity**\n\nEach interval and each candidate value is processed once. The time complexity is $O(n)$ per test case and the memory complexity is $O(n)$. Over all test cases, the time is $O(\\sum n)$.",
          "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> diff(n + 1);\n        for (int k = 1; k <= n; ++k) {\n            ll x;\n            cin >> x;\n\n            ll left = x * k;\n            ll right = (x + 1) * k; // Half-open interval [left, right).\n            if (left < n) {\n                ++diff[static_cast<int>(left)];\n                --diff[static_cast<int>(min<ll>(n, right))];\n            }\n        }\n\n        vector<int> answer;\n        int covered = 0;\n        for (int y = 0; y < n; ++y) {\n            covered += diff[y];\n            if (covered == 0) {\n                answer.push_back(y);\n            }\n        }\n\n        cout << answer.size() << '\\n';\n        for (int y : answer) {\n            cout << y << ' ';\n        }\n        cout << '\\n';\n    }\n}\n"
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          "text": "**Analyzing mex properties for quotient sets**",
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          "tool": "StructuredOutput",
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              "hints": [
                {
                  "order": 1,
                  "content": "For a fixed divisor $k$, the MEX condition says that the quotient value $a_k$ itself must be absent. Work backward: which values of $y$ satisfy $\\left\\lfloor y/k\\right\\rfloor=a_k$?"
                },
                {
                  "order": 2,
                  "content": "Those $y$ form one interval: $a_k k\\le y<(a_k+1)k$. Every value in this interval is forbidden, because including even one of them would make the MEX at $k$ different."
                },
                {
                  "order": 3,
                  "content": "Take $B$ to contain every number from $0$ to $n-1$ that is not forbidden by any $k$. This automatically keeps every observed $a_k$ absent. The remaining question is why adding all these survivors does not lose any smaller quotient."
                },
                {
                  "order": 4,
                  "content": "Use the promise that some valid hidden set exists; call it $A$. It cannot contain a forbidden value, so $A\\subseteq B$. Thus, for every $q<a_k$, the witness value from $A$ with $\\lfloor y/k\\rfloor=q$ is still in $B$, while $a_k$ remains absent."
                },
                {
                  "order": 5,
                  "content": "There are $n$ forbidden intervals, so mark their union with a difference array. Treat $[a_k k,(a_k+1)k)$ as half-open, clip it to $[0,n)$, and use `long long` for the products."
                }
              ],
              "editorial": "Do not try to reconstruct the hidden set exactly. The promise that a solution exists lets us build a canonical, possibly larger set.\n\n**Key observation**\n\nFor a fixed $k$, a value $y$ produces the quotient $a_k$ exactly when\n$$\n\\left\\lfloor\\frac{y}{k}\\right\\rfloor=a_k\n\\quad\\Longleftrightarrow\\quad\n a_k k\\le y<(a_k+1)k.\n$$\nTherefore every integer in the interval\n$$\n[a_k k,(a_k+1)k)\n$$\nmust be absent from any valid set. Otherwise $a_k$ would appear among the quotients, so its MEX could not be $a_k$.\n\nCall the union of these intervals forbidden, and construct\n$$\nB=\\{0,1,\\ldots,n-1\\}\\setminus\\text{(forbidden values)}.\n$$\n\n**Why this works**\n\nLet $A$ be the hidden valid set whose existence is guaranteed.\n\nFirst, $A$ contains no forbidden value. If $y\\in A$ belonged to the interval for some $k$, then $\\lfloor y/k\\rfloor=a_k$ would occur, contradicting $f(A,k)=a_k$. Hence $A\\subseteq B$.\n\nFix any $k$.\n\n- No element of $B$ produces quotient $a_k$, because all such elements were marked forbidden.\n- Since $f(A,k)=a_k$, every value $q=0,1,\\ldots,a_k-1$ occurs as $\\lfloor y/k\\rfloor$ for some $y\\in A$. Because $A\\subseteq B$, all those quotient values also occur for $B$.\n\nThus $B$ contains every quotient below $a_k$ and does not contain quotient $a_k$. Its MEX is exactly $a_k$.\n\nThe important trick is that adding survivors cannot hurt: the only dangerous quotient, $a_k$, has been explicitly excluded. Trying to guess the original set is unnecessary busywork.\n\n**Implementation**\n\nFor each $k$, use the half-open forbidden interval $[L,R)$ with\n$$\nL=a_k k,\\qquad R=(a_k+1)k.\n$$\nOnly values in $[0,n-1]$ matter. If $L<n$, add $+1$ at `L` and $-1$ at `min(n, R)` in a difference array. A prefix sum tells whether each value is covered by at least one forbidden interval.\n\nUse 64-bit arithmetic for $L$ and $R$. The input guarantee makes the construction valid, so no impossibility case needs to be handled. If every value is forbidden, output an empty set followed by an empty second line.\n\n**Correctness**\n\nThe interval characterization shows that every value excluded from $B$ is precisely a possible creator of some required MEX value, so $a_k$ is absent from the quotient set for every $k$. The hidden witness $A$ contains no excluded value, hence $A\\subseteq B$; therefore all quotient values smaller than each $a_k$ that occur for $A$ still occur for $B$. These two facts are exactly the definition of MEX, so $f(B,k)=a_k$ for every $k$.\n\n**Complexity**\n\nEach interval and each candidate value is processed once. The time complexity is $O(n)$ per test case and the memory complexity is $O(n)$. Over all test cases, the time is $O(\\sum n)$.",
              "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> diff(n + 1);\n        for (int k = 1; k <= n; ++k) {\n            ll x;\n            cin >> x;\n\n            ll left = x * k;\n            ll right = (x + 1) * k; // Half-open interval [left, right).\n            if (left < n) {\n                ++diff[static_cast<int>(left)];\n                --diff[static_cast<int>(min<ll>(n, right))];\n            }\n        }\n\n        vector<int> answer;\n        int covered = 0;\n        for (int y = 0; y < n; ++y) {\n            covered += diff[y];\n            if (covered == 0) {\n                answer.push_back(y);\n            }\n        }\n\n        cout << answer.size() << '\\n';\n        for (int y : answer) {\n            cout << y << ' ';\n        }\n        cout << '\\n';\n    }\n}\n"
            },
            "output": "Structured output captured successfully.",
            "metadata": {
              "valid": true
            },
            "title": "Structured Output",
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              "start": 1789604873322,
              "end": 1789604873324
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