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    "slug": "tidy-harbor",
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    "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Noo88J",
    "title": "Nudge content generation",
    "agent": "nudge-generation",
    "model": {
      "id": "gpt-5.6-luna",
      "providerID": "openai",
      "variant": "max"
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  "messages": [
    {
      "info": {
        "role": "user",
        "time": {
          "created": 1789604547495
        },
        "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": [
                  "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"
                  }
                ],
                "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_0acbe4ba7001r1rEqKAsbbadbe",
        "sessionID": "ses_f5341b467ffeJZ9jNabQTz97zx"
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      "parts": [
        {
          "type": "text",
          "text": "Generate Codeforces learning content: five progressive hints, an original editorial, and AC-quality C++.\n\nFollow the editorial-first research workflow in the system instructions. Use the supplied statement to verify that the source solves this exact problem.\n\nWrite clean Markdown with LaTeX as needed. Use quick and clever humor when appropriate. Tell it like it is (don't sugar-coat responses), and use very casual language. You are fully allowed to swear, just don't overdo it like a sailor (be natural). Deconstruct any false assumptions.\n\nGenerate content for Codeforces problem 2262A2: \"Floor of MEX (Hard Version)\".\n\nKnown problem metadata:\n- Contest ID: 2262\n- Index: A2\n- Name: Floor of MEX (Hard Version)\n- Rating: unknown / unrated\n- Tags: dp, implementation\n- Problem URL: https://codeforces.com/contest/2262/problem/A2\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\">A2. Floor of MEX (Hard 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 hard version of the problem. The difference between the versions is that in this version, you need to count how many sets satisfy 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$$$.</p><p>Your task is to count how many sets (including the empty set) $$$B \\subseteq \\{0,1,\\ldots,n-1\\}$$$ such that $$$f(B,k)=a_k$$$ for every $$$1 \\le k \\le n$$$. Since the number can be very large, output it modulo $$$10^9+7$$$. <span class=\"tex-font-style-bf\">Farmer John will only give you arrays where such a $$$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 testcase, output one number, the number of sets $$$B$$$ that satisfy the constraints modulo $$$10^9 + 7$$$.</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\">6</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\">1</div></pre></div></div></div><div class=\"note\"><div class=\"section-title\">Note</div><p>For the first testcase, the $$$6$$$ subsets are $$$$$$\\{1, 2, 5\\}, \\{1, 3, 5\\}, \\{1, 2, 3, 5\\}, \\{1, 2, 4, 5\\}, \\{1, 3, 4, 5\\}, \\{1, 2, 3, 4, 5\\}.$$$$$$</p><p>For the second testcase, the only subset that satisfies $$$f(B,k) = a_k$$$ for every $$$1 \\le k \\le n$$$ is $$$\\{0, 1\\}$$$.</p></div>\n</problem-statement>\n\n\nSource lookup status:\n<source-lookup-status>\n- Tutorial (en): https://codeforces.com/blog/entry/156688 loaded (200 OK; title: \"Codeforces Round 1120 (Div 1, Div 2) Editorial - Codeforces\"; mentions 2262A2)\n</source-lookup-status>\n\nGenerate:\n1. Five progressive hints, from a gentle nudge to the key insight.\n2. A self-contained editorial that teaches the solution clearly, including the key insight, algorithm, correctness argument, complexity, and necessary implementation details. Explain non-obvious steps without padding or a discovery-process transcript.\n3. A complete C++26 solution that gets AC on Codeforces.\n\nFormatting & Style Rules:\n- Hints and editorial: valid Markdown with LaTeX as needed (e.g., $dp[i]$, $$\\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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        "finish": "tool-calls",
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        {
          "type": "reasoning",
          "text": "**Fetching official tutorial for CF 2262A2**",
          "time": {
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        {
          "type": "tool",
          "tool": "webfetch",
          "callID": "call_2PHF3ivYixB9FbcWwfaLRC3S",
          "state": {
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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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-------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------- | -------------------------------------------- |\n\n→ Top contributors \n\n| #  | User                                                                                     | Contrib. |\n| -- | ---------------------------------------------------------------------------------------- | -------- |\n| 1  | [Qingyu](https://codeforces.com/profile/Qingyu \"Legendary Grandmaster Qingyu\")           | 157      |\n| 2  | [maspy](https://codeforces.com/profile/maspy \"Legendary Grandmaster maspy\")              | 150      |\n| 2  | [nik\\_exists](https://codeforces.com/profile/nik%5Fexists \"Expert nik_exists\")           | 150      |\n| 4  | [Um\\_nik](https://codeforces.com/profile/Um%5Fnik \"Legendary Grandmaster Um_nik\")        | 144      |\n| 5  | [Errichto](https://codeforces.com/profile/Errichto \"International Grandmaster Errichto\") | 139      |\n| 6  | [adamant](https://codeforces.com/profile/adamant \"Master adamant\")                       | 136      |\n| 7  | [AmShZ](https://codeforces.com/profile/AmShZ \"Grandmaster AmShZ\")                        | 134      |\n| 8  | [maroonrk](https://codeforces.com/profile/maroonrk \"Legendary Grandmaster maroonrk\")     | 133      |\n| 8  | [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\") | 133      |\n| 10 | [DNR](https://codeforces.com/profile/DNR \"Master DNR\")                                   | 128      |\n\n|  | [View all →](https://codeforces.com/top-contributed) |\n|  | ---------------------------------------------------- |\n\n→ Find user \n\nHandle: \n\n→ Recent actions \n\n* [Down\\_bad\\_4\\_haiender288](https://codeforces.com/profile/Down%5Fbad%5F4%5Fhaiender288 \"Pupil Down_bad_4_haiender288\") → [new manmade horrors (barely within comprehension) just dropped](https://codeforces.com/blog/entry/156826) 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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; 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. 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            "output": "<task id=\"ses_f534174e2ffed6op1ARTlIFqst\" state=\"completed\">\n<task_result>\n**Reduction**\n\nLet \\(U=\\{0,1,\\dots,n-1\\}\\). For fixed \\(k\\), define\n\n\\[\nI_{k,q}=[qk,(q+1)k-1]\\cap U.\n\\]\n\nEvery \\(y\\in I_{k,q}\\) satisfies \\(\\lfloor y/k\\rfloor=q\\). Therefore, if \\(a_k=d\\),\n\n\\[\nf(B,k)=d\n\\]\n\nis equivalent to:\n\n- \\(B\\cap I_{k,q}\\neq\\varnothing\\) for every \\(0\\le q<d\\);\n- \\(B\\cap I_{k,d}=\\varnothing\\).\n\nThe first intervals are the **good intervals**. \\(I_{k,d}\\) is the **bad interval**.\n\nThus:\n\n1. Every point in the union of all bad intervals is forced out of \\(B\\).\n2. Every good interval must contain at least one selected point.\n3. Every point not forced out is independently selectable, subject only to the interval-hitting constraints.\n\nThe bound\n\n\\[\na_k\\le \\left\\lfloor\\frac{n-1}{k}\\right\\rfloor+1\n\\]\n\nis necessary because \\(q=a_k-1\\) must have a nonempty interval inside \\(U\\).\n\n---\n\n**Building the available points**\n\nLet\n\n\\[\nF=\\bigcup_k I_{k,a_k},\\qquad G=U\\setminus F.\n\\]\n\nOnly points in \\(G\\) can belong to \\(B\\). The union \\(F\\) is computed with a difference array:\n\n```text\nfor every bad interval [L, R]:\n    diff[L]++\n    diff[R + 1]--\n\nscan x = 0 ... n - 1\n    x is selectable iff current coverage is zero\n```\n\nIntervals are clipped to \\(U\\). The official code instead leaves good-interval right endpoints possibly larger than \\(n-1\\), which is also safe because it only iterates candidate points below \\(n\\).\n\n---\n\n**Why redundant good intervals can be discarded**\n\nIf two good intervals satisfy\n\n\\[\nI\\subseteq J,\n\\]\n\nthen hitting \\(I\\) automatically hits \\(J\\). Therefore \\(J\\) imposes no additional condition and can be discarded.\n\nIt is enough to retain a containment-minimal family of intervals. This does not necessarily mean globally minimum cardinality; it means every discarded interval contains a retained interval.\n\nThe reference code sorts intervals by increasing length. It then stores retained intervals in a set ordered by left endpoint:\n\n```cpp\nauto it = kept.lower_bound({L, 0});\nif (it != kept.end() && it->second <= R)\n    discard [L, R];\nelse\n    keep [L, R];\n```\n\nWhy is checking only the first interval with left endpoint at least \\(L\\) sufficient?\n\n- Previously retained intervals have length no greater than the current interval.\n- Hence a previously retained interval cannot properly contain the current one.\n- Retained intervals never contain one another.\n- Consequently, when retained intervals are ordered by left endpoint, their right endpoints are also increasing.\n- Therefore, the first retained interval with left endpoint at least \\(L\\) has the smallest right endpoint among all such intervals.\n\nIf that interval ends no later than \\(R\\), it is contained in the current interval, so the current interval is redundant.\n\nAfter pruning, write the retained intervals as\n\n\\[\n[L_0,R_0], [L_1,R_1],\\dots,[L_{m-1},R_{m-1}]\n\\]\n\nwith\n\n\\[\nL_0<L_1<\\dots<L_{m-1}\n\\quad\\text{and}\\quad\nR_0<R_1<\\dots<R_{m-1}.\n\\]\n\n---\n\n**DP invariant**\n\nProcess selectable points \\(x\\in G\\) from left to right.\n\nFor a partial subset, define its state as\n\n\\[\nj=\\#\\{q:L_q\\le \\text{last selected point}\\}.\n\\]\n\nState \\(0\\) means no selected point has reached the first interval's left endpoint. State \\(m\\) means the last selected point is at or beyond the final interval's left endpoint.\n\n`dp[j]` counts active partial subsets in state \\(j\\). The reference code stores the sum of all active states in `poss`.\n\nFor the current point \\(x\\):\n\n1. Expire every interval with \\(R_q<x\\). Such an interval can no longer be hit.\n2. If the current first unexpired interval is \\(q\\), then `dp[q]` represents subsets whose last selected point is before \\(L_q\\). They missed interval \\(q\\), so remove them.\n3. Let `r` be the number of intervals with \\(L_q\\le x\\).\n4. Selecting \\(x\\) moves every active subset into state `r`:\n\n   \\[\n   dp[r]\\mathrel{+}=poss.\n   \\]\n\n5. Not selecting \\(x\\) leaves every active subset where it is. Hence after considering both choices:\n\n   \\[\n   poss\\mathrel{:=}2\\cdot poss.\n   \\]\n\nThe removal is correct because the intervals have increasing right endpoints. Any state larger than \\(q\\) either selected a point inside interval \\(q\\), or was created after interval \\(q\\) had already expired from a subset that had previously hit it.\n\nAt the end, the answer is `dp[m]`.\n\nEvery valid set must hit the final interval, so its last selected point must be at least \\(L_{m-1}\\), putting it in state \\(m\\). Conversely, every state-\\(m\\) subset surviving the expiration process has hit every interval.\n\nIf \\(m=0\\), `dp[0]` simply counts every subset of \\(G\\), including the empty set.\n\n---\n\n**Audit of the reference code**\n\nThe official-looking implementation is correct under the problem's guarantee that the input array is realizable.\n\n- The indexing is correct: `i = 1..n` represents divisor \\(k\\), while candidate values are `0..n-1`.\n- Good intervals use `j = 0..a[i]-1`.\n- Bad intervals correctly use `[a[i] * i, (a[i] + 1) * i - 1]`.\n- `R < x` is the correct expiration test. A point equal to `R` can still hit the interval.\n- `L <= x` is the correct start test. A point equal to `L` can hit the interval.\n- `dp` has size `number_of_intervals + 1`, and `r` can legitimately equal that size.\n- Returning `dp[m]`, rather than the sum of all DP states, is correct.\n- The `min(n, ...)` difference-array indexing is intentional. `diff[n]` is a sentinel and need not be scanned.\n- All-zero input is handled correctly and produces answer `1`.\n- The maximal case \\(a_k=\\lfloor(n-1)/k\\rfloor+1\\) is handled correctly even though the final good interval may extend beyond \\(n-1\\).\n\nEngineering issues worth fixing:\n\n- The reference code checks feasibility while reading `b[i]` and immediately returns. On malformed input, that leaves the rest of the test case unread and corrupts subsequent test cases. The problem guarantee makes this unreachable, but reading the entire array first is cleaner.\n- The products use `int`. They are safe after the local feasibility check under the stated constraints, but `long long` makes the boundary reasoning robust.\n- The one-sided `set.lower_bound` test is correct only because intervals were sorted by length and the retained set is an antichain. This dependency should be documented.\n- Clipping good intervals to `[0,n-1]` makes the implementation easier to audit, although leaving them unclipped is also correct.\n\nLet\n\n\\[\nM=\\sum_{k=1}^n a_k.\n\\]\n\nFor realizable inputs,\n\n\\[\nM=O(n\\log n).\n\\]\n\nThe complexity is therefore:\n\n\\[\nO(M\\log M+n)=O(n\\log^2 n)\n\\]\n\ntime and \\(O(M+n)\\) memory.\n\n---\n\n**Pseudocode**\n\n```text\nread n and a[1..n]\n\ngood = empty list\ndiff[0..n] = 0\n\nfor k = 1..n:\n    d = a[k]\n\n    if d > floor((n - 1) / k) + 1:\n        answer = 0\n\n    bad = [d*k, (d+1)*k - 1]\n    clip bad to [0, n-1]\n    add bad to diff\n\n    for q = 0..d-1:\n        good.push([q*k, (q+1)*k - 1] clipped to [0, n-1])\n\nfree = points with zero bad-interval coverage\n\nsort good by increasing length\n\nkept = ordered set by left endpoint\nfor [L, R] in good:\n    first = kept.lower_bound([L, -infinity])\n    if first exists and first.R <= R:\n        continue\n    insert [L, R]\n\nneed = kept in left-endpoint order\n\ndp[0] = 1\nactive = 1\nexpired = 0\nstarted = 0\n\nfor x in free in increasing order:\n    while expired < m and need[expired].R < x:\n        active -= dp[expired]\n        expired++\n\n    while started < m and need[started].L <= x:\n        started++\n\n    dp[started] += active\n    active *= 2\n\nanswer = dp[m]\n```\n\n---\n\n**Clean C++17 implementation**\n\n```cpp\n#include <bits/stdc++.h>\nusing namespace std;\n\nconstexpr int MOD = 1'000'000'007;\nusing Interval = pair<int, int>;\n\nint solve_case(const vector<int>& a) {\n    int n = (int)a.size();\n\n    vector<Interval> good;\n    vector<int> diff(n + 1, 0);\n\n    for (int k = 1; k <= n; ++k) {\n        int d = a[k - 1];\n\n        if (d > (n - 1) / k + 1)\n            return 0;\n\n        // Bad interval: [d*k, (d+1)*k - 1].\n        long long badL = 1LL * d * k;\n        long long badR = 1LL * (d + 1) * k - 1;\n\n        if (badL < n) {\n            int L = (int)badL;\n            int R = (int)min<long long>(n - 1, badR);\n\n            ++diff[L];\n            --diff[R + 1];\n        }\n\n        // Good intervals for quotients 0, 1, ..., d-1.\n        for (int q = 0; q < d; ++q) {\n            int L = q * k;\n            int R = (int)min<long long>(\n                n - 1, 1LL * (q + 1) * k - 1\n            );\n\n            good.push_back({L, R});\n        }\n    }\n\n    vector<int> freeValues;\n    vector<int> freePrefix(n + 1, 0);\n\n    int coverage = 0;\n    for (int x = 0; x < n; ++x) {\n        coverage += diff[x];\n\n        bool free = (coverage == 0);\n        freePrefix[x + 1] = freePrefix[x] + free;\n\n        if (free)\n            freeValues.push_back(x);\n    }\n\n    sort(good.begin(), good.end(),\n         [](const Interval& x, const Interval& y) {\n             int lenX = x.second - x.first;\n             int lenY = y.second - y.first;\n\n             if (lenX != lenY)\n                 return lenX < lenY;\n\n             return x < y;\n         });\n\n    set<Interval> kept;\n\n    for (auto [L, R] : good) {\n        auto it = kept.lower_bound({L, -1});\n\n        if (it != kept.end() && it->second <= R)\n            continue;\n\n        kept.insert({L, R});\n    }\n\n    vector<Interval> need(kept.begin(), kept.end());\n\n    // Optional robustness check for non-realizable input.\n    for (auto [L, R] : need) {\n        if (freePrefix[R + 1] == freePrefix[L])\n            return 0;\n    }\n\n    int m = (int)need.size();\n\n    vector<int> dp(m + 1, 0);\n    dp[0] = 1;\n\n    long long active = 1;\n    int expired = 0;\n    int started = 0;\n\n    for (int x : freeValues) {\n        while (expired < m && need[expired].second < x) {\n            active += MOD - dp[expired];\n            active %= MOD;\n            ++expired;\n        }\n\n        while (started < m && need[started].first <= x)\n            ++started;\n\n        dp[started] = (dp[started] + active) % MOD;\n        active = (2 * active) % MOD;\n    }\n\n    return dp[m];\n}\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);\n        for (int& x : a)\n            cin >> x;\n\n        cout << solve_case(a) << '\\n';\n    }\n}\n```\n\nNo files were edited.\n</task_result>\n</task>",
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            }
          },
          "id": "prt_0accb304c001HkDPxwJGIhSA0T",
          "sessionID": "ses_f5341b467ffeJZ9jNabQTz97zx",
          "messageID": "msg_0acca74dc001tkaO82LU1PYVI1"
        },
        {
          "type": "text",
          "text": "The reduction is clean: each `k` contributes one forbidden quotient block and several required blocks. I’m validating the interval-pruning invariant and sweep DP against brute-force enumeration before writing the final content.",
          "time": {
            "start": 1789605392485,
            "end": 1789605393275
          },
          "metadata": {
            "openai": {
              "itemId": "msg_069e8b731cc6a257016aab36106d3887d2acb45473ed85d57e",
              "phase": "commentary"
            }
          },
          "id": "prt_0accb3065001vTNIfeh9f9JL4n",
          "sessionID": "ses_f5341b467ffeJZ9jNabQTz97zx",
          "messageID": "msg_0acca74dc001tkaO82LU1PYVI1"
        },
        {
          "type": "tool",
          "tool": "apply_patch",
          "callID": "call_n9m4gLCXpMpvc2Yu1iPBTr8s",
          "state": {
            "status": "completed",
            "input": {
              "patchText": "*** Begin Patch\n*** Add File: verify.cpp\n+#include <bits/stdc++.h>\n+using namespace std;\n+\n+using ll = long long;\n+\n+const int MOD = 1'000'000'007;\n+\n+int solve_case(const vector<int>& a) {\n+    int n = (int)a.size();\n+    for (int k = 1; k <= n; ++k) {\n+        if (a[k - 1] > (n - 1) / k + 1) return 0;\n+    }\n+\n+    vector<pair<int, int>> good;\n+    vector<int> diff(n + 1);\n+\n+    for (int k = 1; k <= n; ++k) {\n+        int d = a[k - 1];\n+        ll badL = 1LL * d * k;\n+        ll badR = 1LL * (d + 1) * k - 1;\n+        if (badL < n) {\n+            int l = (int)badL;\n+            int r = (int)min<ll>(n - 1, badR);\n+            ++diff[l];\n+            --diff[r + 1];\n+        }\n+        for (int q = 0; q < d; ++q) {\n+            int l = q * k;\n+            int r = (int)min<ll>(n - 1, 1LL * (q + 1) * k - 1);\n+            good.push_back({l, r});\n+        }\n+    }\n+\n+    vector<int> freeValues;\n+    int cover = 0;\n+    for (int x = 0; x < n; ++x) {\n+        cover += diff[x];\n+        if (cover == 0) freeValues.push_back(x);\n+    }\n+\n+    sort(good.begin(), good.end(), [](auto x, auto y) {\n+        int lx = x.second - x.first;\n+        int ly = y.second - y.first;\n+        if (lx != ly) return lx < ly;\n+        return x < y;\n+    });\n+\n+    set<pair<int, int>> kept;\n+    for (auto [l, r] : good) {\n+        auto it = kept.lower_bound({l, -1});\n+        if (it == kept.end() || it->second > r) kept.insert({l, r});\n+    }\n+\n+    vector<pair<int, int>> need(kept.begin(), kept.end());\n+    int m = (int)need.size();\n+    vector<int> dp(m + 1);\n+    dp[0] = 1;\n+    int active = 1;\n+    int expired = 0, started = 0;\n+\n+    for (int x : freeValues) {\n+        while (expired < m && need[expired].second < x) {\n+            active += MOD - dp[expired];\n+            if (active >= MOD) active -= MOD;\n+            ++expired;\n+        }\n+        while (started < m && need[started].first <= x) ++started;\n+        dp[started] += active;\n+        if (dp[started] >= MOD) dp[started] -= MOD;\n+        active += active;\n+        if (active >= MOD) active -= MOD;\n+    }\n+    return dp[m];\n+}\n+\n+int brute(const vector<int>& a) {\n+    int n = (int)a.size();\n+    int ans = 0;\n+    for (int mask = 0; mask < (1 << n); ++mask) {\n+        bool ok = true;\n+        for (int k = 1; k <= n && ok; ++k) {\n+            vector<bool> have(n + 1);\n+            for (int y = 0; y < n; ++y) if (mask >> y & 1) {\n+                have[y / k] = true;\n+            }\n+            int mex = 0;\n+            while (have[mex]) ++mex;\n+            if (mex != a[k - 1]) ok = false;\n+        }\n+        ans += ok;\n+    }\n+    return ans;\n+}\n+\n+int main() {\n+    mt19937 rng(1234567);\n+    for (int n = 1; n <= 10; ++n) {\n+        vector<int> a(n);\n+        for (int mask = 0; mask < (1 << n); ++mask) {\n+            for (int k = 1; k <= n; ++k) {\n+                vector<bool> have(n + 1);\n+                for (int y = 0; y < n; ++y) if (mask >> y & 1) have[y / k] = true;\n+                int mex = 0;\n+                while (have[mex]) ++mex;\n+                a[k - 1] = mex;\n+            }\n+            int got = solve_case(a), want = brute(a);\n+            if (got != want) {\n+                cerr << \"bad n=\" << n << \" mask=\" << mask << \" got=\" << got << \" want=\" << want << \"\\n\";\n+                for (int x : a) cerr << x << ' ';\n+                cerr << '\\n';\n+                return 1;\n+            }\n+        }\n+        cerr << \"n \" << n << \" ok\\n\";\n+    }\n+}\n*** End Patch"
            },
            "output": "Success. Updated the following files:\nA private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Noo88J/verify.cpp",
            "metadata": {
              "diff": "Index: /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Noo88J/verify.cpp\n===================================================================\n--- /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Noo88J/verify.cpp\n+++ /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Noo88J/verify.cpp\n@@ -0,0 +1,117 @@\n+#include <bits/stdc++.h>\n+using namespace std;\n+\n+using ll = long long;\n+\n+const int MOD = 1'000'000'007;\n+\n+int solve_case(const vector<int>& a) {\n+    int n = (int)a.size();\n+    for (int k = 1; k <= n; ++k) {\n+        if (a[k - 1] > (n - 1) / k + 1) return 0;\n+    }\n+\n+    vector<pair<int, int>> good;\n+    vector<int> diff(n + 1);\n+\n+    for (int k = 1; k <= n; ++k) {\n+        int d = a[k - 1];\n+        ll badL = 1LL * d * k;\n+        ll badR = 1LL * (d + 1) * k - 1;\n+        if (badL < n) {\n+            int l = (int)badL;\n+            int r = (int)min<ll>(n - 1, badR);\n+            ++diff[l];\n+            --diff[r + 1];\n+        }\n+        for (int q = 0; q < d; ++q) {\n+            int l = q * k;\n+            int r = (int)min<ll>(n - 1, 1LL * (q + 1) * k - 1);\n+            good.push_back({l, r});\n+        }\n+    }\n+\n+    vector<int> freeValues;\n+    int cover = 0;\n+    for (int x = 0; x < n; ++x) {\n+        cover += diff[x];\n+        if (cover == 0) freeValues.push_back(x);\n+    }\n+\n+    sort(good.begin(), good.end(), [](auto x, auto y) {\n+        int lx = x.second - x.first;\n+        int ly = y.second - y.first;\n+        if (lx != ly) return lx < ly;\n+        return x < y;\n+    });\n+\n+    set<pair<int, int>> kept;\n+    for (auto [l, r] : good) {\n+        auto it = kept.lower_bound({l, -1});\n+        if (it == kept.end() || it->second > r) kept.insert({l, r});\n+    }\n+\n+    vector<pair<int, int>> need(kept.begin(), kept.end());\n+    int m = (int)need.size();\n+    vector<int> dp(m + 1);\n+    dp[0] = 1;\n+    int active = 1;\n+    int expired = 0, started = 0;\n+\n+    for (int x : freeValues) {\n+        while (expired < m && need[expired].second < x) {\n+            active += MOD - dp[expired];\n+            if (active >= MOD) active -= MOD;\n+            ++expired;\n+        }\n+        while (started < m && need[started].first <= x) ++started;\n+        dp[started] += active;\n+        if (dp[started] >= MOD) dp[started] -= MOD;\n+        active += active;\n+        if (active >= MOD) active -= MOD;\n+    }\n+    return dp[m];\n+}\n+\n+int brute(const vector<int>& a) {\n+    int n = (int)a.size();\n+    int ans = 0;\n+    for (int mask = 0; mask < (1 << n); ++mask) {\n+        bool ok = true;\n+        for (int k = 1; k <= n && ok; ++k) {\n+            vector<bool> have(n + 1);\n+            for (int y = 0; y < n; ++y) if (mask >> y & 1) {\n+                have[y / k] = true;\n+            }\n+            int mex = 0;\n+            while (have[mex]) ++mex;\n+            if (mex != a[k - 1]) ok = false;\n+        }\n+        ans += ok;\n+    }\n+    return ans;\n+}\n+\n+int main() {\n+    mt19937 rng(1234567);\n+    for (int n = 1; n <= 10; ++n) {\n+        vector<int> a(n);\n+        for (int mask = 0; mask < (1 << n); ++mask) {\n+            for (int k = 1; k <= n; ++k) {\n+                vector<bool> have(n + 1);\n+                for (int y = 0; y < n; ++y) if (mask >> y & 1) have[y / k] = true;\n+                int mex = 0;\n+                while (have[mex]) ++mex;\n+                a[k - 1] = mex;\n+            }\n+            int got = solve_case(a), want = brute(a);\n+            if (got != want) {\n+                cerr << \"bad n=\" << n << \" mask=\" << mask << \" got=\" << got << \" want=\" << want << \"\\n\";\n+                for (int x : a) cerr << x << ' ';\n+                cerr << '\\n';\n+                return 1;\n+            }\n+        }\n+        cerr << \"n \" << n << \" ok\\n\";\n+    }\n+}\n\n",
              "files": [
                {
                  "filePath": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Noo88J/verify.cpp",
                  "relativePath": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Noo88J/verify.cpp",
                  "type": "add",
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              "diff": "Index: /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Noo88J/verify.cpp\n===================================================================\n--- /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Noo88J/verify.cpp\n+++ /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Noo88J/verify.cpp\n@@ -1,117 +0,0 @@\n-#include <bits/stdc++.h>\n-using namespace std;\n-\n-using ll = long long;\n-\n-const int MOD = 1'000'000'007;\n-\n-int solve_case(const vector<int>& a) {\n-    int n = (int)a.size();\n-    for (int k = 1; k <= n; ++k) {\n-        if (a[k - 1] > (n - 1) / k + 1) return 0;\n-    }\n-\n-    vector<pair<int, int>> good;\n-    vector<int> diff(n + 1);\n-\n-    for (int k = 1; k <= n; ++k) {\n-        int d = a[k - 1];\n-        ll badL = 1LL * d * k;\n-        ll badR = 1LL * (d + 1) * k - 1;\n-        if (badL < n) {\n-            int l = (int)badL;\n-            int r = (int)min<ll>(n - 1, badR);\n-            ++diff[l];\n-            --diff[r + 1];\n-        }\n-        for (int q = 0; q < d; ++q) {\n-            int l = q * k;\n-            int r = (int)min<ll>(n - 1, 1LL * (q + 1) * k - 1);\n-            good.push_back({l, r});\n-        }\n-    }\n-\n-    vector<int> freeValues;\n-    int cover = 0;\n-    for (int x = 0; x < n; ++x) {\n-        cover += diff[x];\n-        if (cover == 0) freeValues.push_back(x);\n-    }\n-\n-    sort(good.begin(), good.end(), [](auto x, auto y) {\n-        int lx = x.second - x.first;\n-        int ly = y.second - y.first;\n-        if (lx != ly) return lx < ly;\n-        return x < y;\n-    });\n-\n-    set<pair<int, int>> kept;\n-    for (auto [l, r] : good) {\n-        auto it = kept.lower_bound({l, -1});\n-        if (it == kept.end() || it->second > r) kept.insert({l, r});\n-    }\n-\n-    vector<pair<int, int>> need(kept.begin(), kept.end());\n-    int m = (int)need.size();\n-    vector<int> dp(m + 1);\n-    dp[0] = 1;\n-    int active = 1;\n-    int expired = 0, started = 0;\n-\n-    for (int x : freeValues) {\n-        while (expired < m && need[expired].second < x) {\n-            active += MOD - dp[expired];\n-            if (active >= MOD) active -= MOD;\n-            ++expired;\n-        }\n-        while (started < m && need[started].first <= x) ++started;\n-        dp[started] += active;\n-        if (dp[started] >= MOD) dp[started] -= MOD;\n-        active += active;\n-        if (active >= MOD) active -= MOD;\n-    }\n-    return dp[m];\n-}\n-\n-int brute(const vector<int>& a) {\n-    int n = (int)a.size();\n-    int ans = 0;\n-    for (int mask = 0; mask < (1 << n); ++mask) {\n-        bool ok = true;\n-        for (int k = 1; k <= n && ok; ++k) {\n-            vector<bool> have(n + 1);\n-            for (int y = 0; y < n; ++y) if (mask >> y & 1) {\n-                have[y / k] = true;\n-            }\n-            int mex = 0;\n-            while (have[mex]) ++mex;\n-            if (mex != a[k - 1]) ok = false;\n-        }\n-        ans += ok;\n-    }\n-    return ans;\n-}\n-\n-int main() {\n-    mt19937 rng(1234567);\n-    for (int n = 1; n <= 10; ++n) {\n-        vector<int> a(n);\n-        for (int mask = 0; mask < (1 << n); ++mask) {\n-            for (int k = 1; k <= n; ++k) {\n-                vector<bool> have(n + 1);\n-                for (int y = 0; y < n; ++y) if (mask >> y & 1) have[y / k] = true;\n-                int mex = 0;\n-                while (have[mex]) ++mex;\n-                a[k - 1] = mex;\n-            }\n-            int got = solve_case(a), want = brute(a);\n-            if (got != want) {\n-                cerr << \"bad n=\" << n << \" mask=\" << mask << \" got=\" << got << \" want=\" << want << \"\\n\";\n-                for (int x : a) cerr << x << ' ';\n-                cerr << '\\n';\n-                return 1;\n-            }\n-        }\n-        cerr << \"n \" << n << \" ok\\n\";\n-    }\n-}\n\n",
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                  "patch": "Index: /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Noo88J/verify.cpp\n===================================================================\n--- /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Noo88J/verify.cpp\n+++ /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Noo88J/verify.cpp\n@@ -1,117 +0,0 @@\n-#include <bits/stdc++.h>\n-using namespace std;\n-\n-using ll = long long;\n-\n-const int MOD = 1'000'000'007;\n-\n-int solve_case(const vector<int>& a) {\n-    int n = (int)a.size();\n-    for (int k = 1; k <= n; ++k) {\n-        if (a[k - 1] > (n - 1) / k + 1) return 0;\n-    }\n-\n-    vector<pair<int, int>> good;\n-    vector<int> diff(n + 1);\n-\n-    for (int k = 1; k <= n; ++k) {\n-        int d = a[k - 1];\n-        ll badL = 1LL * d * k;\n-        ll badR = 1LL * (d + 1) * k - 1;\n-        if (badL < n) {\n-            int l = (int)badL;\n-            int r = (int)min<ll>(n - 1, badR);\n-            ++diff[l];\n-            --diff[r + 1];\n-        }\n-        for (int q = 0; q < d; ++q) {\n-            int l = q * k;\n-            int r = (int)min<ll>(n - 1, 1LL * (q + 1) * k - 1);\n-            good.push_back({l, r});\n-        }\n-    }\n-\n-    vector<int> freeValues;\n-    int cover = 0;\n-    for (int x = 0; x < n; ++x) {\n-        cover += diff[x];\n-        if (cover == 0) freeValues.push_back(x);\n-    }\n-\n-    sort(good.begin(), good.end(), [](auto x, auto y) {\n-        int lx = x.second - x.first;\n-        int ly = y.second - y.first;\n-        if (lx != ly) return lx < ly;\n-        return x < y;\n-    });\n-\n-    set<pair<int, int>> kept;\n-    for (auto [l, r] : good) {\n-        auto it = kept.lower_bound({l, -1});\n-        if (it == kept.end() || it->second > r) kept.insert({l, r});\n-    }\n-\n-    vector<pair<int, int>> need(kept.begin(), kept.end());\n-    int m = (int)need.size();\n-    vector<int> dp(m + 1);\n-    dp[0] = 1;\n-    int active = 1;\n-    int expired = 0, started = 0;\n-\n-    for (int x : freeValues) {\n-        while (expired < m && need[expired].second < x) {\n-            active += MOD - dp[expired];\n-            if (active >= MOD) active -= MOD;\n-            ++expired;\n-        }\n-        while (started < m && need[started].first <= x) ++started;\n-        dp[started] += active;\n-        if (dp[started] >= MOD) dp[started] -= MOD;\n-        active += active;\n-        if (active >= MOD) active -= MOD;\n-    }\n-    return dp[m];\n-}\n-\n-int brute(const vector<int>& a) {\n-    int n = (int)a.size();\n-    int ans = 0;\n-    for (int mask = 0; mask < (1 << n); ++mask) {\n-        bool ok = true;\n-        for (int k = 1; k <= n && ok; ++k) {\n-            vector<bool> have(n + 1);\n-            for (int y = 0; y < n; ++y) if (mask >> y & 1) {\n-                have[y / k] = true;\n-            }\n-            int mex = 0;\n-            while (have[mex]) ++mex;\n-            if (mex != a[k - 1]) ok = false;\n-        }\n-        ans += ok;\n-    }\n-    return ans;\n-}\n-\n-int main() {\n-    mt19937 rng(1234567);\n-    for (int n = 1; n <= 10; ++n) {\n-        vector<int> a(n);\n-        for (int mask = 0; mask < (1 << n); ++mask) {\n-            for (int k = 1; k <= n; ++k) {\n-                vector<bool> have(n + 1);\n-                for (int y = 0; y < n; ++y) if (mask >> y & 1) have[y / k] = true;\n-                int mex = 0;\n-                while (have[mex]) ++mex;\n-                a[k - 1] = mex;\n-            }\n-            int got = solve_case(a), want = brute(a);\n-            if (got != want) {\n-                cerr << \"bad n=\" << n << \" mask=\" << mask << \" got=\" << got << \" want=\" << want << \"\\n\";\n-                for (int x : a) cerr << x << ' ';\n-                cerr << '\\n';\n-                return 1;\n-            }\n-        }\n-        cerr << \"n \" << n << \" ok\\n\";\n-    }\n-}\n",
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            {
              "order": 1,
              "content": "Freeze one divisor $k$. The values `floor(y/k)` come from consecutive blocks $I_{k,q}=[qk,(q+1)k-1]$ (clipped to $0,\\ldots,n-1$). If $a_k=d$, which blocks must contain an element of $B$, and which block must contain none?"
            },
            {
              "order": 2,
              "content": "Turn that observation into constraints on individual values. The block for quotient $d$ is forbidden, so take the union of all such bad blocks. Every value outside that union is optional; the only remaining requirements are that every good block for $q=0,\\ldots,d-1$ is hit."
            },
            {
              "order": 3,
              "content": "Do not feed every good interval into the DP. If one interval is contained in another, hitting the smaller already hits the larger. Keep only a containment-minimal family; after sorting intervals by length, a set ordered by left endpoint can discard any current interval that contains an already kept one."
            },
            {
              "order": 4,
              "content": "After pruning, the retained intervals have increasing left and right endpoints. Sweep selectable values $x$ from left to right. Let state $j$ mean that the latest selected value has passed exactly $j$ retained left endpoints (state $0$ is the empty/no-selection case). What happens to state $j$ when its interval's right endpoint is already less than $x$?"
            },
            {
              "order": 5,
              "content": "Maintain $dp[j]$ and `active`, the number of still-viable subsets. At a selectable $x$, remove $dp[j]$ for every newly expired interval, let $r$ be the number of left endpoints at most $x$, do $dp[r] += active$ for choosing $x`, then double `active` for choosing or skipping $x$. The answer is $dp[m]$, where $m$ is the number of retained intervals; build selectable values with a difference array over bad blocks."
            }
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          "editorial": "The floor function looks annoying, but it is just partitioning the universe into intervals.\n\n**Turn MEX into interval constraints**\n\nLet $U=\\{0,1,\\ldots,n-1\\}$. For a fixed $k$ and quotient $q$, define\n\n$$I_{k,q}=[qk,(q+1)k-1]\\cap U.$$\n\nEvery $y$ in this interval has `floor(y/k) = q`. Therefore, if $a_k=d$, then\n\n- every interval $I_{k,q}$ for $0 \\le q < d$ must contain at least one element of $B$;\n- the interval $I_{k,d}$ must contain no element of $B$.\n\nThe second interval is the **bad interval** for $k$. Every value in the union of all bad intervals is forced out of $B$. Call every other value **free**.\n\nAfter removing the forced values, the original problem is exactly this:\n\n- choose any subset of the free values;\n- hit every **good interval** $I_{k,q}$ with $q< a_k$.\n\nValues producing quotients larger than $a_k$ do not matter: MEX only cares about the first missing quotient. For $a_k=0$, there are no good intervals, and the quotient-zero interval is simply bad.\n\nThe input is guaranteed to be realizable. As a small defensive check, a feasible $a_k$ must satisfy\n$$a_k \\le \\left\\lfloor\\frac{n-1}{k}\\right\\rfloor+1,$$\nwhich is the number of quotient blocks that actually meet $U$.\n\n**Remove redundant good intervals**\n\nThere can be many good intervals. If a good interval $X$ is contained in another good interval $Y$, then hitting $X$ automatically hits $Y$, so $Y$ can be discarded.\n\nWe keep a containment-minimal family as follows:\n\n1. Sort all good intervals by nondecreasing length.\n2. Store kept intervals ordered by their left endpoint.\n3. For a current $[L,R]$, find the first kept interval whose left endpoint is at least $L$. If its right endpoint is at most $R$, the current interval contains a kept interval and is redundant. Otherwise, keep the current interval.\n\nWhy is checking only that one interval enough? Previously processed intervals are no longer than the current one, so a previous interval cannot strictly contain the current one. The kept intervals therefore never contain one another. When ordered by left endpoint, their right endpoints are also increasing, so the first possible left endpoint has the smallest possible right endpoint.\n\nLet the retained intervals be\n$$[L_0,R_0], [L_1,R_1], \\ldots, [L_{m-1},R_{m-1}],$$\nordered by left endpoint. Both the $L_i$ and the $R_i$ are increasing.\n\n**Sweep DP**\n\nOnly free values can be selected, so process them from left to right.\n\nFor a selected value $x$, define its state as the number of retained left endpoints it has passed:\n$$state(x)=|\\{i:L_i\\le x\\}|.$$\nThe empty subset has state $0$.\n\nLet $dp[j]$ count generated subsets whose latest selected free value has state $j$. Let `active` be the number of those subsets that have not already missed an interval whose right endpoint has passed. All counts are modulo $10^9+7$.\n\nSuppose the current free value is $x$.\n\n- While $R_i<x$, interval $i$ can no longer be hit by a future value. Among the still-relevant states, exactly state $i$ represents subsets whose latest selected value has not reached $L_i$, so those subsets missed interval $i$. Subtract $dp[i]$ from `active`. Since the $R_i$ are increasing, this is handled by one pointer.\n- Let $r$ be the number of retained intervals with $L_i\\le x$. Choosing $x$ creates one new subset from every active subset, and its latest-value state is $r$. Thus set `dp[r] += active`.\n- Skipping $x$ leaves every active subset unchanged, while choosing it adds one new branch for each. Therefore `active` doubles.\n\nThe old subsets are already represented in their old $dp$ states, so the transition only needs to add the choosing branch.\n\nAt the end, the answer is $dp[m]$. A valid subset must hit the last retained interval, so it must select some value at least $L_{m-1}$ and reach state $m$. Conversely, every subset counted in state $m$ survived all expiration checks and therefore hit every retained interval, hence every original good interval. If $m=0$, this naturally counts every subset of the free values, including the empty set.\n\n**Correctness argument**\n\nFor each $k$, the interval characterization is equivalent to $f(B,k)=a_k$: all smaller quotients occur, and quotient $a_k$ does not. Thus removing bad values and requiring every good interval to be hit is equivalent to satisfying every $k$ simultaneously.\n\nDiscarding a good interval only happens when a kept interval is contained inside it. Any set hitting the kept interval also hits the discarded one, so pruning preserves exactly the valid subsets.\n\nDuring the sweep, `active` contains exactly the choices of processed free values that have not missed an interval already passed by the sweep. Expiring state $i$ removes precisely the subsets that never selected a value in interval $i$; selecting or skipping the current free value enumerates its two possible choices without duplication. Hence the DP counts every subset satisfying the retained intervals exactly once. The final state $m$ condition is exactly the requirement to hit the last interval, so $dp[m]$ is the required answer.\n\n**Complexity**\n\nLet $M=\\sum_k a_k$ be the number of generated good intervals. For a realizable input,\n$$M=O\\left(\\sum_{k=1}^n\\left(\\frac{n}{k}+1\\right)\\right)=O(n\\log n).$$\n\nGenerating intervals and sweeping take $O(M+n)$. Sorting and set operations take $O(M\\log M)$ time, with $O(M+n)$ memory. Thus the overall bound is $O(M\\log M+n)$, or $O(n\\log^2 n)$ in the worst case.",
          "solution": "#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nconst int MOD = 1'000'000'007;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nvoid solve() {\n    int n;\n    cin >> n;\n\n    vector<int> a(n);\n    for (int& x : a) cin >> x;\n\n    for (int k = 1; k <= n; ++k) {\n        if (a[k - 1] > (n - 1) / k + 1) {\n            cout << 0 << '\\\\n';\n            return;\n        }\n    }\n\n    vector<pair<int, int>> good;\n    vector<int> diff(n + 1, 0);\n\n    for (int k = 1; k <= n; ++k) {\n        int d = a[k - 1];\n\n        ll badL = 1LL * d * k;\n        ll badR = 1LL * (d + 1) * k - 1;\n        if (badL < n) {\n            int L = (int)badL;\n            int R = (int)min<ll>(n - 1, badR);\n            ++diff[L];\n            --diff[R + 1];\n        }\n\n        for (int q = 0; q < d; ++q) {\n            ll L = 1LL * q * k;\n            ll R = min<ll>(n - 1, 1LL * (q + 1) * k - 1);\n            good.push_back({(int)L, (int)R});\n        }\n    }\n\n    vector<int> freeValues;\n    int coverage = 0;\n    for (int x = 0; x < n; ++x) {\n        coverage += diff[x];\n        if (coverage == 0) freeValues.push_back(x);\n    }\n\n    sort(good.begin(), good.end(), [](const auto& x, const auto& y) {\n        int lenX = x.second - x.first;\n        int lenY = y.second - y.first;\n        if (lenX != lenY) return lenX < lenY;\n        return x < y;\n    });\n\n    set<pair<int, int>> kept;\n    for (auto [L, R] : good) {\n        auto it = kept.lower_bound({L, -1});\n        if (it == kept.end() || it->second > R) {\n            kept.insert({L, R});\n        }\n    }\n\n    vector<pair<int, int>> need(kept.begin(), kept.end());\n    int m = (int)need.size();\n\n    vector<int> dp(m + 1, 0);\n    dp[0] = 1;\n    ll active = 1;\n    int expired = 0;\n    int started = 0;\n\n    for (int x : freeValues) {\n        while (expired < m && need[expired].second < x) {\n            active = (active - dp[expired] + MOD) % MOD;\n            ++expired;\n        }\n\n        while (started < m && need[started].first <= x) {\n            ++started;\n        }\n\n        dp[started] = (dp[started] + active) % MOD;\n        active = 2 * active % MOD;\n    }\n\n    cout << dp[m] << '\\\\n';\n}\n\nint main() {\n    setIO();\n\n    int t;\n    cin >> t;\n    while (t--) solve();\n}\n"
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          "tool": "StructuredOutput",
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              "hints": [
                {
                  "order": 1,
                  "content": "Freeze one divisor $k$. The values `floor(y/k)` come from consecutive blocks $I_{k,q}=[qk,(q+1)k-1]$ (clipped to $0,\\ldots,n-1$). If $a_k=d$, which blocks must contain an element of $B$, and which block must contain none?"
                },
                {
                  "order": 2,
                  "content": "Turn that observation into constraints on individual values. The block for quotient $d$ is forbidden, so take the union of all such bad blocks. Every value outside that union is optional; the only remaining requirements are that every good block for $q=0,\\ldots,d-1$ is hit."
                },
                {
                  "order": 3,
                  "content": "Do not feed every good interval into the DP. If one interval is contained in another, hitting the smaller already hits the larger. Keep only a containment-minimal family; after sorting intervals by length, a set ordered by left endpoint can discard any current interval that contains an already kept one."
                },
                {
                  "order": 4,
                  "content": "After pruning, the retained intervals have increasing left and right endpoints. Sweep selectable values $x$ from left to right. Let state $j$ mean that the latest selected value has passed exactly $j$ retained left endpoints (state $0$ is the empty/no-selection case). What happens to state $j$ when its interval's right endpoint is already less than $x$?"
                },
                {
                  "order": 5,
                  "content": "Maintain $dp[j]$ and `active`, the number of still-viable subsets. At a selectable $x$, remove $dp[j]$ for every newly expired interval, let $r$ be the number of left endpoints at most $x$, do $dp[r] += active$ for choosing $x`, then double `active` for choosing or skipping $x$. The answer is $dp[m]$, where $m$ is the number of retained intervals; build selectable values with a difference array over bad blocks."
                }
              ],
              "editorial": "The floor function looks annoying, but it is just partitioning the universe into intervals.\n\n**Turn MEX into interval constraints**\n\nLet $U=\\{0,1,\\ldots,n-1\\}$. For a fixed $k$ and quotient $q$, define\n\n$$I_{k,q}=[qk,(q+1)k-1]\\cap U.$$\n\nEvery $y$ in this interval has `floor(y/k) = q`. Therefore, if $a_k=d$, then\n\n- every interval $I_{k,q}$ for $0 \\le q < d$ must contain at least one element of $B$;\n- the interval $I_{k,d}$ must contain no element of $B$.\n\nThe second interval is the **bad interval** for $k$. Every value in the union of all bad intervals is forced out of $B$. Call every other value **free**.\n\nAfter removing the forced values, the original problem is exactly this:\n\n- choose any subset of the free values;\n- hit every **good interval** $I_{k,q}$ with $q< a_k$.\n\nValues producing quotients larger than $a_k$ do not matter: MEX only cares about the first missing quotient. For $a_k=0$, there are no good intervals, and the quotient-zero interval is simply bad.\n\nThe input is guaranteed to be realizable. As a small defensive check, a feasible $a_k$ must satisfy\n$$a_k \\le \\left\\lfloor\\frac{n-1}{k}\\right\\rfloor+1,$$\nwhich is the number of quotient blocks that actually meet $U$.\n\n**Remove redundant good intervals**\n\nThere can be many good intervals. If a good interval $X$ is contained in another good interval $Y$, then hitting $X$ automatically hits $Y$, so $Y$ can be discarded.\n\nWe keep a containment-minimal family as follows:\n\n1. Sort all good intervals by nondecreasing length.\n2. Store kept intervals ordered by their left endpoint.\n3. For a current $[L,R]$, find the first kept interval whose left endpoint is at least $L$. If its right endpoint is at most $R$, the current interval contains a kept interval and is redundant. Otherwise, keep the current interval.\n\nWhy is checking only that one interval enough? Previously processed intervals are no longer than the current one, so a previous interval cannot strictly contain the current one. The kept intervals therefore never contain one another. When ordered by left endpoint, their right endpoints are also increasing, so the first possible left endpoint has the smallest possible right endpoint.\n\nLet the retained intervals be\n$$[L_0,R_0], [L_1,R_1], \\ldots, [L_{m-1},R_{m-1}],$$\nordered by left endpoint. Both the $L_i$ and the $R_i$ are increasing.\n\n**Sweep DP**\n\nOnly free values can be selected, so process them from left to right.\n\nFor a selected value $x$, define its state as the number of retained left endpoints it has passed:\n$$state(x)=|\\{i:L_i\\le x\\}|.$$\nThe empty subset has state $0$.\n\nLet $dp[j]$ count generated subsets whose latest selected free value has state $j$. Let `active` be the number of those subsets that have not already missed an interval whose right endpoint has passed. All counts are modulo $10^9+7$.\n\nSuppose the current free value is $x$.\n\n- While $R_i<x$, interval $i$ can no longer be hit by a future value. Among the still-relevant states, exactly state $i$ represents subsets whose latest selected value has not reached $L_i$, so those subsets missed interval $i$. Subtract $dp[i]$ from `active`. Since the $R_i$ are increasing, this is handled by one pointer.\n- Let $r$ be the number of retained intervals with $L_i\\le x$. Choosing $x$ creates one new subset from every active subset, and its latest-value state is $r$. Thus set `dp[r] += active`.\n- Skipping $x$ leaves every active subset unchanged, while choosing it adds one new branch for each. Therefore `active` doubles.\n\nThe old subsets are already represented in their old $dp$ states, so the transition only needs to add the choosing branch.\n\nAt the end, the answer is $dp[m]$. A valid subset must hit the last retained interval, so it must select some value at least $L_{m-1}$ and reach state $m$. Conversely, every subset counted in state $m$ survived all expiration checks and therefore hit every retained interval, hence every original good interval. If $m=0$, this naturally counts every subset of the free values, including the empty set.\n\n**Correctness argument**\n\nFor each $k$, the interval characterization is equivalent to $f(B,k)=a_k$: all smaller quotients occur, and quotient $a_k$ does not. Thus removing bad values and requiring every good interval to be hit is equivalent to satisfying every $k$ simultaneously.\n\nDiscarding a good interval only happens when a kept interval is contained inside it. Any set hitting the kept interval also hits the discarded one, so pruning preserves exactly the valid subsets.\n\nDuring the sweep, `active` contains exactly the choices of processed free values that have not missed an interval already passed by the sweep. Expiring state $i$ removes precisely the subsets that never selected a value in interval $i$; selecting or skipping the current free value enumerates its two possible choices without duplication. Hence the DP counts every subset satisfying the retained intervals exactly once. The final state $m$ condition is exactly the requirement to hit the last interval, so $dp[m]$ is the required answer.\n\n**Complexity**\n\nLet $M=\\sum_k a_k$ be the number of generated good intervals. For a realizable input,\n$$M=O\\left(\\sum_{k=1}^n\\left(\\frac{n}{k}+1\\right)\\right)=O(n\\log n).$$\n\nGenerating intervals and sweeping take $O(M+n)$. Sorting and set operations take $O(M\\log M)$ time, with $O(M+n)$ memory. Thus the overall bound is $O(M\\log M+n)$, or $O(n\\log^2 n)$ in the worst case.",
              "solution": "#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nconst int MOD = 1'000'000'007;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nvoid solve() {\n    int n;\n    cin >> n;\n\n    vector<int> a(n);\n    for (int& x : a) cin >> x;\n\n    for (int k = 1; k <= n; ++k) {\n        if (a[k - 1] > (n - 1) / k + 1) {\n            cout << 0 << '\\\\n';\n            return;\n        }\n    }\n\n    vector<pair<int, int>> good;\n    vector<int> diff(n + 1, 0);\n\n    for (int k = 1; k <= n; ++k) {\n        int d = a[k - 1];\n\n        ll badL = 1LL * d * k;\n        ll badR = 1LL * (d + 1) * k - 1;\n        if (badL < n) {\n            int L = (int)badL;\n            int R = (int)min<ll>(n - 1, badR);\n            ++diff[L];\n            --diff[R + 1];\n        }\n\n        for (int q = 0; q < d; ++q) {\n            ll L = 1LL * q * k;\n            ll R = min<ll>(n - 1, 1LL * (q + 1) * k - 1);\n            good.push_back({(int)L, (int)R});\n        }\n    }\n\n    vector<int> freeValues;\n    int coverage = 0;\n    for (int x = 0; x < n; ++x) {\n        coverage += diff[x];\n        if (coverage == 0) freeValues.push_back(x);\n    }\n\n    sort(good.begin(), good.end(), [](const auto& x, const auto& y) {\n        int lenX = x.second - x.first;\n        int lenY = y.second - y.first;\n        if (lenX != lenY) return lenX < lenY;\n        return x < y;\n    });\n\n    set<pair<int, int>> kept;\n    for (auto [L, R] : good) {\n        auto it = kept.lower_bound({L, -1});\n        if (it == kept.end() || it->second > R) {\n            kept.insert({L, R});\n        }\n    }\n\n    vector<pair<int, int>> need(kept.begin(), kept.end());\n    int m = (int)need.size();\n\n    vector<int> dp(m + 1, 0);\n    dp[0] = 1;\n    ll active = 1;\n    int expired = 0;\n    int started = 0;\n\n    for (int x : freeValues) {\n        while (expired < m && need[expired].second < x) {\n            active = (active - dp[expired] + MOD) % MOD;\n            ++expired;\n        }\n\n        while (started < m && need[started].first <= x) {\n            ++started;\n        }\n\n        dp[started] = (dp[started] + active) % MOD;\n        active = 2 * active % MOD;\n    }\n\n    cout << dp[m] << '\\\\n';\n}\n\nint main() {\n    setIO();\n\n    int t;\n    cin >> t;\n    while (t--) solve();\n}\n"
            },
            "output": "Structured output captured successfully.",
            "metadata": {
              "valid": true
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            "title": "Structured Output",
            "time": {
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              "end": 1789605631501
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