{
  "info": {
    "id": "ses_f4dad599bffeUKldhhBo3T2D03",
    "slug": "lucky-otter",
    "projectID": "global",
    "directory": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-xZ6Xpc",
    "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-xZ6Xpc",
    "title": "Nudge content generation",
    "agent": "nudge-generation",
    "model": {
      "id": "gpt-6-astra",
      "providerID": "openai",
      "variant": "low"
    },
    "version": "1.18.30",
    "summary": {
      "additions": 0,
      "deletions": 0,
      "files": 0
    },
    "cost": 0,
    "tokens": {
      "input": 37728,
      "output": 3760,
      "reasoning": 443,
      "cache": {
        "read": 25728,
        "write": 0
      }
    },
    "time": {
      "created": 1789698156132,
      "updated": 1789698296879
    }
  },
  "messages": [
    {
      "info": {
        "role": "user",
        "time": {
          "created": 1789698156151
        },
        "agent": "nudge-generation",
        "model": {
          "providerID": "openai",
          "modelID": "gpt-6-astra",
          "variant": "low"
        },
        "system": "You create Codeforces learning content for Nudge. Your job is to understand the best approach for the problem and teach it clearly in your own words, using existing solutions to inform your reasoning.\n\nStart with a brief, targeted web search for existing solutions so relevant editorials are in context before substantial independent problem-solving. Use any supplied tutorial links first; otherwise search by contest ID and problem index. Prefer the official tutorial/editorial, and read relevant explanations rather than relying on search snippets. Look at accessible alternatives when useful, especially if the official explanation is missing, unclear, or unnecessarily complicated. Accepted submissions can also help clarify an approach. Respect supplied source lookup statuses and avoid repeatedly retrying blocked sources. If no usable source is accessible, proceed from the supplied statement.\n\nTreat existing solutions as a starting point, not a constraint. If an approach is clear and suitable, build on it without unnecessary re-derivation. If an explanation has gaps, seems incorrect, or uses a harder method than necessary, reason through it and derive a better or simpler approach when worthwhile. Choose a correct approach that is easy to teach and fits the constraints. Verify it against the supplied statement and edge cases before writing the hints, editorial, and C++ solution.\n\nThe editorial must be self-contained: explain the key insight, algorithm, why it works, complexity, and implementation details a learner needs, filling in steps the source skips. Keep research and verification concise; spend the output on teaching the solution, not narrating the research process. Follow the user prompt and output schema exactly.",
        "format": {
          "type": "json_schema",
          "schema": {
            "type": "object",
            "properties": {
              "status": {
                "type": "string",
                "enum": [
                  "success",
                  "unsolvable"
                ],
                "description": "Set to 'success' when the supplied statement defines a solvable programming problem. Set to 'unsolvable' only if the statement is fundamentally incomplete, contradictory, or depends on an inaccessible required resource."
              },
              "reason": {
                "anyOf": [
                  {
                    "type": "string",
                    "pattern": "^[^\\u0000]*$",
                    "minLength": 1,
                    "maxLength": 2000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'unsolvable', provide a detailed failure message to show users. Name the concrete statement/resource blocker and any relevant source lookup/access status, such as missing official tutorial, 403/Cloudflare challenge, 404, or unavailable accepted submissions. Otherwise null."
              },
              "hints": {
                "anyOf": [
                  {
                    "type": "array",
                    "minItems": 5,
                    "maxItems": 5,
                    "description": "Exactly 5 progressive hints, each building on the last.",
                    "items": {
                      "type": "object",
                      "properties": {
                        "order": {
                          "type": "integer",
                          "minimum": 1,
                          "maximum": 5,
                          "description": "Hint number, 1 through 5."
                        },
                        "content": {
                          "type": "string",
                          "pattern": "^[^\\u0000]*$",
                          "minLength": 1,
                          "maxLength": 10000,
                          "description": "Markdown hint text."
                        }
                      },
                      "required": [
                        "order",
                        "content"
                      ],
                      "additionalProperties": false
                    }
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', provide exactly 5 progressive hints. Otherwise null."
              },
              "editorial": {
                "anyOf": [
                  {
                    "type": "string",
                    "pattern": "^[^\\u0000]*$",
                    "minLength": 1,
                    "maxLength": 100000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', a prose editorial explaining the solution in Nudge's own voice, with no links, citations, or source/research notes. Otherwise null."
              },
              "solution": {
                "anyOf": [
                  {
                    "type": "string",
                    "pattern": "^[^\\u0000]*$",
                    "minLength": 1,
                    "maxLength": 200000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', a complete C++ solution. Otherwise null."
              }
            },
            "required": [
              "status",
              "reason",
              "hints",
              "editorial",
              "solution"
            ],
            "additionalProperties": false,
            "title": "problem_response",
            "description": "Submit the generated content for the problem, or report if it is unsolvable."
          },
          "retryCount": 2
        },
        "id": "msg_0b252a677001wdK3bGintBCWRE",
        "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03"
      },
      "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 2248F: \"Matrix Elimination\" (rated 2500).\n\nKnown problem metadata:\n- Contest ID: 2248\n- Index: F\n- Name: Matrix Elimination\n- Rating: 2500 (Codeforces difficulty rating)\n- Tags: binary search, greedy, math\n- Problem URL: https://codeforces.com/contest/2248/problem/F\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\">F. Matrix Elimination</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>You are given an integer $$$k$$$ and a matrix $$$v$$$ with $$$n$$$ rows and $$$m$$$ columns. The element in row $$$i$$$ and column $$$j$$$ is denoted by $$$v_{i,j}$$$.</p><p>A cell $$$(x, y)$$$ is called a <span class=\"tex-font-style-it\">peak</span> if its value is greater than or equal to the sum of the values in all other cells in row $$$x$$$ and column $$$y$$$. Formally, $$$(x, y)$$$ is a peak if</p><p>$$$$$$ v_{x,y} \\ge \\sum_{\\substack{1 \\le i \\le n \\\\i \\neq x}} v_{i,y} + \\sum_{\\substack{1 \\le j \\le m \\\\j \\neq y}} v_{x,j}. $$$$$$</p><p>You may perform the following operation any number of times (possibly zero):</p><ul> <li> Choose four integers $$$x_l$$$, $$$x_r$$$, $$$y_l$$$, and $$$y_r$$$ ($$$1 \\le x_l \\le x_r \\le n$$$, $$$1 \\le y_l \\le y_r \\le m$$$). </li><li> Subtract $$$1$$$ from $$$v_{i,j}$$$ for every $$$x_l \\le i \\le x_r$$$ and $$$y_l \\le j \\le y_r$$$. </li></ul><p>Find the minimum number of operations required to obtain a matrix with at least $$$k$$$ peaks.</p></div><div class=\"input-specification\"><div class=\"section-title\">Input</div><p>Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 10^4$$$). The description of the test cases follows.</p><p>The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \\le n, m \\le 10^5$$$, $$$1 \\le k \\le n \\cdot m$$$, $$$n \\cdot m \\le 10^5$$$)&nbsp;— the number of rows, the number of columns, and the required number of peaks.</p><p>The $$$i$$$-th of the next $$$n$$$ lines contains $$$m$$$ integers $$$v_{i,1}, v_{i,2}, \\ldots, v_{i,m}$$$ ($$$-10^9 \\le v_{i,j} \\le 10^9$$$).</p><p>It is guaranteed that the sum of $$$n \\cdot m$$$ over all test cases does not exceed $$$10^5$$$.</p></div><div class=\"output-specification\"><div class=\"section-title\">Output</div><p>For each test case, output a single integer&nbsp;— the minimum number of operations required to obtain at least $$$k$$$ peaks.</p><p>If there is no solution, print a single integer $$$-1$$$.</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\">16</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">3 3 7</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">-1 -30 7</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">6 -3 22</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">1 -18 16</div><div class=\"test-example-line test-example-line-even test-example-line-2\">3 1 2</div><div class=\"test-example-line test-example-line-even test-example-line-2\">100000000</div><div class=\"test-example-line test-example-line-even test-example-line-2\">300000000</div><div class=\"test-example-line test-example-line-even test-example-line-2\">200000000</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">1 3 1</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">2 3 4</div><div class=\"test-example-line test-example-line-even test-example-line-4\">1 3 1</div><div class=\"test-example-line test-example-line-even test-example-line-4\">1 2 3</div><div class=\"test-example-line test-example-line-odd test-example-line-5\">1 2 2</div><div class=\"test-example-line test-example-line-odd test-example-line-5\">5 7</div><div class=\"test-example-line test-example-line-even test-example-line-6\">5 2 8</div><div class=\"test-example-line test-example-line-even test-example-line-6\">531959596 -61416172</div><div class=\"test-example-line test-example-line-even test-example-line-6\">425363565 672913308</div><div class=\"test-example-line test-example-line-even test-example-line-6\">981527099 451180253</div><div class=\"test-example-line test-example-line-even test-example-line-6\">-161687136 898803495</div><div class=\"test-example-line test-example-line-even test-example-line-6\">-388356105 897723313</div><div class=\"test-example-line test-example-line-odd test-example-line-7\">1 1 1</div><div class=\"test-example-line test-example-line-odd test-example-line-7\">0</div><div class=\"test-example-line test-example-line-even test-example-line-8\">1 1 1</div><div class=\"test-example-line test-example-line-even test-example-line-8\">-5</div><div class=\"test-example-line test-example-line-odd test-example-line-9\">1 4 3</div><div class=\"test-example-line test-example-line-odd test-example-line-9\">-8 -4 4 -11</div><div class=\"test-example-line test-example-line-even test-example-line-10\">1 9 5</div><div class=\"test-example-line test-example-line-even test-example-line-10\">-12 -4 -13 9 -1 -15 6 -15 -4</div><div class=\"test-example-line test-example-line-odd test-example-line-11\">1 3 3</div><div class=\"test-example-line test-example-line-odd test-example-line-11\">-12 -15 8</div><div class=\"test-example-line test-example-line-even test-example-line-12\">1 5 5</div><div class=\"test-example-line test-example-line-even test-example-line-12\">10 8 -4 0 -4</div><div class=\"test-example-line test-example-line-odd test-example-line-13\">1 3 3</div><div class=\"test-example-line test-example-line-odd test-example-line-13\">12 9 -1</div><div class=\"test-example-line test-example-line-even test-example-line-14\">1 3 3</div><div class=\"test-example-line test-example-line-even test-example-line-14\">-14 11 -6</div><div class=\"test-example-line test-example-line-odd test-example-line-15\">1 6 6</div><div class=\"test-example-line test-example-line-odd test-example-line-15\">-1 0 15 3 1 -14</div><div class=\"test-example-line test-example-line-even test-example-line-16\">1 2 2</div><div class=\"test-example-line test-example-line-even test-example-line-16\">1000000000 -1000000000</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\">2</div><div class=\"test-example-line test-example-line-even test-example-line-2\">100000000</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">1</div><div class=\"test-example-line test-example-line-even test-example-line-4\">0</div><div class=\"test-example-line test-example-line-odd test-example-line-5\">2</div><div class=\"test-example-line test-example-line-even test-example-line-6\">734592698</div><div class=\"test-example-line test-example-line-odd test-example-line-7\">0</div><div class=\"test-example-line test-example-line-even test-example-line-8\">-1</div><div class=\"test-example-line test-example-line-odd test-example-line-9\">0</div><div class=\"test-example-line test-example-line-even test-example-line-10\">0</div><div class=\"test-example-line test-example-line-odd test-example-line-11\">11</div><div class=\"test-example-line test-example-line-even test-example-line-12\">6</div><div class=\"test-example-line test-example-line-odd test-example-line-13\">12</div><div class=\"test-example-line test-example-line-even test-example-line-14\">11</div><div class=\"test-example-line test-example-line-odd test-example-line-15\">7</div><div class=\"test-example-line test-example-line-even test-example-line-16\">2000000000</div></pre></div></div></div><div class=\"note\"><div class=\"section-title\">Note</div><p>For the first test case, you can perform the following two operations: </p><ul> <li> choose $$$(x_l, x_r, y_l, y_r) = (1, 3, 2, 3)$$$; </li><li> choose $$$(x_l, x_r, y_l, y_r) = (2, 3, 1, 3)$$$. </li></ul><p>After these operations, the matrix will be: </p><center> <table class=\"tex-tabular\"><tbody><tr><td class=\"tex-tabular-border-left tex-tabular-text-align-center tex-tabular-border-right tex-tabular-border-top tex-tabular-border-bottom\">$$$\\color{green}{-1}$$$</td><td class=\"tex-tabular-border-left tex-tabular-text-align-center tex-tabular-border-right tex-tabular-border-top tex-tabular-border-bottom\">$$$-31$$$</td><td class=\"tex-tabular-border-left tex-tabular-text-align-center tex-tabular-border-right tex-tabular-border-top tex-tabular-border-bottom\">$$$\\color{green}{6}$$$</td></tr><tr><td class=\"tex-tabular-border-left tex-tabular-text-align-center tex-tabular-border-right tex-tabular-border-top tex-tabular-border-bottom\">$$$5$$$</td><td class=\"tex-tabular-border-left tex-tabular-text-align-center tex-tabular-border-right tex-tabular-border-top tex-tabular-border-bottom\">$$$\\color{green}{-5}$$$</td><td class=\"tex-tabular-border-left tex-tabular-text-align-center tex-tabular-border-right tex-tabular-border-top tex-tabular-border-bottom\">$$$\\color{green}{20}$$$</td></tr><tr><td class=\"tex-tabular-border-left tex-tabular-text-align-center tex-tabular-border-right tex-tabular-border-top tex-tabular-border-bottom\">$$$\\color{green}{0}$$$</td><td class=\"tex-tabular-border-left tex-tabular-text-align-center tex-tabular-border-right tex-tabular-border-top tex-tabular-border-bottom\">$$$\\color{green}{-20}$$$</td><td class=\"tex-tabular-border-left tex-tabular-text-align-center tex-tabular-border-right tex-tabular-border-top tex-tabular-border-bottom\">$$$\\color{green}{14}$$$</td></tr></tbody></table> </center><p>The seven peaks are highlighted in green. For example: </p><ul> <li> The cell $$$(1, 1)$$$ is a peak because $$$-1 \\ge 5 + 0 - 31 + 6 = -20$$$; </li><li> The cell $$$(3, 1)$$$ is a peak because $$$0 \\ge -20 + 14 - 1 + 5 = -2$$$; </li><li> The cell $$$(2, 1)$$$ is not a peak because $$$5  \\lt  -5 + 20 - 1 + 0 = 14$$$. </li></ul><p>It can be shown that fewer than two operations cannot create seven peaks, so the answer is $$$2$$$.</p><p>In the eighth test case, the only cell has value $$$-5$$$. A cell in a $$$1 \\times 1$$$ matrix is a peak if and only if its value is non-negative. Since every operation only decreases its value, it is impossible to make it a peak.</p><p>For the last test case, both cells are peaks exactly when their values are equal. Therefore, the first cell must be decreased from $$$10^9$$$ to $$$-10^9$$$, which requires $$$2 \\cdot 10^9$$$ operations.</p></div>\n</problem-statement>\n\n\nSource lookup status:\n<source-lookup-status>\n- Tutorial (en): https://codeforces.com/blog/entry/155640 loaded (200 OK; title: \"Codeforces Round 1113 (Div. 2) Editorial - Codeforces\"; mentions 2248F)\n</source-lookup-status>\n\nGenerate:\n1. Five progressive hints, from a gentle nudge to the key insight.\n2. A self-contained editorial that teaches the solution clearly, including the key insight, algorithm, correctness argument, complexity, and necessary implementation details. Explain non-obvious steps without padding or a discovery-process transcript.\n3. A complete C++26 solution that gets AC on Codeforces.\n\nFormatting & Style Rules:\n- Hints and editorial: valid Markdown with LaTeX as needed (e.g., $dp[i]$, $$\\sum_{i=1}^{n} a_i$$).\n- Hints and editorial must read like Nudge's own explanation. No research notes, source notes, citations, Markdown links, URLs, or references to editorials/submissions/posts.\n- Solution: raw C++ only, no Markdown fences.\n- Keep C++ short, clean, standard, and single-file. Comments are fine when they clarify the idea.\n\nOutput strictness:\n- Return JSON matching the provided schema exactly.\n- Never include NUL (U+0000) characters in any field. Use literal $ or $$ for math delimiters, never control characters.\n- If the problem is solvable, return `status: \"success\"`, `reason: null`, and fill in `hints`, `editorial`, and `solution`.\n- Return `status: \"unsolvable\"`, a short `reason`, and set `hints`, `editorial`, and `solution` to null only when the supplied statement is fundamentally incomplete, contradictory, or dependent on an inaccessible resource needed to define the task. Missing research, high rating, uncertainty, or \"can't guarantee AC\" is not a reason to return unsolvable.\n- If you return `status: \"unsolvable\"`, the `reason` is shown directly to users. Be concrete: say what exact statement/resource blocker stopped you, and include relevant source lookup status if it mattered (for example: official tutorial missing, Codeforces returned 403/Cloudflare challenge, tutorial page 404, accepted submissions unavailable).\n- Each hint must be an object with `order` (1 through 5) and `content` (the Markdown hint text), not a plain string. The `content` must contain just the hint text, with no \"Hint 1:\" or subtitles. The UI adds those automatically.\n- Do not start the editorial with an \"# Editorial\" heading. The UI already adds that section. Feel free to add other headers as needed though.\n- Use styling in the editorial if needed! Just that first header is no good but any subheaders or bolding or whatever is good!\n\nFor the C++ solution, you MUST use this template and work around it:\n\n```cpp\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n}\n```",
          "id": "prt_0b252a6780013QUa5ZKsZiRvYZ",
          "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03",
          "messageID": "msg_0b252a677001wdK3bGintBCWRE"
        }
      ]
    },
    {
      "info": {
        "parentID": "msg_0b252a677001wdK3bGintBCWRE",
        "role": "assistant",
        "mode": "nudge-generation",
        "agent": "nudge-generation",
        "variant": "low",
        "path": {
          "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-xZ6Xpc",
          "root": "/"
        },
        "cost": 0,
        "tokens": {
          "total": 10689,
          "input": 10657,
          "output": 32,
          "reasoning": 0,
          "cache": {
            "write": 0,
            "read": 0
          }
        },
        "modelID": "gpt-6-astra",
        "providerID": "openai",
        "time": {
          "created": 1789698156285,
          "completed": 1789698165230
        },
        "finish": "tool-calls",
        "id": "msg_0b252a6fd001bEvtt88FQSK0af",
        "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03"
      },
      "parts": [
        {
          "type": "step-start",
          "id": "prt_0b252b4d10010DybwywgD7OxSY",
          "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03",
          "messageID": "msg_0b252a6fd001bEvtt88FQSK0af"
        },
        {
          "type": "tool",
          "tool": "webfetch",
          "callID": "call_fIPgPpNIn6kR5GlN81jobo7d",
          "state": {
            "status": "completed",
            "input": {
              "url": "https://codeforces.com/blog/entry/155640",
              "format": "markdown"
            },
            "output": "---\nimage: https://codeforces.com/codeforces.org/s/64117/images/codeforces-telegram-square-1024x1024.png\ntitle: Codeforces Round 1113 (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/64117/images/codeforces-logo-with-telegram-555x130.png \"Codeforces\")](https://codeforces.com/) \n\n[![In English](//codeforces.com/codeforces.org/s/64117/images/flags/24/gb.png \"In English\")](https://codeforces.com/blog/entry/155640?locale=en) [![По-русски](//codeforces.com/codeforces.org/s/64117/images/flags/24/ru.png \"По-русски\")](https://codeforces.com/blog/entry/155640?locale=ru) \n\n[Enter](https://codeforces.com/enter?back=%2Fblog%2Fentry%2F155640)| [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* [Rating](https://codeforces.com/ratings)\n* [Edu](https://codeforces.com/edu/courses)\n* [API](https://codeforces.com/apiHelp)\n* [Calendar](https://codeforces.com/calendar)\n* [Help](https://codeforces.com/help)\n  \n  \n→ Pay attention \n\nBefore contest  \n[Codeforces Round 1122 (Div. 3)](https://codeforces.com/contests/2266)  \n4 days  \n[Register now »](https://codeforces.com/contestRegistration/2266) \n\n→ Top rated \n\n| #  | User                                                                                          | Rating |\n| -- | --------------------------------------------------------------------------------------------- | ------ |\n| 1  | [jiangly](https://codeforces.com/profile/jiangly \"Legendary Grandmaster jiangly\")             | 3810   |\n| 2  | [Kevin114514](https://codeforces.com/profile/Kevin114514 \"Legendary Grandmaster Kevin114514\") | 3655   |\n| 3  | [Benq](https://codeforces.com/profile/Benq \"Legendary Grandmaster Benq\")                      | 3650   |\n| 4  | [maroonrk](https://codeforces.com/profile/maroonrk \"Legendary Grandmaster maroonrk\")          | 3427   |\n| 5  | [strapple](https://codeforces.com/profile/strapple \"Legendary Grandmaster strapple\")          | 3413   |\n| 6  | [Um\\_nik](https://codeforces.com/profile/Um%5Fnik \"Legendary Grandmaster Um_nik\")             | 3387   |\n| 7  | [heuristica](https://codeforces.com/profile/heuristica \"Legendary Grandmaster heuristica\")    | 3322   |\n| 8  | [tourist](https://codeforces.com/profile/tourist \"Legendary Grandmaster tourist\")             | 3301   |\n| 9  | [turmax](https://codeforces.com/profile/turmax \"Legendary Grandmaster turmax\")                | 3297   |\n| 10 | [squareOf105](https://codeforces.com/profile/squareOf105 \"Legendary Grandmaster squareOf105\") | 3262   |\n\n| [Countries](https://codeforces.com/ratings/countries) \\| [Cities](https://codeforces.com/ratings/cities) | [Organizations](https://codeforces.com/ratings/organizations) | [View all →](https://codeforces.com/ratings) |\n| -------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------- | -------------------------------------------- |\n\n→ Top contributors \n\n| #  | User                                                                                     | Contrib. |\n| -- | ---------------------------------------------------------------------------------------- | -------- |\n| 1  | [Qingyu](https://codeforces.com/profile/Qingyu \"Legendary Grandmaster Qingyu\")           | 157      |\n| 2  | [nik\\_exists](https://codeforces.com/profile/nik%5Fexists \"Expert nik_exists\")           | 150      |\n| 2  | [maspy](https://codeforces.com/profile/maspy \"Legendary Grandmaster maspy\")              | 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| 9  | [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\") | 132      |\n| 10 | [qwexd](https://codeforces.com/profile/qwexd \"Master qwexd\")                             | 129      |\n\n|  | [View all →](https://codeforces.com/top-contributed) |\n|  | ---------------------------------------------------- |\n\n→ Find user \n\nHandle: \n\n→ Recent actions \n\n* [prabowo](https://codeforces.com/profile/prabowo \"Grandmaster prabowo\") → [Invitation to TOKI Indonesian NOI Open Contest 2026](https://codeforces.com/blog/entry/156765) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\")\n* [root.svp](https://codeforces.com/profile/root.svp \"Newbie root.svp\") → [An question about the comunity of codeforces ](https://codeforces.com/blog/entry/156870) ![Text created or updated](//codeforces.com/codeforces.org/s/64117/images/icons/x-update-12x12.png \"Text created or updated\")\n* [WorldWarV](https://codeforces.com/profile/WorldWarV \"Candidate Master WorldWarV\") → [Codeforces Round 1122 (Div. 3)](https://codeforces.com/blog/entry/156834) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\")\n* [AmShZ](https://codeforces.com/profile/AmShZ \"Grandmaster AmShZ\") → [Repovive Premier Round 8 — $300 Prize Pool](https://codeforces.com/blog/entry/156827) ![Text created or updated](//codeforces.com/codeforces.org/s/64117/images/icons/x-update-12x12.png \"Text created or updated\")\n* [turmax](https://codeforces.com/profile/turmax \"Legendary Grandmaster turmax\") → [Are we ready for AGI?](https://codeforces.com/blog/entry/156668) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\")\n* [destructive\\_criticism](https://codeforces.com/profile/destructive%5Fcriticism \"Master destructive_criticism\") → [2026 Turkish Team Selection Problems Are Now Public!](https://codeforces.com/blog/entry/156803) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\")\n* [mariza\\_CY](https://codeforces.com/profile/mariza%5FCY \"Candidate Master mariza_CY\") → [Balkan OI 2026 teams](https://codeforces.com/blog/entry/153535) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\")\n* [qwexd](https://codeforces.com/profile/qwexd \"Master qwexd\") → [Codeforces Round 1121 (Div. 2)](https://codeforces.com/blog/entry/156653) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\")\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) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\")\n* [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\") → [Educational Codeforces Round 194 - Editorial](https://codeforces.com/blog/entry/156529) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\")\n* [AlirezaBest](https://codeforces.com/profile/AlirezaBest \"Pupil AlirezaBest\") → [Is this thing regular?](https://codeforces.com/blog/entry/156867) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\")\n* [ashutosh2711](https://codeforces.com/profile/ashutosh2711 \"Newbie ashutosh2711\") → [I’m Still Bad at CP, But Here Is What 30 Days of Failing Taught Me](https://codeforces.com/blog/entry/156866) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\")\n* [brunomont](https://codeforces.com/profile/brunomont \"Master brunomont\") → [tgen::miniblog(5): random graphs with large diameter](https://codeforces.com/blog/entry/156865) ![Text created or updated](//codeforces.com/codeforces.org/s/64117/images/icons/x-update-12x12.png \"Text created or updated\")\n* [DuyMinh3005](https://codeforces.com/profile/DuyMinh3005 \"Pupil DuyMinh3005\") → [Is this acceptable?](https://codeforces.com/blog/entry/155682) ![Text created or updated](//codeforces.com/codeforces.org/s/64117/images/icons/x-update-12x12.png \"Text created or updated\")\n* [UNDERTAKER](https://codeforces.com/profile/UNDERTAKER \"Grandmaster UNDERTAKER\") → [Looking for contests to host — we'll adapt our judge to your problems](https://codeforces.com/blog/entry/156823) ![Text created or updated](//codeforces.com/codeforces.org/s/64117/images/icons/x-update-12x12.png \"Text created or updated\")\n* [Ovi\\_404](https://codeforces.com/profile/Ovi%5F404 \"Pupil Ovi_404\") → [Finally hit 1000 problems solved on Codeforces](https://codeforces.com/blog/entry/156830) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\")\n* [yse](https://codeforces.com/profile/yse \"Candidate Master yse\") → [Codeforces Round 1114 (Div. 3) — Editorial](https://codeforces.com/blog/entry/155666) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\")\n* [M.Bemanali](https://codeforces.com/profile/M.Bemanali \"Newbie M.Bemanali\") → [QueueForces AGAIN???](https://codeforces.com/blog/entry/156822) ![Text created or updated](//codeforces.com/codeforces.org/s/64117/images/icons/x-update-12x12.png \"Text created or updated\")\n* [BitByBit123](https://codeforces.com/profile/BitByBit123 \"Master BitByBit123\") → [Suspicious Master yql0991666](https://codeforces.com/blog/entry/155862) ![Text created or updated](//codeforces.com/codeforces.org/s/64117/images/icons/x-update-12x12.png \"Text created or updated\")\n* [lumibons](https://codeforces.com/profile/lumibons \"International Grandmaster lumibons\") → [IOI 2027 Call for Tasks](https://codeforces.com/blog/entry/156793) ![Text created or updated](//codeforces.com/codeforces.org/s/64117/images/icons/x-update-12x12.png \"Text created or updated\")\n* [Grizoo](https://codeforces.com/profile/Grizoo \"Candidate Master Grizoo\") → [\\[GYM\\] Aleppo Collegiate Programming Contest 2026](https://codeforces.com/blog/entry/156707) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\")\n* [MikeMirzayanov](https://codeforces.com/profile/MikeMirzayanov \"Headquarters, MikeMirzayanov\") → [Rule Restricting the Use of AI \\[revision 2024-09-14\\]](https://codeforces.com/blog/entry/133941) ![Text created or updated](//codeforces.com/codeforces.org/s/64117/images/icons/x-update-12x12.png \"Text created or updated\") ![Necropost](//codeforces.com/codeforces.org/s/64117/images/icons/hourglass.png \"Necropost\")\n* [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") → [Codeforces Round 1120 (Div 1, Div 2) Editorial](https://codeforces.com/blog/entry/156688) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\")\n* [borgar02](https://codeforces.com/profile/borgar02 \"Master borgar02\") → [IATI 2022 editorial](https://codeforces.com/blog/entry/109545) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\") ![Necropost](//codeforces.com/codeforces.org/s/64117/images/icons/hourglass.png \"Necropost\")\n* [Kerolloz](https://codeforces.com/profile/Kerolloz \"Newbie Kerolloz\") → [Codeforces JSON-Problem-Scraper API](https://codeforces.com/blog/entry/75074) ![Text created or updated](//codeforces.com/codeforces.org/s/64117/images/icons/x-update-12x12.png \"Text created or updated\") ![Necropost](//codeforces.com/codeforces.org/s/64117/images/icons/hourglass.png \"Necropost\")\n\n|  | [Detailed →](https://codeforces.com/recent-actions) |\n|  | --------------------------------------------------- |\n\n* [szdytom](https://codeforces.com/profile/szdytom)\n* [Blog](https://codeforces.com/blog/szdytom)\n* [Teams](https://codeforces.com/teams/with/szdytom)\n* [Submissions](https://codeforces.com/submissions/szdytom)\n* [Groups](https://codeforces.com/groups/with/szdytom)\n* [Contests](https://codeforces.com/contests/with/szdytom)\n* [Problemsetting](https://codeforces.com/contests/writer/szdytom)\n\n### [szdytom's blog](https://codeforces.com/blog/szdytom)\n\n[ Codeforces Round 1113 (Div. 2) Editorial ](https://codeforces.com/blog/entry/155640) \n\nBy [szdytom](https://codeforces.com/profile/szdytom \"Master szdytom\"), 7 weeks ago, ![In English](//codeforces.com/codeforces.org/s/64117/images/flags/24/gb.png \"In English\") \n\n[2248A - You Delete, I Delete](https://codeforces.com/contest/2248/problem/A \"Codeforces Round 1113 (Div. 2)\")\n\n**Hint1**\n\nAlice deletes a `0`, Bob deletes a `1`. Which occurrence should each of them choose?\n\n**Hint2**\n\nDeleting a `0` makes the string lexicographically larger. Does the position of that `0` matter?\n\n**Hint3**\n\nAfter Alice's move, Bob sees one string and deletes a `1`. What is his best move?\n\n**Solution**\n\nDeleting a `0` at an earlier position is always better for Alice. Indeed, compare deleting the `0`s at positions $$$i \\\\lt j$$$: the two results agree before position $$$i$$$, and at position $$$i$$$ the first result has `1` while the second still has `0`. So Alice deletes the **first** `0`.\n\nAfter that, Bob wants to minimize the string. By the symmetric argument he deletes the **first** `1` of the remaining string, which is the first `1` of the original string, since deleting a `0` never changes the order of the `1`s.\n\nHence the answer is the original string with its first `0` and its first `1` removed.\n\n**Complexity:** $$$O(n)$$$ per test case.\n\n**Code**\n\n```\n#include <cstdio>\n#include <cstring>\nconst int N = 105;\nchar str[N];\nvoid solve() {\n    scanf(\"%s\", str + 1);\n    int n = strlen(str + 1);\n    bool c0 = 0, c1 = 0;\n    for (int i = 1; i <= n; i++) {\n        if (!c0 && str[i] == '0') {\n            c0 = true;\n            continue;\n        }\n        if (!c1 && str[i] == '1') {\n            c1 = true;\n            continue;\n        }\n        putchar(str[i]);\n    }\n    puts(\"\");\n}\nint main() {\n    int t = 0;\n    scanf(\"%d\", &t);\n    while (t--) solve();\n    return 0;\n}\n\n```\n\n**Code (Brute Force)**\n\n```\n#include<bits/stdc++.h>\nusing namespace std;\n\n#define ll long long\n#define nline \"\\n\"\n#define sz(x) (int)x.size()\n\nvoid solve(){\n\tstring s; cin>>s;\n\tstring ans;\n\tfor(int i=0;i<sz(s);i++){\n\t\tif(s[i]=='1'){\n\t\t\tcontinue;\n\t\t}\n\t\tstring t=s;\n\t\tt.erase(t.begin()+i);\n\t\tstring cur;\n\t\tfor(int j=0;j<sz(t);j++){\n\t\t\tif(t[j]=='0'){\n\t\t\t\tcontinue;\n\t\t\t}\n\t\t\tstring now=t;\n\t\t\tnow.erase(now.begin()+j);\n\t\t\tif((cur.empty()) or (now<cur)){\n\t\t\t\tcur=now;\n\t\t\t}\n\t\t}\n\t\tif((ans.empty()) or (ans<cur)){\n\t\t\tans=cur;\n\t\t}\n\t}\n\tcout<<ans<<nline;\n\treturn;\n}\n\nint main(){\n\tios_base::sync_with_stdio(false);\n\tcin.tie(NULL);\n\tll test_cases=1;\n\tcin>>test_cases;\n\twhile(test_cases--){\n\t\tsolve();\n\t}\n}\n\n```\n\n[2248B - Merge to Match](https://codeforces.com/contest/2248/problem/B \"Codeforces Round 1113 (Div. 2)\")\n\n**Hint1**\n\nSort both arrays. Each operation reduces the number of elements by exactly one.\n\n**Hint2**\n\nAll values are distinct, so no final element can be an untouched original element. What lower bound on $$$n$$$ does this give?\n\n**Hint3**\n\nAfter sorting, in which interval of $$$a$$$ must the $$$i$$$-th smallest value $$$b\\_i$$$ lie?\n\n**Solution**\n\nSort both arrays. An operation merges two elements $$$x \\\\le y$$$ into any $$$z \\\\in \\[x,y\\]$$$, so think of each final element as the merge of a group of original elements whose min/max bracket its value.\n\n**Necessity.**\n\n* **$$$n \\\\ge 2m$$$.** The values in $$$b$$$ are distinct from those in $$$a$$$, so a final element consisting of a single original element is impossible. Every final element needs at least two originals.\n* **$$$a\\_i \\\\le b\\_i \\\\le a\\_{n-m+i}$$$.** The $$$i$$$ smallest final values are produced by $$$i$$$ groups containing at least $$$i$$$ original elements, all $$$\\\\le b\\_i$$$. Only $$$i-1$$$ original elements are smaller than $$$a\\_i$$$, so $$$b\\_i \\\\ge a\\_i$$$. The symmetric argument on the largest values gives $$$b\\_i \\\\le a\\_{n-m+i}$$$.\n\n**Sufficiency.** If the conditions hold, pair $$$a\\_i$$$ with $$$a\\_{n-m+i}$$$ for each $$$i$$$ (the pairs are disjoint since $$$n \\\\ge 2m$$$) and merge each pair into $$$b\\_i$$$. The remaining middle elements can be attached to any group one by one: merging the current group value $$$b\\_1$$$ with an extra element $$$e$$$ yields $$$b\\_1$$$ again, because $$$b\\_1$$$ lies between them.\n\n**Complexity:** $$$O((n+m)\\\\log(n+m))$$$ per test case.\n\n**Code**\n\n```\nusing namespace std;\n\nint main() {\n\tint T;\n\tscanf(\"%d\", &T);\n\tfor (int _ = 0; _ < T; ++_) {\n\t\tint n, m;\n\t\tscanf(\"%d %d\", &n, &m);\n\t\tvector<int> a(n), b(m);\n\t\tfor (int &x : a) {\n\t\t    scanf(\"%d\", &x);\n\t\t}\n\t\tfor (int &x : b) {\n\t\t\tscanf(\"%d\", &x);\n\t\t}\n\t\tsort(a.begin(), a.end());\n\t\tsort(b.begin(), b.end());\n\n\t\tif (n < 2 * m) {\n\t\t\tputs(\"NO\");\n\t\t\tcontinue;\n\t\t}\n\n\t\tint i = 0;\n\t\twhile (i < m && a[i] < b[i] && b[i] < a[n - m + i]) ++i;\n\t\tputs(i < m ? \"no\" : \"Yes\");\n\t}\n\treturn 0;\n}\n\n```\n\n[2248C - Maximize the Score](https://codeforces.com/contest/2248/problem/C \"Codeforces Round 1113 (Div. 2)\")\n\n**Hint1**\n\nIf a value is chosen when some elements between its two occurrences were already deleted, move its deletion earlier. When is this never worse? Think about $$$(a+b)^2 \\\\ge a^2 + b^2$$$.\n\n**Hint2**\n\nConclude that the process is equivalent to partitioning the array into disjoint blocks: each block is either a single position, or the whole interval between two equal values at its ends.\n\n**Hint3**\n\nDP over the prefix. Consider the block containing the last position.\n\n**Solution**\n\nCall an operation **full** if it deletes the entire original interval between the two occurrences of the chosen value.\n\n**Lemma.** There is an optimal strategy in which every operation is full.\n\nSuppose a value is chosen with current span $$$p \\\\dots q$$$, while an interval $$$J$$$ strictly inside its original span was deleted earlier. Swap the two deletions: the value now covers the whole original interval. If the two original lengths are $$$A$$$ and $$$B$$$ ($$$A \\\\ge B$$$), the score changes from at most $$$(A-B)^2 + B^2$$$ to $$$A^2$$$, and $$$(A-B)^2 + B^2 \\\\le A^2$$$. So the total never decreases, and repeating this exchange makes every operation full.\n\nNow the deleted intervals are disjoint, and the whole process is exactly: partition the array into blocks, each being a single position or the interval between the two equal values at its ends, and collect $$$\\\\sum (\\\\text{block length})^2$$$.\n\nLet $$$dp\\[i\\]$$$ be the best score for the prefix $$$a\\_1 \\\\dots a\\_i$$$. The block containing position $$$i$$$ is either the singleton $$${i}$$$, or the interval $$$\\[l,i\\]$$$ where $$$l$$$ is the first occurrence of $$$a\\_i$$$:\n\n$$$ dp\\[i\\] = \\\\max\\\\big(dp\\[i-1\\] + 1,\\\\; dp\\[l-1\\] + (i-l+1)^2\\\\big). $$$\n\nThe singleton option never overestimates: if the mate of $$$a\\_i$$$ is still present in the optimal strategy of the prefix, deleting $$$\\[l,i\\]$$$ instead is at least as good, since $$$(i-l+1)^2 \\\\ge (i-l)^2 + 1$$$.\n\nThe answer is $$$dp\\[2n\\]$$$.\n\n**Complexity:** $$$O(n)$$$ per test case.\n\n**Code**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nint main() {\n\tint T;\n\tscanf(\"%d\", &T);\n\tfor (int _ = 0; _ < T; ++_) {\n\t\tint n;\n\t\tscanf(\"%d\", &n);\n\t\tvector<int> a(2 * n);\n\t\tfor (int &x : a) {\n\t\t\tscanf(\"%d\", &x);\n\t\t}\n\n\t\tvector<int> lp(2 * n, -1);\n\t\tvector<int> occ(n, -1);\n\t\tfor (int i = 0; i < 2 * n; ++i) {\n\t\t\tif (occ[a[i] - 1] != -1) {\n\t\t\t\tlp[i] = occ[a[i] - 1];\n\t\t\t} else {\n\t\t\t\tocc[a[i] - 1] = i;\n\t\t\t}\n\t\t}\n\n\t\tvector<long long> dp(2 * n);\n\t\tdp[0] = 1;\n\t\tfor (int i = 1; i < 2 * n; ++i) {\n\t\t\tdp[i] = dp[i - 1] + 1;\n\t\t\tif (lp[i] != -1) {\n\t\t\t\tdp[i] = max(dp[i], (lp[i] == 0 ? 0 : dp[lp[i] - 1]) + 1LL * (i - lp[i] + 1) * (i - lp[i] + 1));\n\t\t\t}\n\t\t}\n\t\tprintf(\"%lld\\n\", dp[2 * n - 1]);\n\t}\n}\n\n```\n\n[2248D - Good Pair Queries](https://codeforces.com/contest/2248/problem/D \"Codeforces Round 1113 (Div. 2)\")\n\n**Hint1**\n\nClassify each position by the pair $$$(s\\_i, t\\_i)$$$ and count the four types: $$$(0,0)$$$, $$$(0,1)$$$, $$$(1,0)$$$, $$$(1,1)$$$.\n\n**Hint2**\n\nFind a quantity that only decreases during the game, and equals $$$0$$$ at the end.\n\n**Hint3**\n\nWhich pairs of columns can be deleted together in one operation? When can all columns be paired up?\n\n**Solution**\n\nFor a substring pair let $$$x=N(0,1)$$$, $$$y=N(1,0)$$$, $$$u=N(0,0)$$$, $$$v=N(1,1)$$$, where $$$N(0,1)$$$ denotes the number occurrences of $$$(0,1)$$$.\n\n**Claim.** The pair is good iff $$$|x-y| \\\\le u+v$$$.\n\n**Necessity**: Take any allowed operation. Let the deleted columns have counts $$$u\\_0, v\\_0, x\\_0, y\\_0$$$ and total $$$k = u\\_0+v\\_0+x\\_0+y\\_0$$$. \n\n**Case $$$c=0$$$:**  \nIn the deleted subsequence of $$$s$$$, character $$$0$$$ appears $$$u\\_0+x\\_0$$$ times; in $$$t$$$ it appears $$$u\\_0+y\\_0$$$ times.  \nMode condition: \n\n$$$ u\\_0+x\\_0 \\\\ge \\\\lceil k/2 \\\\rceil,\\\\qquad u\\_0+y\\_0 \\\\ge \\\\lceil k/2 \\\\rceil . $$$\n\nAdding them: $$$2u\\_0 + x\\_0 + y\\_0 \\\\ge k = u\\_0+v\\_0+x\\_0+y\\_0$$$ $$$\\\\Rightarrow$$$ $$$u\\_0 \\\\ge v\\_0$$$.  \nSubtracting one from the other gives $$$|x\\_0 - y\\_0| \\\\le u\\_0 - v\\_0 \\\\le u\\_0 + v\\_0$$$. \n\n**Case $$$c=1$$$:** similarly $$$v\\_0 \\\\ge u\\_0$$$ and $$$|x\\_0 - y\\_0| \\\\le v\\_0 - u\\_0 \\\\le u\\_0 + v\\_0$$$. \n\nThus **every operation** satisfies $$$|x\\_0 - y\\_0| \\\\le u\\_0+v\\_0$$$.\n\nNow define the potential $$$\\\\Phi = |x-y| - (u+v)$$$.  \nAfter one operation the change is \n\n$$$ \\\\Phi' - \\\\Phi = \\\\bigl(|x-x\\_0 - (y-y\\_0)| - |x-y|\\\\bigr) + (u\\_0+v\\_0). $$$\n\nBy the triangle inequality, $$$|x-y| - |(x-y)-(x\\_0-y\\_0)| \\\\le |x\\_0-y\\_0|$$$.  \nTherefore \n\n$$$ \\\\Phi' - \\\\Phi = (u\\_0+v\\_0) - \\\\bigl(|x-y| - |(x-y)-(x\\_0-y\\_0)|\\\\bigr) \\\\ge u\\_0+v\\_0 - |x\\_0-y\\_0| \\\\ge 0\\. $$$\n\nSo $$$\\\\Phi$$$ never decreases.  \nThe empty pair has $$$\\\\Phi = 0$$$. Hence initially $$$\\\\Phi \\\\le 0$$$, i.e. $$$|x-y| \\\\le u+v$$$.\n\n**Sufficiency**: We can empty the strings by repeatedly deleting columns in groups of size 1 or 2.\n\nA single column can be deleted alone **iff** it is pure: $$$(0,0)$$$ (mode $$$0$$$) or $$$(1,1)$$$ (mode $$$1$$$). Two columns can be deleted together **iff** they are **not** two identical mixed columns:\n\n* $$$(0,0)$$$ pairs with anything.\n* $$$(1,1)$$$ pairs with anything.\n* $$$(0,1)$$$ pairs with $$$(1,0)$$$.\n\nSo we can pair $$$\\\\min(x,y)$$$ copies of $$$(0,1)$$$ with $$$(1,0)$$$, and the remaining $$$|x-y|$$$ mixed columns (all of the same type) are paired with pure columns $$$(0,0)$$$ or $$$(1,1)$$$. This is possible exactly because $$$u+v \\\\ge |x-y|$$$.\n\nThe leftover $$$u+v-|x-y|$$$ pure columns are paired arbitrarily among themselves. If the total number of columns is odd, exactly one pure column remains and can be deleted alone.\n\nAll these pairs / singletons are valid operations. After their deletion the strings become empty, so the pair is good.\n\nPrecompute prefix sums for the four column types. For a query $$$\\[l,r\\]$$$ obtain the four counts $$$u,v,x,y$$$ in constant time. Answer yes if $$$|x-y| \\\\le u+v$$$, otherwise no.\n\n**Complexity:** $$$O(n+q)$$$ per test case.\n\n**Code**\n\n```\n#include <cstdio>\n#include <cstring>\n#include <cctype>\n#include <algorithm>\n#include <iostream>\n#include <queue>\ntypedef long long LL;\nusing namespace std;\ninline int read() {\n    int num = 0 ,f = 1; char c = getchar();\n    while (!isdigit(c)) f = c == '-' ? -1 : f ,c = getchar();\n    while (isdigit(c)) num = (num << 1) + (num << 3) + (c ^ 48) ,c = getchar();\n    return num * f;\n}\nconst int N = 2e5 + 5;\nchar s[N] ,t[N];\nint sum[4][N] ,n ,q;\ninline void fuling_solve() {\n    n = read() ,q = read();\n    scanf(\"%s%s\" ,s + 1 ,t + 1);\n    for (int j = 0; j < 4; j++) sum[j][0] = 0;\n    for (int i = 1; i <= n; i++) {\n        int id = (s[i] ^ 48) * 2 + (t[i] ^ 48);\n        for (int j = 0; j < 4; j++) sum[j][i] = sum[j][i - 1] + (id == j);\n    }\n    while (q--) {\n        int l = read() ,r = read();\n        int a = sum[0][r] - sum[0][l - 1];\n        int b = sum[1][r] - sum[1][l - 1];\n        int c = sum[2][r] - sum[2][l - 1];\n        int d = sum[3][r] - sum[3][l - 1];\n        puts(abs(b - c) <= a + d ? \"YES\" : \"NO\");\n    }\n}\nsigned main() {\n    int t = read();\n    while (t--) fuling_solve();\n    return 0;\n}\n\n```\n\n**Bonus Question**\n\nWe have consider to change this problem to a harder version: Find number of pairs $$$(l, r) (1 \\\\le l \\\\le r \\\\le n)$$$ such that $$$(s\\[l, r\\], t\\[l, r\\])$$$ is good.\n\nCan you solve this too?\n\n[2248E - Excuse for Breaks](https://codeforces.com/contest/2248/problem/E \"Codeforces Round 1113 (Div. 2)\")\n\n**Hint1**\n\nLet $$$S\\_i = F(I(i))$$$. Since the counter resets at $$$n$$$, $$$S\\_{i+n} = S\\_i + S\\_n$$$.\n\n**Hint2**\n\nIf $$$S\\_x + S\\_y \\\\gt S\\_{x+y+1}$$$ for some $$$x,y$$$, the array $$$I(x),\\[0\\],I(y)$$$ already wins. Conversely, if no such pair exists, every array loses.\n\n**Hint3**\n\nFor fixed $$$y$$$, the function $$$S\\_x + S\\_y - S\\_{x+y+1}$$$ changes only when $$$x+1$$$ or $$$x+y+2$$$ is a reward point. Where is its maximum attained?\n\n**Solution**\n\nLet $$$S\\_i = F(I(i))$$$ ($$$S\\_0 = 0$$$). The counter resets exactly at $$$n$$$, so $$$S\\_{i+n} = S\\_i + S\\_n$$$; also $$$F$$$ is additive across a `0`: if a string ends with `0`, the next part starts from $$$c=0$$$.\n\nDefine\n\n$$$ M = \\\\max\\_{x,y \\\\ge 1} \\\\big(S\\_x + S\\_y - S\\_{x+y+1}\\\\big). $$$\n\n**Claim.** $$$F(s) \\\\le S\\_{|s|} + \\\\max(M,0)$$$ for every binary array $$$s$$$, with equality for $$$s = I(x),\\[0\\],I(y)$$$ at a maximizing pair. Hence the answer is `YES` iff $$$M \\\\gt 0$$$.\n\nInduct on the number of runs. If $$$s$$$ has no `0`, then $$$F(s) = S\\_{|s|}$$$. Otherwise write $$$s = I(x),\\[0\\],t$$$ with $$$x$$$ leading ones. Then\n\n$$$F(s) = S\\_x + F(t) \\\\le S\\_x + S\\_{|t|} + \\\\max(M,0) \\\\le S\\_{|s|} + \\\\max(M,0),$$$\n\nbecause $$$S\\_x + S\\_{|t|} - S\\_{x+|t|+1} \\\\le M$$$.\n\n**Only reward thresholds matter.** Let $$$g(x,y) = S\\_x + S\\_y - S\\_{x+y+1}$$$. By periodicity $$$g$$$ is $$$n$$$-periodic in each variable. Fix $$$y$$$:\n\n$$$g(x+1,y) - g(x,y) = (S\\_{x+1}-S\\_x) - (S\\_{x+y+2}-S\\_{x+y+1}),$$$\n\nwhich is positive only when $$$x+1$$$ is a reward point $$$p\\_i$$$. On the circle of residues modulo $$$n$$$, the maximum is attained on a plateau whose left end is congruent to some $$$p\\_i$$$ — so some $$$x = p\\_i$$$ attains it. The same argument in $$$y$$$ gives\n\n$$$ M = \\\\max\\_{1 \\\\le i,j \\\\le m} \\\\big(S\\_{p\\_i} + S\\_{p\\_j} - S\\_{p\\_i+p\\_j+1}\\\\big). $$$\n\n(If $$$m=0$$$, then $$$S\\_x = dx$$$ and $$$M = -d \\\\le 0$$$.)\n\n**Algorithm.** All needed values of $$$S$$$ follow from the periodic formula $$$S\\_i = \\\\lfloor i/n\\\\rfloor S\\_n + S\\_{i\\\\bmod n}$$$. For each pair $$$(i,j)$$$, locate $$$p\\_i+p\\_j+1$$$ among the sorted points $$${p\\_1,\\\\dots,p\\_m, n+p\\_1,\\\\dots,n+p\\_m}$$$ with a two-pointer scan; if $$$S\\_{p\\_i}+S\\_{p\\_j} \\\\gt S\\_{p\\_i+p\\_j+1}$$$ for some pair, print `YES`.\n\n**Complexity:** $$$O(m^2)$$$ per test case ($$$\\\\sum m \\\\le 2000$$$).\n\n**Code**\n\n```\nimport sys\ninput = sys.stdin.readline\n\nclass Point:\n    def __init__(self, t, v):\n        self.t = t\n        self.v = v\n\nsrw = [Point(0, 0) for _ in range(4100)]\n\nT = int(input())\nfor _ in range(T):\n    n, m, d = map(int, input().split())\n    for i in range(1, m+1):\n        t, v = map(int, input().split())\n        srw[i].t = t\n        srw[i].v = v + srw[i-1].v\n    for i in range(1, m+1):\n        srw[m+i] = Point(n + srw[i].t, srw[m].v + srw[i].v)\n    ok = False\n    for i in range(1, m+1):\n        k = i\n        for j in range(1, m+1):\n            st = srw[i].t + 1 + srw[j].t\n            while k+1 <= 2*m and srw[k+1].t <= st:\n                k += 1\n            if srw[i].v + srw[i].t*d + srw[j].v + srw[j].t*d > srw[k].v + st*d:\n                ok = True\n    print(\"Yes\" if ok else \"No\")\n    for i in range(1, 2*m+1):\n        srw[i] = Point(0, 0)\n\n```\n\n[2248F - Matrix Elimination](https://codeforces.com/contest/2248/problem/F \"Codeforces Round 1113 (Div. 2)\")\n\n**Hint1**\n\nDefine the deficit $$$g\\_{x,y} = \\\\text{rowSum}\\_x + \\\\text{colSum}\\_y - 3v\\_{x,y}$$$ of a cell. A cell is a peak iff its deficit is $$$\\\\le 0$$$. How much can one operation reduce a deficit?\n\n**Hint2**\n\nFor $$$n,m \\\\ge 2$$$, every operation reduces any deficit by at most $$$n+m-3$$$, and full-matrix operations achieve this bound for every cell at once. So the problem reduces to a $$$k$$$-th smallest value.\n\n**Hint3**\n\nIn one dimension a cell is a peak iff $$$2a\\_i \\\\ge S$$$ (total sum). Only \"full\" operations and operations that skip one endpoint matter, and the two endpoint types are never both useful. The rest is a couple of linear inequalities solved in $$$O(1)$$$.\n\n**Solution**\n\nLet $$$R\\_x = \\\\sum\\_j v\\_{x,j}$$$ and $$$C\\_y = \\\\sum\\_i v\\_{i,y}$$$. The peak condition is equivalent to\n\n$$$ g\\_{x,y} := R\\_x + C\\_y - 3v\\_{x,y} \\\\le 0\\. $$$\n\nIf an operation covers $$$a$$$ cells of row $$$x$$$ and $$$b$$$ cells of column $$$y$$$, it changes the deficit by $$$-(a+b-3)$$$ when it covers $$$(x,y)$$$, and by $$$-(a+b)$$$ otherwise. For $$$n,m \\\\ge 2$$$ this means the deficit is reduced by at most $$$n+m-3$$$ per operation: full coverage gives $$$a+b-3 \\\\le n+m-3$$$, while an operation avoiding $$$(x,y)$$$ can cover cells of only one of the two lines, contributing at most $$$\\\\max(n-1,m-1) \\\\le n+m-3$$$.\n\n**Case $$$n,m \\\\ge 2$$$.** After $$$T$$$ operations, cell $$$(x,y)$$$ can be a peak only if $$$g\\_{x,y} \\\\le T(n+m-3)$$$. Conversely, $$$T$$$ full-matrix operations make every cell with $$$\\\\lceil g\\_{x,y}/(n+m-3)\\\\rceil \\\\le T$$$ a peak simultaneously. Hence the answer is the $$$k$$$-th smallest value of\n\n$$$ \\\\text{need}\\_{x,y} = \\\\max\\\\left(0,\\\\ \\\\left\\\\lceil \\\\frac{R\\_x + C\\_y - 3v\\_{x,y}}{n+m-3}\\\\right\\\\rceil\\\\right). $$$\n\n**Case $$$n=1$$$ or $$$m=1$$$.** Rotate to a row of length $$$L=\\\\max(n,m)$$$, sum $$$S$$$.  \nPeak condition: $$$a\\_i \\\\ge S-a\\_i \\\\iff h\\_i := S-2a\\_i \\\\le 0$$$.\n\nAn operation on a segment of length $$$\\\\ell$$$ subtracts 1 from each element inside.\n\n* If $$$i$$$ is inside: $$$h\\_i$$$ decreases by $$$\\\\ell-2$$$\n* If $$$i$$$ is outside: $$$h\\_i$$$ decreases by $$$\\\\ell$$$\n\nOnly three segment types are useful (others are dominated):\n\n1. **Full** $$$\\[1,L\\]$$$: every $$$h\\_i$$$ reduced by $$$L-2$$$.\n2. **Skip‑first** $$$\\[2,L\\]$$$: $$$h\\_1$$$ reduced by $$$L-1$$$, others by $$$L-3$$$.\n3. **Skip‑last** $$$\\[1,L-1\\]$$$: $$$h\\_L$$$ reduced by $$$L-1$$$, others by $$$L-3$$$.\n\n**Only full operations**: Same as the general formula with $$$n+m-3 = L-2$$$:  \nAnswer = $$$k$$$-th smallest $$$\\\\left\\\\lceil \\\\frac{S-2a\\_i}{L-2} \\\\right\\\\rceil$$$ (with $$$\\\\max(0,\\\\cdot)$$$). (This is exactly the `xlim` part in the code).\n\n**Full + one skip type, with an endpoint among the peaks**: Assume the chosen endpoint is $$$p$$$ (either 1 or $$$L$$$), using $$$x$$$ full and $$$y$$$ skip‑$p$ operations ($$$T=x+y$$$). Take the other $$$k-1$$$ peaks as the cells with the **largest** $$$a\\_i$$$ (smallest $$$h\\_i$$$) among the remaining $$$L-1$$$ cells; let $$$v\\_k$$$ be the smallest of these. Reductions: - For $$$p$$$: $$$(L-2)x + (L-1)y = (L-2)T + y$$$ - For the other $$$k-1$$$ cells: $$$(L-2)x + (L-3)y = (L-2)T - y$$$\n\nRequirements:\n\n$$$ \\\\begin{aligned} (L-2)T + y &\\\\ge S - 2a\\_p \\\\quad &(1)\\\\\\\\ (L-2)T - y &\\\\ge S - 2v\\_k \\\\quad &(2)\\\\\\\\ 0 \\\\le y &\\\\le T . \\\\end{aligned} $$$\n\nFrom the constraints:\n\n* (2) with $$$y=0$$$ gives $$$T \\\\ge \\\\lceil (S-2v\\_k)/(L-2) \\\\rceil$$$\n* (1) with $$$y=T$$$ gives $$$T \\\\ge \\\\lceil (S-2a\\_p)/(L-1) \\\\rceil$$$\n* Adding (1) and (2): $$$2(L-2)T \\\\ge 2S-2a\\_p-2v\\_k \\\\iff T \\\\ge \\\\lceil (S-a\\_p-v\\_k)/(L-2) \\\\rceil$$$\n\nSet \n\n$$$ z = \\\\max\\\\left( \\\\lceil \\\\frac{S-a\\_p-v\\_k}{L-2} \\\\rceil,\\\\; \\\\lceil \\\\frac{S-2a\\_p}{L-1} \\\\rceil \\\\right) $$$\n\n(If the first bound $$$\\\\lceil (S-2v\\_k)/(L-2) \\\\rceil$$$ were strictly larger, then $$$y=0$$$ is optimal and Strategy of only full operations already covers it.)\n\nNow we must check whether there exists an integer $$$y$$$ satisfying the box constraints:\n\n$$$ \\\\max\\\\bigl(0,\\\\; S-2a\\_p - (L-2)z\\\\bigr) \\\\;\\\\le\\\\; y \\\\;\\\\le\\\\; \\\\min\\\\bigl(z,\\\\; (L-2)z - (S-2v\\_k)\\\\bigr). $$$\n\nIf this holds, $$$T=z$$$ is feasible and optimal for this endpoint; otherwise the endpoint strategy cannot beat the \"only full operations\" strategy.\n\nThe code’s `solveL` implements exactly this: it computes $$$z$$$, tests feasibility, and returns $$$z$$$ if valid or $$$\\\\infty$$$ otherwise.  \nThe final answer is $$$\\\\min(\\\\text{Strategy full},\\\\; \\\\text{solveL for left endpoint},\\\\; \\\\text{solveL for right endpoint})$$$.\n\n**Complexity:** $$$O(nm)$$$ per test case.\n\n**Code**\n\n```\nimport math\n\ndef ceil_div(a, b):\n    return (a + b - 1) // b\n\ndef solveL(v, v1, s, k):\n    n = len(v) + 1\n    if k == 1:\n        return max(0, ceil_div(s - 2 * v1, n - 1))\n    v.sort(reverse=True)\n    vk = v[k - 2]\n    z = max(ceil_div(s - v1 - vk, n - 2), ceil_div(s - 2 * v1, n - 1))\n    if min(z, (n - 2) * z - s + 2 * vk) >= max(0, s - (n - 2) * z - 2 * v1):\n        return z\n    return math.inf\n\nT = int(input())\nfor _ in range(T):\n    n, m, k = map(int, input().split())\n    s = 0\n    u, rs, cs = [], [0] * n, [0] * m\n    v = []\n    for _ in range(n):\n        row = list(map(int, input().split()))\n        v.append(row)\n        s += sum(row)\n        rs[_] = sum(row)\n        for j in range(m):\n            cs[j] += row[j]\n            u.append(row[j])\n\n    if n + m <= 3:\n        u.sort()\n        if k == 1:\n            if n + m == 2 and u[0] < 0:\n                print(-1)\n            else:\n                print(0)\n        else:\n            print(abs(u[0] - u[1]))\n        continue\n\n    xlim = []\n    for i in range(n):\n        for j in range(m):\n            xlim.append(max(0, ceil_div(rs[i] + cs[j] - 3 * v[i][j], n + m - 3)))\n\n    xlim.sort()\n    ans = xlim[k - 1]\n\n    if m == 1 or n == 1:\n        ans = min(ans, solveL(u[:-1], u[-1], s, k))\n        ans = min(ans, solveL(u[1:], u[0], s, k))\n    print(ans)\n\n```\n\n[2248G - No Balance Left](https://codeforces.com/contest/2248/problem/G \"Codeforces Round 1113 (Div. 2)\")\n\n**Hint1**\n\nA purchase is a transaction: spend an achievable amount $$$x$$$, gain rebate $$$r(x)$$$, net change $$$r(x)-x$$$. The achievable spends form a knapsack.\n\n**Hint2**\n\nBelow the smallest spend that yields a net increase, the balance only decreases; reachability is a DP on a DAG.\n\n**Hint3**\n\nAbove that threshold the balance can grow without bound, and every transaction changes it by a multiple of a single gcd. What determines whether $$$0$$$ is reachable?\n\n**Hint4**\n\nIf the smallest threshold does not exceed the cheapest product, every purchase is discounted and no positive balance can ever become $$$0$$$.\n\n**Solution**\n\nLet $$$V = \\\\max(s, a\\_m)$$$ and compute the set $$$C$$$ of spend amounts achievable up to $$$V$$$ with an unbounded knapsack over the prices (a bitset, $$$O(V^2/64)$$$). The rebate for a spend $$$x$$$ is $$$r(x) = \\\\max{b\\_j : a\\_j \\\\le x}$$$, obtained by a prefix maximum.\n\nFor every achievable $$$x$$$:\n\n* if $$$r(x) \\\\lt x$$$, the net decrease is $$$d = x - r(x)$$$; keep $$$L\\[d\\] = \\\\min{x}$$$ — the cheapest way to lose exactly $$$d$$$;\n* if $$$r(x) \\\\gt x$$$, the purchase increases the balance; let $$$\\\\text{mnip}$$$ be the smallest such spend.\n\n**Below $$$\\\\text{mnip}$$$.** Every affordable transaction has $$$r(x) \\\\le x$$$, so the balance only decreases. Define $$$f\\[0\\] = 1$$$ and\n\n$$$ f\\[i\\] = \\\\bigvee\\_{d \\\\le i} \\\\big(L\\[d\\] \\\\le i \\\\text{ and } f\\[i-d\\]\\\\big). $$$\n\nThis is a DAG (all transitions go down); using the minimum spend for each decrease $$$d$$$ is optimal because a cheaper transaction is only more flexible. A shifted bitset computes all $$$f\\[i\\]$$$ in $$$O(V^2/64)$$$.\n\n**At or above $$$\\\\text{mnip}$$$.** A net-increasing purchase is affordable and can be repeated, so the balance can be made arbitrarily large. Every transaction changes the balance by a multiple of\n\n$$$ g = \\\\gcd(I,\\\\ D,\\\\ c\\_1, \\\\dots, c\\_n,\\\\ b\\_m), $$$\n\nso the residue of the balance modulo $$$g$$$ is invariant, and from a sufficiently large balance every value in the same residue class is reachable (Frobenius). Any path to $$$0$$$ must enter the decreasing region: for the last purchase, $$$b - x + r(x) = 0$$$ with $$$x \\\\le b$$$ forces $$$r(x) = 0$$$ and $$$x = b \\\\lt a\\_1$$$. Therefore a balance $$$i \\\\ge \\\\text{mnip}$$$ can reach $$$0$$$ iff $$$i \\\\equiv 0 \\\\pmod g$$$ and some positive balance below $$$\\\\text{mnip}$$$ is reachable.\n\nThe second condition is automatic when $$$a\\_1 \\\\gt c\\_{\\\\min}$$$: the spend $$$c\\_{\\\\min}$$$ (the cheapest product) has rebate $$$0$$$, so $$$f\\[c\\_{\\\\min}\\] = 1$$$ and $$$c\\_{\\\\min} \\\\lt \\\\text{mnip}$$$. Conversely, if $$$a\\_1 \\\\le c\\_{\\\\min}$$$, every purchase has $$$r(x) \\\\ge b\\_1 \\\\ge 1$$$, so a purchase of cost $$$x \\\\le h$$$ leaves balance $$$h - x + r(x) \\\\gt 0$$$ — no positive balance can ever become $$$0$$$.\n\nHere $$$I$$$ and $$$D$$$ are the net increases/decreases found above, and the costs and $$$b\\_m$$$ stand in for the spends beyond $$$V$$$: for $$$x \\\\ge a\\_m$$$ the net change is $$$x - b\\_m$$$, and since all achievable spends are multiples of $$$\\\\gcd(c\\_1,\\\\dots,c\\_n)$$$, the whole tail contributes exactly $$$\\\\gcd(c\\_1,\\\\dots,c\\_n, b\\_m)$$$.\n\n**Complexity:** $$$O(V^2/64)$$$ time and $$$O(V)$$$ memory, with $$$V = \\\\max(s, a\\_m) \\\\le 125\\\\,000$$$.\n\n**Code**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nconst int maxV = 125000 + 5;\n\nint Lgcd(const vector<int> &v) {\n\treturn accumulate(v.begin(), v.end(), 0, [](int a, int b) { return gcd(a, b); });\n}\n\nint main() {\n\tint n, m, s;\n\tscanf(\"%d %d %d\", &n, &m, &s);\n\tvector<int> c(n);\n\tvector<pair<int, int>> ab(m);\n\tfor (auto &x : c)\n\t\tscanf(\"%d\", &x);\n\tfor (auto &[a, b] : ab)\n\t\tscanf(\"%d %d\", &a, &b);\n\tab.insert(ab.begin(), make_pair(0, 0));\n\tm += 1;\n\n\tbitset<maxV> p;\n\tp.set(0);\n\tfor (int x : c)\n\t\tp.set(x);\n\n\tconst int V = max(s, ab.back().first);\n\tfor (int i = 1; i <= V; ++i) {\n\t\tif (p[i])\n\t\t\tp |= (p << i);\n\t}\n\n\tvector<int> l(V + 1, s + 1);\n\tl[0] = 0;\n\tint mnip = s + 1;\n\tvector<int> I, D;\n\tfor (int i = 1, abp = 0; i <= V; ++i) {\n\t\tif (!p[i])\n\t\t\tcontinue;\n\t\twhile (abp + 1 < m && ab[abp + 1].first <= i)\n\t\t\tabp += 1;\n\n\t\tauto [a, b] = ab[abp];\n\t\tif (b <= i) {\n\t\t\tl[i - b] = min(l[i - b], i);\n\t\t\tD.push_back(i - b);\n\t\t} else {\n\t\t\tI.push_back(b - i);\n\t\t\tmnip = min(mnip, i);\n\t\t}\n\t}\n\n\tvector<pair<int, int>> lv;\n\tfor (int i = 1; i <= V; ++i) {\n\t\tif (l[i] <= s)\n\t\t\tlv.emplace_back(l[i], i);\n\t}\n\tconst int ln = lv.size();\n\n\tsort(lv.begin(), lv.end());\n\tsort(c.begin(), c.end());\n\tsort(I.begin(), I.end());\n\tsort(D.begin(), D.end());\n\n\tbitset<maxV> f, L;\n\tf.set(0);\n\tfor (int i = 1, lp = 0; i < mnip; ++i) {\n\t\twhile (lp < ln && lv[lp].first <= i) {\n\t\t\tL.set(maxV - lv[lp].second);\n\t\t\tlp += 1;\n\t\t}\n\n\t\tf[i] = ((f & (L >> (maxV - i)))._Find_first() < V);\n\t}\n\n\tint Ig = Lgcd(I), Dg = Lgcd(D), cg = Lgcd(c);\n\tint ugcd = gcd(gcd(Ig, Dg), gcd(cg, ab.back().second));\n\n\tfor (int i = 1; i < mnip; ++i)\n\t\tputs(f[i] ? \"YES\" : \"NO\");\n\tfor (int i = mnip; i <= s; ++i)\n\t\tputs((i % ugcd == 0 && ab[1].first > c[0]) ? \"YES\" : \"NO\");\n\treturn 0;\n}\n\n```\n\nI promise I'll improve this soon. I know its unclear and skipped some steps for now.\n\nAlso check out:\n\n* <https://codeforces.com/blog/entry/155549?#comment-1382258>\n* <https://codeforces.com/blog/entry/155640?#comment-1382379>\n\n![](//codeforces.com/codeforces.org/s/64117/images/icons/paperclip-16x16.png) Tutorial of [Codeforces Round 1113 (Div. 2)](https://codeforces.com/contest/2248) \n\n* [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/voteup-gray.png \"Vote: I like it\")](#)\n* +154\n* [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/votedown-gray.png \"Vote: I do not like it\")](#)\n\n* [![Author](//codeforces.com/codeforces.org/s/64117/images/blog/user_16x16.png \"Author\")](https://codeforces.com/profile/szdytom) [szdytom ](https://codeforces.com/profile/szdytom)\n* ![Publication date](//codeforces.com/codeforces.org/s/64117/images/blog/date_16x16.png \"Publication date\") 7 weeks ago\n* [![Comments](//codeforces.com/codeforces.org/s/64117/images/blog/comments_16x16.png \"Comments\")](https://codeforces.com/blog/entry/155640#comments) [83 ](https://codeforces.com/blog/entry/155640#comments)\n\n  \n![Comments](//codeforces.com/codeforces.org/s/64117/images/icons/comments-48x48.png \"Comments\") Comments (83) \n\n[Write comment?](#) \n\n| » [szdytom](https://codeforces.com/profile/szdytom \"Master szdytom\") | 7 weeks ago, [show](#) # |\n| -------------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/1198980/avatar/905c3a9d5029cd8e.jpg) ](https://codeforces.com/profile/szdytom) [szdytom](https://codeforces.com/profile/szdytom \"Master szdytom\") | 7 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155640?#comment-1382276 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +1 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) _Auto comment: topic has been updated by [szdytom](https://codeforces.com/profile/szdytom \"Master szdytom\") ([previous revision](https://codeforces.com/topic/156282/en6), [new revision](https://codeforces.com/topic/156282/en7), [compare](https://codeforces.com/topic/156282/diff/en6/en7))._ [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |\n\n| » [rdor](https://codeforces.com/profile/rdor \"Pupil rdor\") | 7 weeks ago, [show (+1)](#) # |\n| ---------------------------------------------------------- | ----------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/4921754/avatar/6939fd62eb19593f.jpg) ](https://codeforces.com/profile/rdor) [rdor](https://codeforces.com/profile/rdor \"Pupil rdor\") | 7 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155640?#comment-1382280 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +56 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) is it just me or B felt harder than C and D? [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [overi](https://codeforces.com/profile/overi \"Newbie overi\") | 7 weeks ago, [show](#) # |\n| ---------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/overi) [overi](https://codeforces.com/profile/overi \"Newbie overi\") | 7 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155640?#comment-1382291 \"Link to comment\") [^](#comment-1382280 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) agreed [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |\n* | » » [petryshynserhii361](https://codeforces.com/profile/petryshynserhii361 \"Specialist petryshynserhii361\") | 7 weeks ago, [show (+1)](#) # |\n| ----------------------------------------------------------------------------------------------------------- | ----------------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/petryshynserhii361) [petryshynserhii361](https://codeforces.com/profile/petryshynserhii361 \"Specialist petryshynserhii361\") | 7 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155640?#comment-1382301 \"Link to comment\") [^](#comment-1382280 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Not really for me, i actually struggled more with C because my dumb ahh couldnt realise it was just a dp for what felt like eternity. Also congrats on becoming specialist! [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n  * | » » » [rdor](https://codeforces.com/profile/rdor \"Pupil rdor\") | 7 weeks ago, [show](#) # |\n| -------------------------------------------------------------- | ------------------------ |\n\n| » » » [ ![](https://codeforces.com/userpic.codeforces.org/4921754/avatar/6939fd62eb19593f.jpg) ](https://codeforces.com/profile/rdor) [rdor](https://codeforces.com/profile/rdor \"Pupil rdor\") | 7 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155640?#comment-1382306 \"Link to comment\") [^](#comment-1382301 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +1 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) thanks!! :D [→](#) [Reply](#) |\n| ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |\n  * | » » » [JeffreyLee](https://codeforces.com/profile/JeffreyLee \"Expert JeffreyLee\") | 7 weeks ago, [show (+1)](#) # |\n| --------------------------------------------------------------------------------- | ----------------------------- |\n\n| » » » [ ![](https://codeforces.com/userpic.codeforces.org/4029260/avatar/63a4474c60b159c.jpg) ](https://codeforces.com/profile/JeffreyLee) [JeffreyLee](https://codeforces.com/profile/JeffreyLee \"Expert JeffreyLee\") | 7 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155640?#comment-1382388 \"Link to comment\") [^](#comment-1382301 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) \\-8 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I think C and F are fraud problems,if you guess the conclusion and succeed,you will solve them immediately. [→](#) [Reply](#) |\n| ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n    * | » » » » [silverfish](https://codeforces.com/profile/silverfish \"Specialist silverfish\") | 7 weeks ago, [show](#) # |\n| --------------------------------------------------------------------------------------- | ------------------------ |\n\n\n...157878 bytes truncated...\n\nThe tool call succeeded but the output was truncated. Full output saved to: /Users/zayd/.local/share/opencode/tool-output/tool_0b252c9e1001KvCeBb4uNVotIA\nUse the Task tool to have explore agent process this file with Grep and Read (with offset/limit). Do NOT read the full file yourself - delegate to save context.",
            "metadata": {
              "truncated": true,
              "outputPath": "/Users/zayd/.local/share/opencode/tool-output/tool_0b252c9e1001KvCeBb4uNVotIA"
            },
            "title": "https://codeforces.com/blog/entry/155640 (text/markdown; charset=utf-8)",
            "time": {
              "start": 1789698160568,
              "end": 1789698165221
            }
          },
          "metadata": {
            "openai": {
              "itemId": "fc_0ed6462526416c01016aaca06fc22c87d0addc36ad30c77fad"
            }
          },
          "id": "prt_0b252b4d300134QC8DFM4USNhr",
          "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03",
          "messageID": "msg_0b252a6fd001bEvtt88FQSK0af"
        },
        {
          "reason": "tool-calls",
          "type": "step-finish",
          "tokens": {
            "total": 10689,
            "input": 10657,
            "output": 32,
            "reasoning": 0,
            "cache": {
              "write": 0,
              "read": 0
            }
          },
          "cost": 0,
          "id": "prt_0b252c9ea001Cffs0in7eHCVDR",
          "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03",
          "messageID": "msg_0b252a6fd001bEvtt88FQSK0af"
        }
      ]
    },
    {
      "info": {
        "parentID": "msg_0b252a677001wdK3bGintBCWRE",
        "role": "assistant",
        "mode": "nudge-generation",
        "agent": "nudge-generation",
        "variant": "low",
        "path": {
          "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-xZ6Xpc",
          "root": "/"
        },
        "cost": 0,
        "tokens": {
          "total": 26730,
          "input": 26046,
          "output": 471,
          "reasoning": 213,
          "cache": {
            "write": 0,
            "read": 0
          }
        },
        "modelID": "gpt-6-astra",
        "providerID": "openai",
        "time": {
          "created": 1789698165233,
          "completed": 1789698188742
        },
        "finish": "tool-calls",
        "id": "msg_0b252c9f10010WANMPUsFShntt",
        "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03"
      },
      "parts": [
        {
          "type": "step-start",
          "id": "prt_0b252d237001qF6cq8DqLdQkqv",
          "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03",
          "messageID": "msg_0b252c9f10010WANMPUsFShntt"
        },
        {
          "type": "reasoning",
          "text": "**Verifying interval coverage formulas**",
          "time": {
            "start": 1789698167353,
            "end": 1789698175098
          },
          "metadata": {
            "openai": {
              "itemId": "rs_08a9a462da2054a0016aaca077459487d090d53d5277a05e3d"
            }
          },
          "id": "prt_0b252d239001avDaPhFZk9wUr1",
          "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03",
          "messageID": "msg_0b252c9f10010WANMPUsFShntt"
        },
        {
          "type": "reasoning",
          "text": "**Testing coverage formulas**",
          "time": {
            "start": 1789698175100,
            "end": 1789698175331
          },
          "metadata": {
            "openai": {
              "itemId": "rs_08a9a462da2054a0016aaca077459487d090d53d5277a05e3d",
              "reasoningEncryptedContent": "gAAAAABqrKB-zswcdLi7gg7CnI3-Tti7uROimrxU9EvTDjYn1Eu4DLnsUk57fK7rE5ognJ48pdcXeQg373rwxDEd1LpCc-QYfWhyU4u-YdDFJWaU78qHE97cWu7HO9usnXPAG-VkonVIey1fRPA4jRhQrpsf0pXjM8xTiHHImiWLm1X-qLA-8Xkxa4pi69nDZXt7v-vGaaNKDOwsiXhLAXm8JeckqqWHXwNXbUptjSrKtX4PjFgEc6EBMDowBapiY-4jktWNuc0q87T5vzmQud8CV6M2znGj38N90fwTO6nGyca660vJzxvBWo2tbwlez6OlxZJDgZTuLjv9w2XOnNZdt2OHCrXeDkW_cuzbU7YbQFwVzDPITl4FB-tr24bs_DdVGK-z7T0UNjrjTw6yaE7xzNs2H8-0PuxbwPyIyRmPbAEoJ1BEBhZ8GPl9hCV-GfcdbKeGQmR-pFSsOIOXmEtUiK6M_aOvSzzT1g67ZhgTxRr0jwEDHmWDHZ1_Pas5JDfHKN9ZxVqsgrSreaEXBr07qqIosf34ERIvSyBlbFuHUgHrTyMYqvcOZlbQqvtsNpkbGoaCrSQBvMhtIQupXitXPE6iYR6AetWcSI_vRYrbygltF3ZvT3Kdj-r5AcmqcFQsx7lwKFg7G8F9YMPn1C-kKHMwJcYPEC1OQMVU7_6Ron8td-FmX_IF5CfKRYcDmYOknvO8MLsH2e-i_Hyyjx9v5f5M2NeaKL3ctW5BsBiDO8Qk-0cBlMeEzeGQbw6WLmC6LoEmFmFTky5Ywecq_A-wEA1aRZrhU5Cy7te5vgeWS-w6mf0YI4bZaBSKXHDmU5VTkYxQr13_cTGc9_yP4SI81GIbHzQ95_MGHR0QbObYGjAkq1GjIl0fgl7Z0NirANpZbaKxxhjeB-qDi1efuaoUWGmDmjzdlUMrkTdyHbaG8knDg_auT4NhtDCtIPK47pApeoZZGk4qxU3-_1Gd5kHRrICsDsAPhgL5TeIOla-nTmzHs5me8peVUjY4jbrXjeVX_um0wPMZgepr7pC78EIBD_OVjnmZB6ZLPRrH-OXXi9RiAgxoKCRF7Wy4eViEoHB0dFIfVauTnyCbh6FTslRvJJ4qZcXNzQlKPgVlGNeTMyKy5Tm-__Yrr3jpByPgtKFLpTC6bXpyAaGtcrnqZscgw6RJolE5G5e_ZtD2OT0dkJdyReoaLNDvxfDj1tPcpAdK9XfHJvpaEiB28rnYWLX-DLrflOpLlaXefzYlnKf02rxrilFTGOfZ7w8hHgke7O4RZaw6SnLuLKg2LMS0CqqwbqPif6ccAvUKIcaBMdxUkAFROJ1xjvIgX05FzlxKzHdScj5uDPycVD2b2PiMfxCkYs-yMnyOJSHEA5Q9938v2A_f1TWdp26LcIxRNWjCeY8eTLOQ1uHnbQZoMdAlyDGjGlRb8MlT48xwc5wWPiOjXkEJQgu18SCEjw9U7UAGJjEzOFDjpofEeZFxDwk3_ksZXfg0ERkZ37e575vJN39_VkaQ3jjqlr89xGcR2tujvWVfcYDgbG6SjKF1unu5PfhN-2yPFHhffiqupkPLtyv8I7Bt1izO2v8KEY6CX6rDgHVvYwQScih9Zbj5vMEW6jnUAsSmfauE3BAkoIBO4lgrsaoporaIc3tyX6XTpuvSs6GQs33fNN1lqtduJV32BnXZROTxknUfaqJ-KgFPbBF0WGl4r2Jd7KuMlgFJNlYvUuFOKfDIKCMEW3kqvkPETVouitK4PCDnyLiPfdTqUF8OlPPatBjq4a9ggca9lFc9IRrtl9wDERbJiR1eekApPZoM59syBw9GIlUjPPISHJqQhwq9JoFgT8W-l0VPOielY-oSTAwAZy2P8ru2AuV0LO2newKwY-2xry-OvkoHL7p7lkwJNWQt8QK3xBZT2jAUGaRhY6nd5Be-AaDFWYrNRFnVV5DOeOvjEkLupETIoSnCqncltV83P8aO9RghNKfBeadAgnbAiMlAZIKVk8wwgGUGYB1CX8lhqSUR18CnfKe3vpQAGZA3fjf3qr23fdQBV8C8Ah7bJVT-fVEo7G0dTWeutBY9O3NwqIZ0AyM0oyf_VVAo3S3incyg0EuAB04rckdZRL1EYq5LJ5fQ8I53x0G_LJeF1DQYgZierX1LoZ26rXOTaguhfMMeX5lBVS1q2RMt-_tWH1QfBimeihiSiP0bXwv9JfWCTqrFaTtFe3RgiJSIW6WKo5ZY2mPqxjzRcpEtOW5UTZ5XqB0TrsAftteWGmjyJevlHtdNi4x2IJOXDNQk3iPvWDgbEbURnaG6nxekqajcsNIy9OtF9eSw6dV3cv0t_zTqoM187YlnAsr-Ty9T1Bun0wn-RGhZ9KTQWT2vuIGQ3Ruh"
            }
          },
          "id": "prt_0b252f07c001NUzZufuy1ZT4bZ",
          "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03",
          "messageID": "msg_0b252c9f10010WANMPUsFShntt"
        },
        {
          "type": "tool",
          "tool": "bash",
          "callID": "call_YNU5aGSCExgtQSHuuGrqp2i3",
          "state": {
            "status": "completed",
            "input": {
              "command": "python3 - <<'PY'\nfrom itertools import product, combinations_with_replacement\n\ndef solve(a,k):\n    n=len(a); s=sum(a); h=[s-2*x for x in a]\n    c=lambda x,d:max(0,(x+d-1)//d)\n    d=n-2\n    ans=c(sorted(h)[k-1],d)\n    for p in (0,n-1):\n        A=h[p]\n        if k==1: cur=c(A,d+1)\n        else:\n            B=sorted(h[:p]+h[p+1:])[k-2]\n            cur=max(c(B,d),c(A,d+1),c(A+B,2*d))\n        ans=min(ans,cur)\n    return ans\n\ncases=0\nfor n in range(3,6):\n    intervals=[tuple(int(l<=i<=r) for i in range(n)) for l in range(n) for r in range(l,n)]\n    effects=[]\n    for t in range(4):\n        effects.append({tuple(sum(intervals[j][i] for j in comb) for i in range(n)) for comb in combinations_with_replacement(range(len(intervals)),t)})\n    for a in product(range(-1,2),repeat=n):\n        best=[None]*(n+1)\n        for t,es in enumerate(effects):\n            for e in es:\n                b=[x-y for x,y in zip(a,e)]; s=sum(b)\n                cnt=sum(2*x>=s for x in b)\n                for k in range(1,cnt+1):\n                    if best[k] is None: best[k]=t\n        for k in range(1,n+1):\n            got=solve(list(a),k)\n            assert (got if got<=3 else None)==best[k],(a,k,got,best[k])\n            cases+=1\nprint('Exhaustive interval-operation checks passed:',cases)\nPY",
              "timeout": 120000
            },
            "output": "Exhaustive interval-operation checks passed: 1620\n",
            "metadata": {
              "output": "Exhaustive interval-operation checks passed: 1620\n",
              "exit": 0,
              "truncated": false
            },
            "title": "python3 - <<'PY'\nfrom itertools import product, combinations_with_replacement\n\ndef solve(a,k):\n    n=len(a); s=sum(a); h=[s-2*x for x in a]\n    c=lambda x,d:max(0,(x+d-1)//d)\n    d=n-2\n    ans=c(sorted(h)[k-1],d)\n    for p in (0,n-1):\n        A=h[p]\n        if k==1: cur=c(A,d+1)\n        else:\n            B=sorted(h[:p]+h[p+1:])[k-2]\n            cur=max(c(B,d),c(A,d+1),c(A+B,2*d))\n        ans=min(ans,cur)\n    return ans\n\ncases=0\nfor n in range(3,6):\n    intervals=[tuple(int(l<=i<=r) for i in range(n)) for l in range(n) for r in range(l,n)]\n    effects=[]\n    for t in range(4):\n        effects.append({tuple(sum(intervals[j][i] for j in comb) for i in range(n)) for comb in combinations_with_replacement(range(len(intervals)),t)})\n    for a in product(range(-1,2),repeat=n):\n        best=[None]*(n+1)\n        for t,es in enumerate(effects):\n            for e in es:\n                b=[x-y for x,y in zip(a,e)]; s=sum(b)\n                cnt=sum(2*x>=s for x in b)\n                for k in range(1,cnt+1):\n                    if best[k] is None: best[k]=t\n        for k in range(1,n+1):\n            got=solve(list(a),k)\n            assert (got if got<=3 else None)==best[k],(a,k,got,best[k])\n            cases+=1\nprint('Exhaustive interval-operation checks passed:',cases)\nPY",
            "time": {
              "start": 1789698188537,
              "end": 1789698188686
            }
          },
          "metadata": {
            "openai": {
              "itemId": "fc_08a9a462da2054a0016aaca07ec81887d0a014e4c1c2a3a1b4"
            }
          },
          "id": "prt_0b252f164001v99c40418YrNOf",
          "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03",
          "messageID": "msg_0b252c9f10010WANMPUsFShntt"
        },
        {
          "reason": "tool-calls",
          "type": "step-finish",
          "tokens": {
            "total": 26730,
            "input": 26046,
            "output": 471,
            "reasoning": 213,
            "cache": {
              "write": 0,
              "read": 0
            }
          },
          "cost": 0,
          "id": "prt_0b25325c4001PssuAPCYJsUknN",
          "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03",
          "messageID": "msg_0b252c9f10010WANMPUsFShntt"
        }
      ]
    },
    {
      "info": {
        "parentID": "msg_0b252a677001wdK3bGintBCWRE",
        "role": "assistant",
        "mode": "nudge-generation",
        "agent": "nudge-generation",
        "variant": "low",
        "path": {
          "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-xZ6Xpc",
          "root": "/"
        },
        "cost": 0,
        "tokens": {
          "total": 30240,
          "input": 1025,
          "output": 3257,
          "reasoning": 230,
          "cache": {
            "write": 0,
            "read": 25728
          }
        },
        "modelID": "gpt-6-astra",
        "providerID": "openai",
        "time": {
          "created": 1789698188743,
          "completed": 1789698296874
        },
        "finish": "tool-calls",
        "structured": {
          "status": "success",
          "reason": null,
          "hints": [
            {
              "order": 1,
              "content": "Turn the peak condition into a single number. If $R_i$ and $C_j$ are the row and column sums, define $g_{i,j}=R_i+C_j-3v_{i,j}$. A cell is a peak exactly when $g_{i,j}\\le 0$. How much can one rectangle operation reduce this deficit?"
            },
            {
              "order": 2,
              "content": "When $n,m\\ge 2$, no operation can reduce any cell’s deficit by more than $n+m-3$. A full-matrix operation achieves exactly that reduction for **every** cell simultaneously. This gives the answer from the $k$-th smallest deficit. Be careful: this argument changes when one dimension is $1$."
            },
            {
              "order": 3,
              "content": "For a one-dimensional array of length $L$, write $h_i=S-2a_i$, where $S$ is its sum. An interval of length $\\ell$ reduces $h_i$ by $\\ell-2$ inside the interval and by $\\ell$ outside it. For $L\\ge 3$, every interval of length at most $L-2$ is dominated by the full interval. Only three types remain: the full array, all except the first element, and all except the last element."
            },
            {
              "order": 4,
              "content": "One skip-first operation plus one skip-last operation can be replaced by two full-array operations without making any deficit worse. So consider only full operations plus one chosen skip type. If the skipped endpoint is not among your required peaks, full operations alone are at least as good. Otherwise, choose the other $k-1$ peaks as the cells with the smallest initial deficits outside that endpoint."
            },
            {
              "order": 5,
              "content": "Let $d=L-2$, let $A$ be the skipped endpoint’s deficit, and let $B$ be the $(k-1)$-th smallest deficit among the other cells. Using $T$ operations, of which $y$ skip that endpoint, requires $dT+y\\ge A$, $dT-y\\ge B$, and $0\\le y\\le T$. Such an integer $y$ exists exactly when $dT\\ge B$, $(d+1)T\\ge A$, and $2dT\\ge A+B$. These give a direct formula for the minimum $T$. Try both endpoints and compare with full operations only. Handle $k=1$ and lengths $1,2$ separately."
            }
          ],
          "editorial": "The rectangle choices look like the hard part. The trick is to compare operations by how much they improve each cell’s peak condition. Most rectangles turn out to be completely unnecessary.\n\n## 1. Measure how far a cell is from being a peak\n\nLet $R_i$ be the sum of row $i$ and $C_j$ the sum of column $j$. The peak condition is\n\n$$v_{i,j}\\ge R_i+C_j-2v_{i,j}.$$\n\nDefine the **deficit**\n\n$$g_{i,j}=R_i+C_j-3v_{i,j}.$$\n\nA cell is a peak exactly when $g_{i,j}\\le 0$. Smaller deficits are always better, including when matrix entries are negative.\n\n## 2. The genuinely two-dimensional case\n\nSuppose $n,m\\ge 2$, and put $D=n+m-3$.\n\nConsider one operation and a fixed cell $(i,j)$.\n\n- If the rectangle contains the cell and has height $a$ and width $b$, the row sum decreases by $b$, the column sum by $a$, and the cell value by $1$. Its deficit decreases by $a+b-3\\le D$.\n- If the rectangle avoids the cell, it cannot intersect both its row and its column: that would include their intersection. It can therefore reduce the deficit by at most $\\max(n-1,m-1)\\le D$.\n\nSo **no operation improves any deficit by more than $D$**.\n\nNow choose the entire matrix. Every deficit decreases by exactly $D$. This simultaneously achieves the best possible improvement for every cell. No fancy rectangle gymnastics needed.\n\nAfter $T$ operations, a cell can be a peak only if $g_{i,j}\\le DT$, and $T$ full-matrix operations make every such cell a peak.\n\nIf $G$ is the $k$-th smallest deficit, the answer is therefore\n\n$$\\boxed{\\max\\left(0,\\left\\lceil\\frac{G}{D}\\right\\rceil\\right)}.$$\n\n## 3. One dimension needs its own argument\n\nIf one dimension is $1$, view the matrix as an array $a_1,\\dots,a_L$. Rectangles become contiguous intervals.\n\nLet $S=\\sum_i a_i$. The deficit is now\n\n$$h_i=S-2a_i.$$\n\nThis is also exactly what the row-and-column formula computes for a one-dimensional matrix.\n\nFirst handle two tiny cases:\n\n- **$L=1$:** the only cell is a peak iff $a_1\\ge 0$. The answer is $0$ if so, and $-1$ otherwise, since operations only decrease its value.\n- **$L=2$:** at least one cell is always a peak. For $k=1$, answer $0$. For $k=2$, both must be equal, so answer $|a_1-a_2|$: repeatedly decrease the larger cell. One operation changes their difference by at most $1$, so this is optimal.\n\nFrom now on, assume $L\\ge 3$ and write $d=L-2>0$.\n\n### Only three interval types matter\n\nAn operation on an interval of length $\\ell$ reduces a deficit by\n\n- $\\ell-2$ if the cell is inside;\n- $\\ell$ if the cell is outside.\n\nA full-array operation reduces every deficit by $L-2=d$.\n\nIf $\\ell\\le L-2$, both possible reductions are at most $d$. Replacing that operation with a full-array operation makes every deficit no larger. Thus we only need intervals of length $L$ or $L-1$:\n\n| Operation | First endpoint | Interior cells | Last endpoint |\n|---|---:|---:|---:|\n| Full array | $d$ | $d$ | $d$ |\n| Skip first | $d+1$ | $d-1$ | $d-1$ |\n| Skip last | $d-1$ | $d-1$ | $d+1$ |\n\nThe entries are deficit reductions.\n\nA skip-first operation paired with a skip-last operation reduces each endpoint’s deficit by $2d$, and every interior deficit by $2d-2$. Two full-array operations reduce all deficits by $2d$.\n\nTherefore, we can replace opposite skip operations in pairs. **An optimal strategy exists using full operations and at most one skip type.**\n\n## 4. Evaluate the remaining strategies\n\n### Full operations only\n\nLet $H$ be the $k$-th smallest value among all $h_i$. The required number is\n\n$$T_{\\mathrm{full}}=\\max\\left(0,\\left\\lceil\\frac{H}{d}\\right\\rceil\\right).$$\n\n### Full operations plus skipping endpoint $p$\n\nTry $p=1$ and $p=L$ separately.\n\nIf endpoint $p$ is not among a chosen set of $k$ peaks, replacing every skip operation with a full operation only helps that set. The full-only strategy already covers this possibility.\n\nSo we only need to evaluate strategies in which $p$ is one of the required peaks.\n\nLet $A=h_p$. Every other cell receives the same deficit reduction, so the easiest other $k-1$ cells to turn into peaks are those with the smallest initial deficits. For $k\\ge 2$, let $B$ be the $(k-1)$-th smallest deficit after removing endpoint $p$.\n\nSuppose there are $T$ operations in total, of which $y$ skip $p$. Then:\n\n- endpoint $p$ receives reduction $dT+y$;\n- every other cell receives reduction $dT-y$.\n\nThe requirements are\n\n$$dT+y\\ge A,\\qquad dT-y\\ge B,\\qquad 0\\le y\\le T.$$\n\nEquivalently,\n\n$$\\max(0,A-dT)\\le y\\le\\min(T,dT-B).$$\n\nAll bounds are integers. For $T\\ge 0$, this interval is nonempty exactly when\n\n$$dT\\ge B,\\qquad (d+1)T\\ge A,\\qquad 2dT\\ge A+B.$$\n\nTo see sufficiency explicitly, these inequalities ensure that each lower bound, $0$ and $A-dT$, is no greater than either upper bound, $T$ and $dT-B$.\n\nThus this endpoint strategy needs\n\n$$\\boxed{T_p=\\max\\left(0,\\left\\lceil\\frac{B}{d}\\right\\rceil,\\left\\lceil\\frac{A}{d+1}\\right\\rceil,\\left\\lceil\\frac{A+B}{2d}\\right\\rceil\\right)}.$$\n\nFor **$k=1$**, there are no other required peaks. Skipping $p$ every time gives its maximum improvement, $d+1$ per operation, so\n\n$$T_p=\\max\\left(0,\\left\\lceil\\frac{A}{d+1}\\right\\rceil\\right).$$\n\nThe one-dimensional answer is\n\n$$\\min(T_{\\mathrm{full}},T_1,T_L).$$\n\nThere is no need for binary search: the feasibility inequalities give the answer directly.\n\n## 5. Correctness argument\n\nFor $n,m\\ge 2$, every operation reduces each deficit by at most $D$, while a full-matrix operation achieves that bound for all cells. Hence the $k$-th-deficit formula is both a lower bound and achievable.\n\nFor a one-dimensional array of length at least $3$, replacing short intervals with full intervals never worsens any deficit. Replacing pairs of opposite skip operations with two full operations also never worsens any deficit. Consequently, some optimal solution uses only full operations and one skip type.\n\nIf its skipped endpoint is unnecessary, full operations alone achieve at least the same required peaks. Otherwise, we can choose that endpoint and the $k-1$ smallest remaining deficits, since all remaining cells improve equally. The derived inequalities are necessary and sufficient for precisely this strategy, and the formula finds their smallest feasible integer $T$.\n\nTaking the minimum over full-only and both endpoint strategies therefore finds the global optimum. The direct arguments for lengths $1$ and $2$ cover all remaining cases.\n\n## 6. Complexity and implementation\n\nLet $N=nm$. Compute all row sums, column sums, and deficits in $O(N)$ time. Sort the deficits to find the required order statistics. In one dimension, sort the remaining deficits once for each endpoint.\n\nThe total complexity is **$O(N\\log N)$ time and $O(N)$ memory** per test case, comfortably within the total $N\\le 10^5$ limit.\n\nUse `long long`: row sums and deficits can be on the order of $10^{14}$. Since every ceiling in the answer is also maximized with zero, use a helper that returns $0$ for a nonpositive numerator and otherwise returns `(x + d - 1) / d`. This avoids signed-division gotchas entirely.",
          "solution": "#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nll need(ll x, ll d) {\n    return x <= 0 ? 0 : (x + d - 1) / d;\n}\n\nint main() {\n    setIO();\n\n    int t;\n    cin >> t;\n    while (t--) {\n        int n, m, k;\n        cin >> n >> m >> k;\n        int N = n * m;\n        vector<ll> a(N), row(n), col(m);\n        for (int i = 0; i < n; ++i) {\n            for (int j = 0; j < m; ++j) {\n                cin >> a[i * m + j];\n                row[i] += a[i * m + j];\n                col[j] += a[i * m + j];\n            }\n        }\n\n        if (N == 1) {\n            cout << (a[0] >= 0 ? 0 : -1) << '\\n';\n            continue;\n        }\n        if (N == 2) {\n            cout << (k == 1 ? 0LL : abs(a[0] - a[1])) << '\\n';\n            continue;\n        }\n\n        vector<ll> h(N);\n        for (int i = 0; i < n; ++i) {\n            for (int j = 0; j < m; ++j) {\n                int p = i * m + j;\n                h[p] = row[i] + col[j] - 3 * a[p];\n            }\n        }\n\n        ll d = n + m - 3;\n        vector<ll> sorted = h;\n        sort(sorted.begin(), sorted.end());\n        ll ans = need(sorted[k - 1], d);\n\n        if (n == 1 || m == 1) {\n            for (int p : {0, N - 1}) {\n                ll A = h[p];\n                ll cur = need(A, d + 1);\n                if (k >= 2) {\n                    vector<ll> other;\n                    other.reserve(N - 1);\n                    for (int i = 0; i < N; ++i) {\n                        if (i != p) other.push_back(h[i]);\n                    }\n                    sort(other.begin(), other.end());\n                    ll B = other[k - 2];\n                    cur = max({cur, need(B, d), need(A + B, 2 * d)});\n                }\n                ans = min(ans, cur);\n            }\n        }\n\n        cout << ans << '\\n';\n    }\n}\n"
        },
        "id": "msg_0b25325c70016fNiC22vh2LNXM",
        "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03"
      },
      "parts": [
        {
          "type": "step-start",
          "id": "prt_0b25332d8001JSX5gnXb2an69C",
          "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03",
          "messageID": "msg_0b25325c70016fNiC22vh2LNXM"
        },
        {
          "type": "reasoning",
          "text": "**Verifying deficit handling**",
          "time": {
            "start": 1789698192090,
            "end": 1789698201383
          },
          "metadata": {
            "openai": {
              "itemId": "rs_04f2fb0851a1cac7016aaca09007fc87d0ab3e3f1a9ac00bd4",
              "reasoningEncryptedContent": "gAAAAABqrKCZyw6e7XjTQgwuA2LLxOMQ8IFe087W2TfuuD-838qry5cNwuSyUdynB5uvdSFEgv0QBYp0iCRneayERkceqSZE8wBTKyiar-gu6E1hKSPl1jd36HmRkW2QoXZlO3XPHiY1Aq5_aYxEEtV09dQI5w8bvhzWNa834uHsDFxZNS1b543rFbbxQQTOWAZSpC1y3dlfG_X9CBY5Pd9bzrSWOVc7gKltIOOtmAx6-jVzX4Y5uFzIPMD0YdXTLVJTvPyI6xCHZRgSGdaPd7PU3khgmL19JN970Rrx5xhHuA6kOyeb9BrZWu3N1cU0H8N4nNqXcW8cawD-O0hlyXy7pC_Nekp0oqleAz9CEIYXhvLcfRGtnP15hdNlvTh3JaJMKZY5twZ0GNZmcZGaIN44cPt-RhzUQN4IrMYMQkuPgESI7Ms_PdEFFCgtziuS0EU4fdxQHr6JcHpt7lFjId5R-yiLjhDKOsrc9BMyZDI85e0kxRlod7Un9M3XO5gq-eiAQh-j2KIi3RRXuq1zINbgwcdjowHeo0X_lHzsW8xyUxDRXVefWNshUE-50RXo1O8I6xkArN-ayEmCilfku6Clz1j8Kw1_Wr3ejshKDXbn4IoRzDrgA_JddP0rx7pvV3SGN2iiHhkEmEPdOgFvdrQSAUDPbmIhzMwAILJ6rvEJpXdyTnBMeQgGSTe4kaLcXKczOFnHtCX7KxP0FE6MoKELTbW73BPiQT6fj_mthtBomet-oPJNxVwEGWBYLhUPIx5XTT4FYyWvJQLucxv7SlskOe56AZHkdeWIotArX8hNGJas3TwBzFQUo0NanZjXoQOxCGkRDSgPxOEBxiYRp3nUi22EEyjJPVzKrs9tZX0Wt0PdTqwKYS8ejAveEm1f_nHpQTpWGEIvpGQ5TSMzZornJtWUw0C8lhRidZx2aTPMo_cbYi8TcSr3AmcVSSQIsyjp1C08D3HIP3dBRNz_ej-COguE9GxV7cP8hiOHLgOWn70t4G6EV_EYCGQmyrOoQN0OQAi7x1rYau-tYkxAGjvqVjrb78NJ3AiGzc4M3k2GfV__nfNwiuOxxVHXiSlGPFFs5FGbwbum70GseMxpNACfFHDQU3N7R0vaLUokoMtrdXfr_ORU1-d6tyVrFtVOMyljV9FlpIUdvExeQa_mA7xyedOAt5vNKtFmKs7fghPfYCbjbuxnKiu4rg7rm4G-l6mvwL71X0d57xfmxQdo98zgvB4GRUxIdxsUZ5UVEqYKgyStl67nBuVI17GWs9pztC5uYOhG1o9PQZG6LxHmT74lFFttuIENb2Q6vQLEZs4lG4ubo-isa37Fqee4bYchhr6m6VyJ0CoLUleFTsYi9XtXajqCVCN4CsYK9xUvUSkdrhsORrMow4nexl69g6K-YSv52EjbzctCh8bPVptjVW5MPsUJyhY65bodXelfZLeCnh4K4qQE7Q3y39XYGgeSZ3pKlAI0S1xVQPBxaBO25uqOkQReNmvqmE43jDgcySjkb0sVYs6QP7gI13j4BnY4WKpLQYUibZwc8xxDd1x8LGadHZPksIOkTmzW12c4AgisoaUv9hPxNs3kImnfvbbT8DACA1wcuiKrKnDcyZxdFNLFQmbYGXdGLkdBkhdfQ_J4icryeFNKkrHRwASOV29xKp-kgEqy_M9VqgyVkX1lgYKWCIz7orjIUXvnl2TROxCpnY5vOFOS2AVKTI0SHvKkznV-FjfPHUDgrYWUbdnTwFVKFTgpFBALekzjAWpbVY1wrvh6EB3ySmOIG43xhADj_3KWrynTmd6tCA7TXj5KEeifRt7u-_H8EkcWvF-WXoUzYp-BcU0Hy-ZX_qcxIZo0suYVYVMWm1x9z7fJTQp2T9ud9jIrmiZP8Z5JZYzW-6teLLjuDjZBwGuL7vLUDKJym4B1u6Y8gsj-r27sPioLF_YlxuBnCHSYEUfWTtfwozA3LdPE65yIzesoyj7056AfRiUtbZx3KcFJGp8bKJ4nIbhqysQgTxmafk-mzaWsFL0bkhk_DRQPmHHF6w1z9CJVIniXDVGS06O1uhBcLQEyNe8IqtdV3LzRXNZUvRgzK29zxuiGvaHkQB6PCb7wl1FDCCvgKzmveY2qxdGaqKABBgPaZ9xcpczikwdwcBt1Spp_nSCKXA3P_yRtVi3309kDfQUEq-HeQjj4ATo7KEsvHIGVvToiuFX4JQ0RlJF9-DHd5HObFQEXxAwTIIgwc5wUc8LBXjCnfn1uLm5yupM5b1lZi3sC86Stz9tUP1Ak0YV5SBwVTnZPA0SNAxSq3nT7gZ905mBEWuzDIFip2Eu1DnnP6XsVe4t5hM4f-nPS-oy7zWTMfY6VMgdjCAjB5gYxZOwDAktuASz18qkqBu8DTmK2lpdQqa82CaE2MGSnEVsFEcg9V8ejCmEL-5-CYliHjGMT3JWmiC9iDnpqns3mwXMNUQtdbVjnJQ=="
            }
          },
          "id": "prt_0b25332da001PweLk5fbssO2bu",
          "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03",
          "messageID": "msg_0b25325c70016fNiC22vh2LNXM"
        },
        {
          "type": "tool",
          "tool": "StructuredOutput",
          "callID": "call_cILaJphRajAPuTKKF2ECzZ4t",
          "state": {
            "status": "completed",
            "input": {
              "status": "success",
              "reason": null,
              "hints": [
                {
                  "order": 1,
                  "content": "Turn the peak condition into a single number. If $R_i$ and $C_j$ are the row and column sums, define $g_{i,j}=R_i+C_j-3v_{i,j}$. A cell is a peak exactly when $g_{i,j}\\le 0$. How much can one rectangle operation reduce this deficit?"
                },
                {
                  "order": 2,
                  "content": "When $n,m\\ge 2$, no operation can reduce any cell’s deficit by more than $n+m-3$. A full-matrix operation achieves exactly that reduction for **every** cell simultaneously. This gives the answer from the $k$-th smallest deficit. Be careful: this argument changes when one dimension is $1$."
                },
                {
                  "order": 3,
                  "content": "For a one-dimensional array of length $L$, write $h_i=S-2a_i$, where $S$ is its sum. An interval of length $\\ell$ reduces $h_i$ by $\\ell-2$ inside the interval and by $\\ell$ outside it. For $L\\ge 3$, every interval of length at most $L-2$ is dominated by the full interval. Only three types remain: the full array, all except the first element, and all except the last element."
                },
                {
                  "order": 4,
                  "content": "One skip-first operation plus one skip-last operation can be replaced by two full-array operations without making any deficit worse. So consider only full operations plus one chosen skip type. If the skipped endpoint is not among your required peaks, full operations alone are at least as good. Otherwise, choose the other $k-1$ peaks as the cells with the smallest initial deficits outside that endpoint."
                },
                {
                  "order": 5,
                  "content": "Let $d=L-2$, let $A$ be the skipped endpoint’s deficit, and let $B$ be the $(k-1)$-th smallest deficit among the other cells. Using $T$ operations, of which $y$ skip that endpoint, requires $dT+y\\ge A$, $dT-y\\ge B$, and $0\\le y\\le T$. Such an integer $y$ exists exactly when $dT\\ge B$, $(d+1)T\\ge A$, and $2dT\\ge A+B$. These give a direct formula for the minimum $T$. Try both endpoints and compare with full operations only. Handle $k=1$ and lengths $1,2$ separately."
                }
              ],
              "editorial": "The rectangle choices look like the hard part. The trick is to compare operations by how much they improve each cell’s peak condition. Most rectangles turn out to be completely unnecessary.\n\n## 1. Measure how far a cell is from being a peak\n\nLet $R_i$ be the sum of row $i$ and $C_j$ the sum of column $j$. The peak condition is\n\n$$v_{i,j}\\ge R_i+C_j-2v_{i,j}.$$\n\nDefine the **deficit**\n\n$$g_{i,j}=R_i+C_j-3v_{i,j}.$$\n\nA cell is a peak exactly when $g_{i,j}\\le 0$. Smaller deficits are always better, including when matrix entries are negative.\n\n## 2. The genuinely two-dimensional case\n\nSuppose $n,m\\ge 2$, and put $D=n+m-3$.\n\nConsider one operation and a fixed cell $(i,j)$.\n\n- If the rectangle contains the cell and has height $a$ and width $b$, the row sum decreases by $b$, the column sum by $a$, and the cell value by $1$. Its deficit decreases by $a+b-3\\le D$.\n- If the rectangle avoids the cell, it cannot intersect both its row and its column: that would include their intersection. It can therefore reduce the deficit by at most $\\max(n-1,m-1)\\le D$.\n\nSo **no operation improves any deficit by more than $D$**.\n\nNow choose the entire matrix. Every deficit decreases by exactly $D$. This simultaneously achieves the best possible improvement for every cell. No fancy rectangle gymnastics needed.\n\nAfter $T$ operations, a cell can be a peak only if $g_{i,j}\\le DT$, and $T$ full-matrix operations make every such cell a peak.\n\nIf $G$ is the $k$-th smallest deficit, the answer is therefore\n\n$$\\boxed{\\max\\left(0,\\left\\lceil\\frac{G}{D}\\right\\rceil\\right)}.$$\n\n## 3. One dimension needs its own argument\n\nIf one dimension is $1$, view the matrix as an array $a_1,\\dots,a_L$. Rectangles become contiguous intervals.\n\nLet $S=\\sum_i a_i$. The deficit is now\n\n$$h_i=S-2a_i.$$\n\nThis is also exactly what the row-and-column formula computes for a one-dimensional matrix.\n\nFirst handle two tiny cases:\n\n- **$L=1$:** the only cell is a peak iff $a_1\\ge 0$. The answer is $0$ if so, and $-1$ otherwise, since operations only decrease its value.\n- **$L=2$:** at least one cell is always a peak. For $k=1$, answer $0$. For $k=2$, both must be equal, so answer $|a_1-a_2|$: repeatedly decrease the larger cell. One operation changes their difference by at most $1$, so this is optimal.\n\nFrom now on, assume $L\\ge 3$ and write $d=L-2>0$.\n\n### Only three interval types matter\n\nAn operation on an interval of length $\\ell$ reduces a deficit by\n\n- $\\ell-2$ if the cell is inside;\n- $\\ell$ if the cell is outside.\n\nA full-array operation reduces every deficit by $L-2=d$.\n\nIf $\\ell\\le L-2$, both possible reductions are at most $d$. Replacing that operation with a full-array operation makes every deficit no larger. Thus we only need intervals of length $L$ or $L-1$:\n\n| Operation | First endpoint | Interior cells | Last endpoint |\n|---|---:|---:|---:|\n| Full array | $d$ | $d$ | $d$ |\n| Skip first | $d+1$ | $d-1$ | $d-1$ |\n| Skip last | $d-1$ | $d-1$ | $d+1$ |\n\nThe entries are deficit reductions.\n\nA skip-first operation paired with a skip-last operation reduces each endpoint’s deficit by $2d$, and every interior deficit by $2d-2$. Two full-array operations reduce all deficits by $2d$.\n\nTherefore, we can replace opposite skip operations in pairs. **An optimal strategy exists using full operations and at most one skip type.**\n\n## 4. Evaluate the remaining strategies\n\n### Full operations only\n\nLet $H$ be the $k$-th smallest value among all $h_i$. The required number is\n\n$$T_{\\mathrm{full}}=\\max\\left(0,\\left\\lceil\\frac{H}{d}\\right\\rceil\\right).$$\n\n### Full operations plus skipping endpoint $p$\n\nTry $p=1$ and $p=L$ separately.\n\nIf endpoint $p$ is not among a chosen set of $k$ peaks, replacing every skip operation with a full operation only helps that set. The full-only strategy already covers this possibility.\n\nSo we only need to evaluate strategies in which $p$ is one of the required peaks.\n\nLet $A=h_p$. Every other cell receives the same deficit reduction, so the easiest other $k-1$ cells to turn into peaks are those with the smallest initial deficits. For $k\\ge 2$, let $B$ be the $(k-1)$-th smallest deficit after removing endpoint $p$.\n\nSuppose there are $T$ operations in total, of which $y$ skip $p$. Then:\n\n- endpoint $p$ receives reduction $dT+y$;\n- every other cell receives reduction $dT-y$.\n\nThe requirements are\n\n$$dT+y\\ge A,\\qquad dT-y\\ge B,\\qquad 0\\le y\\le T.$$\n\nEquivalently,\n\n$$\\max(0,A-dT)\\le y\\le\\min(T,dT-B).$$\n\nAll bounds are integers. For $T\\ge 0$, this interval is nonempty exactly when\n\n$$dT\\ge B,\\qquad (d+1)T\\ge A,\\qquad 2dT\\ge A+B.$$\n\nTo see sufficiency explicitly, these inequalities ensure that each lower bound, $0$ and $A-dT$, is no greater than either upper bound, $T$ and $dT-B$.\n\nThus this endpoint strategy needs\n\n$$\\boxed{T_p=\\max\\left(0,\\left\\lceil\\frac{B}{d}\\right\\rceil,\\left\\lceil\\frac{A}{d+1}\\right\\rceil,\\left\\lceil\\frac{A+B}{2d}\\right\\rceil\\right)}.$$\n\nFor **$k=1$**, there are no other required peaks. Skipping $p$ every time gives its maximum improvement, $d+1$ per operation, so\n\n$$T_p=\\max\\left(0,\\left\\lceil\\frac{A}{d+1}\\right\\rceil\\right).$$\n\nThe one-dimensional answer is\n\n$$\\min(T_{\\mathrm{full}},T_1,T_L).$$\n\nThere is no need for binary search: the feasibility inequalities give the answer directly.\n\n## 5. Correctness argument\n\nFor $n,m\\ge 2$, every operation reduces each deficit by at most $D$, while a full-matrix operation achieves that bound for all cells. Hence the $k$-th-deficit formula is both a lower bound and achievable.\n\nFor a one-dimensional array of length at least $3$, replacing short intervals with full intervals never worsens any deficit. Replacing pairs of opposite skip operations with two full operations also never worsens any deficit. Consequently, some optimal solution uses only full operations and one skip type.\n\nIf its skipped endpoint is unnecessary, full operations alone achieve at least the same required peaks. Otherwise, we can choose that endpoint and the $k-1$ smallest remaining deficits, since all remaining cells improve equally. The derived inequalities are necessary and sufficient for precisely this strategy, and the formula finds their smallest feasible integer $T$.\n\nTaking the minimum over full-only and both endpoint strategies therefore finds the global optimum. The direct arguments for lengths $1$ and $2$ cover all remaining cases.\n\n## 6. Complexity and implementation\n\nLet $N=nm$. Compute all row sums, column sums, and deficits in $O(N)$ time. Sort the deficits to find the required order statistics. In one dimension, sort the remaining deficits once for each endpoint.\n\nThe total complexity is **$O(N\\log N)$ time and $O(N)$ memory** per test case, comfortably within the total $N\\le 10^5$ limit.\n\nUse `long long`: row sums and deficits can be on the order of $10^{14}$. Since every ceiling in the answer is also maximized with zero, use a helper that returns $0$ for a nonpositive numerator and otherwise returns `(x + d - 1) / d`. This avoids signed-division gotchas entirely.",
              "solution": "#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nll need(ll x, ll d) {\n    return x <= 0 ? 0 : (x + d - 1) / d;\n}\n\nint main() {\n    setIO();\n\n    int t;\n    cin >> t;\n    while (t--) {\n        int n, m, k;\n        cin >> n >> m >> k;\n        int N = n * m;\n        vector<ll> a(N), row(n), col(m);\n        for (int i = 0; i < n; ++i) {\n            for (int j = 0; j < m; ++j) {\n                cin >> a[i * m + j];\n                row[i] += a[i * m + j];\n                col[j] += a[i * m + j];\n            }\n        }\n\n        if (N == 1) {\n            cout << (a[0] >= 0 ? 0 : -1) << '\\n';\n            continue;\n        }\n        if (N == 2) {\n            cout << (k == 1 ? 0LL : abs(a[0] - a[1])) << '\\n';\n            continue;\n        }\n\n        vector<ll> h(N);\n        for (int i = 0; i < n; ++i) {\n            for (int j = 0; j < m; ++j) {\n                int p = i * m + j;\n                h[p] = row[i] + col[j] - 3 * a[p];\n            }\n        }\n\n        ll d = n + m - 3;\n        vector<ll> sorted = h;\n        sort(sorted.begin(), sorted.end());\n        ll ans = need(sorted[k - 1], d);\n\n        if (n == 1 || m == 1) {\n            for (int p : {0, N - 1}) {\n                ll A = h[p];\n                ll cur = need(A, d + 1);\n                if (k >= 2) {\n                    vector<ll> other;\n                    other.reserve(N - 1);\n                    for (int i = 0; i < N; ++i) {\n                        if (i != p) other.push_back(h[i]);\n                    }\n                    sort(other.begin(), other.end());\n                    ll B = other[k - 2];\n                    cur = max({cur, need(B, d), need(A + B, 2 * d)});\n                }\n                ans = min(ans, cur);\n            }\n        }\n\n        cout << ans << '\\n';\n    }\n}\n"
            },
            "output": "Structured output captured successfully.",
            "metadata": {
              "valid": true
            },
            "title": "Structured Output",
            "time": {
              "start": 1789698296786,
              "end": 1789698296790
            }
          },
          "metadata": {
            "openai": {
              "itemId": "fc_04f2fb0851a1cac7016aaca0994bd087d084584fbe822fea64"
            }
          },
          "id": "prt_0b2535737001WGgS2K7vM4gAqc",
          "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03",
          "messageID": "msg_0b25325c70016fNiC22vh2LNXM"
        },
        {
          "reason": "tool-calls",
          "type": "step-finish",
          "tokens": {
            "total": 30240,
            "input": 1025,
            "output": 3257,
            "reasoning": 230,
            "cache": {
              "write": 0,
              "read": 25728
            }
          },
          "cost": 0,
          "id": "prt_0b254cc27001n9au5zAK5g4nIO",
          "sessionID": "ses_f4dad599bffeUKldhhBo3T2D03",
          "messageID": "msg_0b25325c70016fNiC22vh2LNXM"
        }
      ]
    }
  ]
}
