{
  "info": {
    "id": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX",
    "slug": "cosmic-meadow",
    "projectID": "global",
    "directory": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-rwrVSG",
    "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-rwrVSG",
    "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": 19989,
      "output": 3237,
      "reasoning": 30,
      "cache": {
        "read": 37632,
        "write": 0
      }
    },
    "time": {
      "created": 1789696541819,
      "updated": 1789696649956
    }
  },
  "messages": [
    {
      "info": {
        "role": "user",
        "time": {
          "created": 1789696541838
        },
        "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_0b23a048e001tw3mYpro3ozh19",
        "sessionID": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX"
      },
      "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 2255C: \"Even If the World Turns\" (rated 2100).\n\nKnown problem metadata:\n- Contest ID: 2255\n- Index: C\n- Name: Even If the World Turns\n- Rating: 2100 (Codeforces difficulty rating)\n- Tags: communication, constructive algorithms, expression parsing, greedy, interactive, math, number theory\n- Problem URL: https://codeforces.com/contest/2255/problem/C\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\">C. Even If the World Turns</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><div class=\"epigraph\"><div class=\"epigraph-text\"><span class=\"tex-font-style-it\">Now, no matter what anyone says, I am the happiest girl in the world.</span></div><div class=\"epigraph-source\">— <span class=\"tex-font-style-it\">Chtholly</span></div></div><p><span class=\"tex-font-style-it\">This is a run-twice (communication) problem.</span></p><p>At the Fairy Warehouse, Nygglatho prepared an unusual game for Chtholly and Willem.</p><p>They would be separated and unable to exchange a single word. Between them there would be only a black-and-white picture&nbsp;— one that Nygglatho could shift, rotate, reflect, or even invert its colors.</p><p>Nygglatho saw it as a test of how well they understood each other. Chtholly and Willem perhaps saw only another kind of promise. No matter how the world turned, they would still find the same place.</p><p>The two players are Chtholly and Willem. The jury, acting as Nygglatho, first communicates with Chtholly. After Chtholly finishes, the jury communicates with Willem. Chtholly and Willem may agree on the strategy they will use beforehand, but they cannot directly pass information to each other.</p><p>In each test case, Nygglatho prepares a black-and-white picture consisting of $$$n \\times n$$$ cells and chooses a target cell $$$x$$$. The rows and columns are numbered from $$$1$$$ to $$$n$$$.</p><p>The picture contains $$$w$$$ black cells. <span class=\"tex-font-style-bf\">It is guaranteed that $$$\\gcd(n,w)=1$$$.</span></p><p>Nygglatho first shows the picture and the target cell $$$x$$$ to Chtholly. Chtholly must choose two cells and swap their colors. The two cells are allowed to be the same. If they are the same, or if they have the same color, the picture does not change. She cannot send Willem any other information.</p><p>Swapping colors does not move the target cell.</p><p>Nygglatho then secretly transforms the picture. She may perform the following operations any number of times, possibly zero, in any order:</p><ol> <li> Choose two integers $$$d_r$$$ and $$$d_c$$$ ($$$0\\le d_r,d_c \\lt n$$$) and cyclically shifts the picture. Every cell $$$(r,c)$$$ moves to $$$\\left(\\left(r-1+d_r\\right)\\bmod n+1,\\left(c-1+d_c\\right)\\bmod n+1\\right)$$$; </li><li> Rotate the picture clockwise by $$$90^\\circ$$$. One such rotation moves every cell $$$(r,c)$$$ to $$$(c,n+1-r)$$$; </li><li> Reflect the picture across its vertical axis. Such a reflection moves every cell $$$(r,c)$$$ to $$$(r,n+1-c)$$$; </li><li> Invert all colors. This changes every black cell into a white cell and every white cell into a black cell. </li></ol><p>The target cell $$$x$$$ undergoes every cyclic shift, rotation, and reflection in exactly the same way as the picture. Color inversions do not move it.</p><p>Finally, Nygglatho shows only the resulting picture to Willem. Willem must determine the final position of $$$x$$$.</p><p>Your program will be run exactly twice on each test. On the first run, it must act as Chtholly. On the second run, it must act as Willem. No information is preserved between the two runs except for the information passed by the jury according to the rules above.</p><p>The order of the test cases may be changed between the two runs.</p><p><span><span class=\"tex-font-style-bf\">First Run</span></span></p><p>On the first run, you are Chtholly.</p><p><span class=\"tex-font-style-bf\">Input</span></p><p>The first line contains the string <span class=\"tex-font-style-tt\">first</span>, indicating that this is the first run.</p><p>After this line, the remaining input has the following format.</p><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 an integer $$$n$$$ ($$$2 \\le n \\le 800$$$)&nbsp;— the height and width of the picture.</p><p>Each of the next $$$n$$$ lines contains a string of length $$$n$$$. The character $$$\\mathtt{\\#}$$$ denotes a black cell, and the character $$$\\mathtt{.}$$$ denotes a white cell.</p><p>The next line contains two integers $$$r_x$$$ and $$$c_x$$$ ($$$1 \\le r_x,c_x \\le n$$$)&nbsp;— the row and column of the target cell $$$x$$$.</p><p>Let $$$w$$$ be the number of black cells in the picture. It is guaranteed that $$$\\gcd(n,w)=1$$$.</p><p>It is guaranteed that the sum of $$$n^2$$$ over all test cases does not exceed $$$800^2$$$.</p><p><span class=\"tex-font-style-bf\">Output</span></p><p>For each test case, output four integers $$$r_1$$$, $$$c_1$$$, $$$r_2$$$, and $$$c_2$$$ ($$$1 \\le r_1,c_1,r_2,c_2 \\le n$$$)&nbsp;— the two cells whose colors Chtholly chooses to swap. The two cells are allowed to be the same.</p><p><span><span class=\"tex-font-style-bf\">Second Run</span></span></p><p>On the second run, you are Willem.</p><p><span class=\"tex-font-style-bf\">Input</span></p><p>The first line contains the string <span class=\"tex-font-style-tt\">second</span>, indicating that this is the second run.</p><p>After this line, the remaining input has the following format.</p><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 value of $$$t$$$ is the same as on the first run, although the test cases may appear in a different order.</p><p>The first line of each test case contains the integer $$$n$$$ ($$$2 \\le n \\le 800$$$).</p><p>Each of the next $$$n$$$ lines contains a string of length $$$n$$$, describing the resulting picture after Chtholly's swap and Nygglatho's transformations for the corresponding test case from the first run. Again, $$$\\mathtt{\\#}$$$ denotes a black cell and $$$\\mathtt{.}$$$ denotes a white cell.</p><p>It is guaranteed that the sum of $$$n^2$$$ over all test cases does not exceed $$$800^2$$$.</p><p><span class=\"tex-font-style-bf\">Output</span></p><p>For each test case, output two integers $$$r'_x$$$ and $$$c'_x$$$ ($$$1 \\le r'_x,c'_x \\le n$$$)&nbsp;— the row and column of the target cell after all transformations.</p><p><span class=\"tex-font-style-bf\">Hacks are disabled in this problem.</span></p></div><div class=\"sample-tests\"><div class=\"section-title\">Examples</div><div class=\"sample-test\"><div class=\"input\"><div class=\"title\">Input</div><pre>first\n2\n5\n#....\n.#...\n.....\n.....\n.....\n3 4\n5\n.....\n####.\n.....\n#####\n.....\n1 1</pre></div><div class=\"output\"><div class=\"title\">Output</div><pre>1 1 4 1\n1 1 1 1</pre></div><div class=\"input\"><div class=\"title\">Input</div><pre>second\n2\n5\n.....\n####.\n.....\n#####\n.....\n5\n.....\n.....\n....#\n..#..\n.....</pre></div><div class=\"output\"><div class=\"title\">Output</div><pre>1 1\n1 4</pre></div></div></div><div class=\"note\"><div class=\"section-title\">Note</div><p>The first example shows Chtholly's run for two test cases. The second example shows Willem's run for the same two test cases. The outputs shown for Chtholly's run are just one possible set of valid choices.</p><p>In the first test case, Chtholly swaps cells $$$(1,1)$$$ and $$$(4,1)$$$. Afterwards, the two black cells are at $$$(2,2)$$$ and $$$(4,1)$$$.</p><p>In this example, Nygglatho performs the following transformations.</p><ul> <li> she cyclically shifts the picture one row down and two columns to the right; </li><li> she rotates the picture clockwise by $$$90^\\circ$$$ once; </li><li> she reflects the picture across its vertical axis; </li></ul><p>The target moves from $$$(3,4)$$$ to $$$(4,1)$$$ after the cyclic shift, then to $$$(1,2)$$$ after the rotation, and finally to $$$(1,4)$$$ after the reflection. The two black cells finally arrive at $$$(3,5)$$$ and $$$(4,3)$$$, giving exactly the picture in Willem's run. Willem reports the target position $$$(1,4)$$$.</p><p>In the second test case, Chtholly can choose the same cell, in which case nothing changes. Nygglatho can then also choose to do nothing.</p></div>\n</problem-statement>\n\n\nSource lookup status:\n<source-lookup-status>\n- Tutorial (en): https://codeforces.com/blog/entry/155840 loaded (200 OK; title: \"Codeforces Round 1116 (Div. 1, Div. 2) Editorial - Codeforces\"; mentions 2255C)\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_0b23a04900011xS7IAkQ48Qups",
          "sessionID": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX",
          "messageID": "msg_0b23a048e001tw3mYpro3ozh19"
        }
      ]
    },
    {
      "info": {
        "parentID": "msg_0b23a048e001tw3mYpro3ozh19",
        "role": "assistant",
        "mode": "nudge-generation",
        "agent": "nudge-generation",
        "variant": "low",
        "path": {
          "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-rwrVSG",
          "root": "/"
        },
        "cost": 0,
        "tokens": {
          "total": 10008,
          "input": 5368,
          "output": 32,
          "reasoning": 0,
          "cache": {
            "write": 0,
            "read": 4608
          }
        },
        "modelID": "gpt-6-astra",
        "providerID": "openai",
        "time": {
          "created": 1789696541961,
          "completed": 1789696548918
        },
        "finish": "tool-calls",
        "id": "msg_0b23a0509001GLLmzALjR21VnY",
        "sessionID": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX"
      },
      "parts": [
        {
          "type": "step-start",
          "id": "prt_0b23a134d0018q2wjTKPfR2VYM",
          "sessionID": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX",
          "messageID": "msg_0b23a0509001GLLmzALjR21VnY"
        },
        {
          "type": "tool",
          "tool": "webfetch",
          "callID": "call_LZhdFp2oqIJaWdIa6tzmX14a",
          "state": {
            "status": "completed",
            "input": {
              "url": "https://codeforces.com/blog/entry/155840",
              "format": "markdown"
            },
            "output": "---\nimage: https://codeforces.com/codeforces.org/s/70137/images/codeforces-telegram-square-1024x1024.png\ntitle: Codeforces Round 1116 (Div. 1, Div. 2) Editorial - Codeforces\ndescription: Codeforces. Programming competitions and contests, programming community\n---\n\n<!DOCTYPE HTML PUBLIC \"-//W3C//DTD HTML 4.01//EN\"> \n\n  \n[![Codeforces](//codeforces.com/codeforces.org/s/70137/images/codeforces-logo-with-telegram-555x130.png \"Codeforces\")](https://codeforces.com/) \n\n[![In English](//codeforces.com/codeforces.org/s/70137/images/flags/24/gb.png \"In English\")](https://codeforces.com/blog/entry/155840?locale=en) [![По-русски](//codeforces.com/codeforces.org/s/70137/images/flags/24/ru.png \"По-русски\")](https://codeforces.com/blog/entry/155840?locale=ru) \n\n[Enter](https://codeforces.com/enter?back=%2Fblog%2Fentry%2F155840)| [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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/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/70137/images/icons/x-update-12x12.png \"Text created or updated\") ![Necropost](//codeforces.com/codeforces.org/s/70137/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/70137/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/70137/images/icons/comment-12x12.png \"New comment(s)\") ![Necropost](//codeforces.com/codeforces.org/s/70137/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/70137/images/icons/x-update-12x12.png \"Text created or updated\") ![Necropost](//codeforces.com/codeforces.org/s/70137/images/icons/hourglass.png \"Necropost\")\n\n|  | [Detailed →](https://codeforces.com/recent-actions) |\n|  | --------------------------------------------------- |\n\n* [paulzrm](https://codeforces.com/profile/paulzrm)\n* [Blog](https://codeforces.com/blog/paulzrm)\n* [Teams](https://codeforces.com/teams/with/paulzrm)\n* [Submissions](https://codeforces.com/submissions/paulzrm)\n* [Groups](https://codeforces.com/groups/with/paulzrm)\n* [Contests](https://codeforces.com/contests/with/paulzrm)\n* [Problemsetting](https://codeforces.com/contests/writer/paulzrm)\n\n### [paulzrm's blog](https://codeforces.com/blog/paulzrm)\n\n[ Codeforces Round 1116 (Div. 1, Div. 2) Editorial ](https://codeforces.com/blog/entry/155840) \n\nBy [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\"), [history](https://codeforces.com/topic/156483/en2), 6 weeks ago, ![In English](//codeforces.com/codeforces.org/s/70137/images/flags/24/gb.png \"In English\") \n\n[2256A-Three Numbers on the Blackboard](https://codeforces.com/contest/2256/problem/A) Idea: [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\")\n\n**Hint 1**\n\nSort the numbers as $$$a\\\\le b\\\\le c$$$.\n\n**Hint 2**\n\nAfter at least one operation, consider the median.\n\n**Hint 3**\n\nDoing nothing gives $$$c-a$$$; replacing $$$c$$$ with $$$a+b$$$ gives $$$b$$$.\n\n**Tutorial**\n\nSuppose the current numbers are $$$x\\\\le y\\\\le z$$$. An operation keeps two numbers and replaces the third with their sum. Since all values are nonnegative, the new sum is no smaller than either retained number. Thus the new median is the larger retained number, which is at least the old median $$$y$$$. The median never decreases.\n\nAfter an operation that retains $$$u\\\\le v$$$, the numbers are $$$u,v,u+v$$$. Their range is exactly $$$v$$$, also their median. Hence every nonempty sequence of operations ends with a range of at least the initial median $$$b$$$.\n\nWith no operation the range is $$$c-a$$$. Replacing $$$c$$$ with $$$a+b$$$ produces $$$a,b,a+b$$$, whose range is $$$b$$$. Therefore the answer is $$$\\\\min(c-a,b)$$$.\n\nThe time and space complexities are both $$$O(1)$$$.\n\n[2256B-Domino Tiles](https://codeforces.com/contest/2256/problem/B) Idea: [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\")\n\n**Hint 1**\n\nCancel the common term in the inequality between two adjacent dominoes.\n\n**Hint 2**\n\nThe condition is equivalent to $$$s\\_i\\\\ne s\\_{i+2}$$$ for every $$$i$$$.\n\n**Hint 3**\n\nOnce the first two characters are fixed, the rest are determined.\n\n**Tutorial**\n\nAdjacent dominoes have different weights exactly when $$$s\\_i+s\\_{i+1}\\\\ne s\\_{i+1}+s\\_{i+2}$$$.\n\nCanceling $$$s\\_{i+1}$$$ gives $$$s\\_i\\\\ne s\\_{i+2}$$$. Since the alphabet is binary, this is equivalent to $$$s\\_{i+2}=1-s\\_i$$$.\n\nTherefore $$$s\\_1,s\\_2$$$ determine the whole string. Enumerate their four assignments and check whether the implied string agrees with every known character. The answer is at most $$$4$$$, so the modulus does not affect it.\n\nThe time complexity is $$$O(n)$$$ and the space complexity is $$$O(1)$$$.\n\n[2255A-Hot Potatoes at the Fairy Warehouse](https://codeforces.com/contest/2255/problem/A) / [2256C](https://codeforces.com/contest/2256/problem/C) Idea: [Error\\_Yuan](https://codeforces.com/profile/Error%5FYuan \"Grandmaster Error_Yuan\")\n\n**Hint 1**\n\nSplit the 1s on the cycle into maximal runs. Unless a run covers the whole cycle, initially only its last potato can move.\n\n**Hint 2**\n\nPassing too early gives the next opportunity to players of the other team.\n\n**Hint 3**\n\nThe last player of each run waits until the final round. Every initial 10 becomes 01, all other potatoes stay still, and the exact value of $$$k$$$ is irrelevant.\n\n**Tutorial**\n\nThe total score is always the number of potatoes, so the game is zero-sum. Consider a maximal run of 1s followed by an empty position. Every potato except the last one is blocked.\n\nIf the last potato is passed early, it reaches the other team and leaves an empty position behind, enabling the preceding potato. Both newly enabled players belong to the other team, so with another round the opponent may cancel the point just gained. Passing in the final round leaves no response and is optimal.\n\nThus only the last potato of each run moves one step in the final round. If the whole cycle is filled, nothing moves. For an initial position $$$i$$$:\n\n* if $$$s\\_i=1$$$ and $$$s\\_{i+1}=0$$$, the team owning $$$i$$$ scores;\n* if $$$s\\_i=s\\_{i+1}=1$$$, the other team scores.\n\nScan the cycle once. The implementation is zero-indexed: even positions belong to red and odd positions to blue. Their scores are printed in that order.\n\nThe time complexity is $$$O(n)$$$ and the space complexity is $$$O(1)$$$.\n\n[2255B-A Ribbon for Tomorrow](https://codeforces.com/contest/2255/problem/B) / [2256D](https://codeforces.com/contest/2256/problem/D) Idea: [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\")\n\n**Hint 1**\n\nCheck whether a valid reversal can change either run count or the first character.\n\n**Hint 2**\n\nIn $$$x^ay^bx^c$$$, choosing one endpoint in each $$$x$$$-run lets us redistribute the total length of those two runs.\n\n**Hint 3**\n\nOperations redistribute run lengths but do not change the order of run colors.\n\n**Tutorial**\n\nSplit the string into maximal runs. The two endpoints of an operation contain the same character. Internal adjacencies are only reversed, and the endpoint character on either boundary stays unchanged. Hence the numbers of 0-runs and 1-runs are invariant, as are the character counts.\n\nIf the reversal contains the first position, the new first character is the old right endpoint, equal to the old first character. Otherwise it is untouched. Thus the first character is invariant. Since run colors alternate, the run counts and first character determine their entire order.\n\nConversely, consider $$$x^ay^bx^c$$$. A reversal with one endpoint in each $$$x$$$-run can split the total $$$a+c$$$ into any two positive lengths while leaving the middle run unchanged. Repeating this between adjacent runs of the same color realizes any positive composition of all 0s, and independently any positive composition of all 1s. These invariants are therefore sufficient.\n\nIf character $$$x$$$ occurs $$$cnt\\_x$$$ times in $$$seg\\_x$$$ nonempty runs, its run lengths can be chosen in $$$\\\\binom{cnt\\_x-1}{seg\\_x-1}$$$ ways. The two colors are independent, so the answer is $$$\\\\binom{cnt\\_0-1}{seg\\_0-1}\\\\binom{cnt\\_1-1}{seg\\_1-1}$$$.\n\nAn absent character contributes a factor of $$$1$$$.\n\nWe can precompute inverses without binary exponentiation. Let the prime modulus be $$$P$$$. Since every required $$$i \\\\lt P$$$, its inverse exists. From $$$P=\\\\left\\\\lfloor\\\\frac Pi\\\\right\\\\rfloor i+(P\\\\bmod i)$$$ we obtain $$$i^{-1}=-\\\\left\\\\lfloor\\\\frac Pi\\\\right\\\\rfloor(P\\\\bmod i)^{-1}\\\\pmod P$$$.\n\nBecause $$$P\\\\bmod i \\\\lt i$$$, enumerating $$$i$$$ in increasing order computes each inverse from an earlier one. Factorials and inverse factorials follow in the same loop.\n\nEach test case takes $$$O(n)$$$ time. Preprocessing uses $$$O(N)$$$ time and space, where $$$N=10^6$$$.\n\n[2255C-Even If the World Turns](https://codeforces.com/contest/2255/problem/C) / [2256E](https://codeforces.com/contest/2256/problem/E) Idea: [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\")\n\n**Hint 1**\n\nLet $$$w$$$ be the number of black cells and $$$S\\_r,S\\_c$$$ their coordinate sums modulo $$$n$$$. Observe how they change after a shift.\n\n**Hint 2**\n\nSince $$$\\\\gcd(w,n)=1$$$, $$$(w^{-1}S\\_r,w^{-1}S\\_c)$$$ behaves like the center of mass of the picture.\n\n**Hint 3**\n\nSwapping a black cell $$$p$$$ with a white cell $$$p+\\\\delta$$$ increases the coordinate sum by $$$\\\\delta$$$.\n\n**Hint 4**\n\nIn the first run, move the center to the target. In the second run, recompute it from the transformed picture.\n\n**Tutorial**\n\nIdentify coordinates $$$1,2,\\\\ldots,n$$$ with $$$0,1,\\\\ldots,n-1$$$ in $$$\\\\mathbb Z\\_n$$$. Let $$$S=(S\\_r,S\\_c)$$$ be the sum of black-cell coordinates. A shift by $$$d$$$ changes it to $$$S+wd$$$. Define $$$C=w^{-1}S$$$. Then $$$C$$$ changes to $$$C+d$$$, exactly as an ordinary cell does.\n\nEvery rotation or reflection is an affine map $$$p\\\\mapsto Mp+t$$$. It changes the sum to $$$MS+wt$$$ and therefore changes the center to $$$MC+t$$$, again exactly like a cell.\n\nAfter color inversion, $$$w'=n^2-w\\\\equiv-w\\\\pmod n$$$. The coordinate sum of the whole board is $$$0$$$ modulo $$$n$$$, so the new black-cell sum is $$$S'=-S$$$. Hence the new center is $$$(-w)^{-1}(-S)=C$$$. Color inversion does not affect it.\n\nIn the first run, we want the sum after swapping to be $$$wx$$$. Let $$$\\\\delta=wx-S$$$. If $$$\\\\delta=0$$$, swap one cell with itself. Otherwise find a black cell $$$p$$$ for which $$$p+\\\\delta$$$ is white and swap them.\n\nSuch a cell must exist. Otherwise the black-cell set would be invariant under translation by nonzero $$$\\\\delta$$$. Every orbit of this translation has length $$$L \\\\gt 1$$$ with $$$L\\\\mid n$$$. The black cells would be a union of complete orbits, so $$$L\\\\mid w$$$, contradicting $$$\\\\gcd(w,n)=1$$$.\n\nThe swap makes the center equal to the target. All later transformations move them together. In the second run, recompute $$$w,S\\_r,S\\_c$$$ and output $$$w^{-1}S$$$. Since $$$n\\\\le800$$$, enumerate $$$1\\\\le t \\\\lt n$$$ until $$$wt\\\\equiv1\\\\pmod n$$$; no extended Euclidean algorithm is needed.\n\nEach run takes $$$O(n^2)$$$ time and $$$O(n^2)$$$ space.\n\n[2255D-How Long Until Nothing Remains?](https://codeforces.com/contest/2255/problem/D) / [2256F](https://codeforces.com/contest/2256/problem/F) Idea: [Error\\_Yuan](https://codeforces.com/profile/Error%5FYuan \"Grandmaster Error_Yuan\")\n\n**Hint 1**\n\nFix the chosen index in every second and work backward from the final zero array. A current upper bound $$$x$$$ becomes either $$$2x+1$$$ or $$$2x$$$ after one backward step.\n\n**Hint 2**\n\nIf index $$$i$$$ is selected in the seconds in $$$S$$$, it can end at zero after $$$T$$$ seconds exactly when $$$a\\_i\\\\le\\\\sum\\_{s\\\\in S}2^{s-1}$$$.\n\n**Hint 3**\n\nFor fixed $$$T$$$, distribute $$$1,2,4,\\\\ldots,2^{T-1}$$$ among the demands. Process them in decreasing order and always choose the largest remaining demand.\n\n**Tutorial**\n\nFix all choices during the $$$T$$$ seconds and consider position $$$i$$$. Working backward, if values at most $$$x$$$ can already reach zero, then a preceding selected step allows $$$y$$$ exactly when $$$\\\\lfloor y/2\\\\rfloor\\\\le x$$$, or $$$y\\\\le2x+1$$$. A nonselected step gives $$$\\\\lceil y/2\\\\rceil\\\\le x$$$, or $$$y\\\\le2x$$$. Repeating this proves the condition in Hint 2.\n\nThus feasibility in $$$T$$$ seconds is equivalent to assigning every capacity $$$1,2,\\\\ldots,2^{T-1}$$$ to one demand so that demand $$$a\\_i$$$ receives total capacity at least $$$a\\_i$$$.\n\nProcess capacities from largest to smallest. Let the current capacity be $$$p$$$ and the largest remaining demand be $$$x$$$. If $$$x \\\\gt p$$$, all smaller capacities sum to only $$$p-1$$$, so every feasible assignment must give $$$p$$$ to $$$x$$$. If $$$x\\\\le p$$$, let $$$p$$$ finish $$$x$$$. In any feasible assignment giving $$$p$$$ to some $$$y\\\\le x$$$ and a set of smaller capacities to $$$x$$$, swapping those two assignments remains feasible. Hence the greedy rule is correct.\n\nEvery positive number must be selected at least once, since repeated ceiling division alone never reaches zero. Thus $$$T\\\\ge n$$$. Also $$$a\\_i \\\\lt 2^{30}$$$, so $$$T=n+30$$$ is always sufficient. Binary-search $$$T$$$ in this interval.\n\nEvery capacity at least $$$2^{30}$$$ can finish one demand alone. Remove the largest $$$T-30$$$ demands with these capacities. At most $$$30$$$ demands remain; put them in a max-heap and simulate only $$$2^{29},\\\\ldots,1$$$.\n\nSorting takes $$$O(n\\\\log n)$$$. Each check takes $$$O(30\\\\log30)$$$, there are $$$O(\\\\log30)$$$ checks, and the space complexity is $$$O(n)$$$.\n\n[2255E1-What Will Remain at the End? (Easy Version)](https://codeforces.com/contest/2255/problem/E1) Idea: [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\")\n\n**Hint 1**\n\nFor each position, its values over all versions form a sequence. Maintain its sum, maximum prefix, maximum suffix, and maximum subarray when appending a segment.\n\n**Hint 2**\n\nA value is always in $$${-1,0,1}$$$. For each possible initial value, a sequence of operations can store its final value and the summary of the generated history.\n\n**Hint 3**\n\nThis representation is closed under concatenating operation sequences, so it can serve as a lazy segment-tree tag.\n\n**Hint 4**\n\nRange updates only compose tags. A position's history is materialized when that position is queried by pushing its root-to-leaf tags.\n\n**Tutorial**\n\nFor a nonempty sequence segment, store $$$(S,P,Q,M)$$$: its sum, maximum prefix sum, maximum suffix sum, and maximum nonempty subarray sum. Two adjacent summaries merge in $$$O(1)$$$; a crossing maximum subarray is the left maximum suffix followed by the right maximum prefix.\n\nFix one array position. Its value lies in $$$V={-1,0,1}$$$. For an operation segment $$$T$$$, let $$$f\\_T(s)$$$ be the final value from initial state $$$s$$$, and let $$$H\\_T(s)$$$ summarize the values recorded while executing $$$T$$$. There are only three states, so this representation has constant size.\n\nIf $$$A$$$ is followed by $$$B$$$, then $$$f\\_{AB}(s)=f\\_B(f\\_A(s))$$$ and $$$H\\_{AB}(s)=H\\_A(s)\\\\mathbin{\\\\Vert}H\\_B(f\\_A(s))$$$, where $$$\\\\Vert$$$ concatenates two histories and merges their summaries. Tags therefore compose in $$$O(1)$$$.\n\nEach segment-tree node stores the unpushed operation segment applying to its whole interval. A range update composes tags in $$$O(\\\\log n)$$$ nodes. After each operation, a record-current-value event is applied to the whole tree to create the new version.\n\nA query at $$$p$$$ pushes the tags on its root-to-leaf path and materializes all history since the previous query. If the old history has maximum suffix $$$Q\\_0$$$ and maximum subarray $$$M\\_0$$$, while the new segment has $$$(S,P,Q,M)$$$, then $$$M'=\\\\max(M\\_0,M,Q\\_0+P)$$$ and $$$Q'=\\\\max(Q,Q\\_0+S)$$$.\n\nOnly these two old values need to persist at each position.\n\nRecord version $$$0$$$ initially. A type-4 operation at time $$$i$$$ queries versions $$$0$$$ through $$$i-1$$$, so answer it before recording version $$$i$$$, which equals version $$$i-1$$$. Other operations update first and record afterward. Decode online input modulo $$$2^{64}$$$ exactly as specified.\n\nThe total time complexity is $$$O((n+q)\\\\log n)$$$ and the space complexity is $$$O(n)$$$.\n\n[2255E2-What Will Remain at the End? (Hard Version)](https://codeforces.com/contest/2255/problem/E2) Idea: [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\")\n\n**Hint 1**\n\nThe segment tree and historical maximum-subarray framework are unchanged from G1\\. Only the constant-size operation tag changes.\n\n**Hint 2**\n\nBefore the first assignment, after fixing the sign of the initial $$$x$$$, every value is one of $$$-|x|,0,|x|$$$.\n\n**Hint 3**\n\nThe first assignment removes all dependence on $$$x$$$. Split a tag into a coefficient prefix depending on $$$|x|$$$ and a fixed numeric suffix.\n\n**Hint 4**\n\nTo compose $$$A$$$ followed by $$$B$$$, distinguish whether an assignment has occurred in either segment.\n\n**Tutorial**\n\n**Solution 1**\n\nWithout assignments, negation changes only the sign and $$$\\\\max(x,0)$$$ only replaces a negative value with zero. For each of the three initial signs, store the summary of the coefficient sequence and the final coefficient in $$${-1,0,1}$$$. Substituting the real input multiplies the sequence by $$$|x|$$$.\n\nThe first assignment is a dividing point. Before it, values still depend on $$$x$$$; from it onward, they are fixed. Represent an operation segment by:\n\n* a coefficient prefix depending on $$$|x|$$$, stored for all three initial signs;\n* a fixed numeric suffix and the final actual value.\n\nThis representation has constant size. It is also closed under concatenation.\n\nIf $$$A$$$ contains an assignment, its output is constant. Feeding that constant into $$$B$$$ fixes all history produced by $$$B$$$, so concatenate the fixed histories.\n\nIf neither segment contains an assignment, compose their three coefficient states exactly as in G1.\n\nIf only $$$B$$$ contains an assignment, append the part of $$$B$$$ before its first assignment to each coefficient prefix of $$$A$$$. The fixed suffix of $$$B$$$ becomes the fixed suffix of the result. Every case takes $$$O(1)$$$.\n\nAssignment, negation, $$$\\\\max(x,0)$$$, and recording a version all have direct tags of this form. The remaining segment-tree logic and version ordering are identical to G1.\n\nThe total time complexity is $$$O((n+q)\\\\log n)$$$ and the space complexity is $$$O(n)$$$.\n\n**Solution 2**\n\nThere is also an amortized solution that performs operation $$$3$$$ directly on the segment tree. Store the current minimum $$$mn$$$ and maximum $$$mx$$$ in every node. Range assignment and range negation use ordinary lazy tags. For operation $$$3$$$:\n\n* if $$$mn\\\\ge0$$$, do nothing;\n* if $$$mx\\\\le0$$$, assign zero to the whole node;\n* otherwise the node contains both positive and negative values, so recurse into its children.\n\nThus operation $$$3$$$ never has to appear in a historical tag. It is materialized as several range assignments to zero.\n\nCall a segment-tree node bad if $$$mn \\\\lt 0 \\\\lt mx$$$.\n\nLet the potential $$$\\\\Phi$$$ be the number of bad nodes. When a fully covered node is negated, $$$(mn,mx)$$$ becomes $$$(-mx,-mn)$$$, so the bad status of that node and every node below it is preserved. Assignment can only remove bad nodes. A range assignment or negation recomputes nodes only on its two boundary paths, so it creates at most $$$O(\\\\log n)$$$ new bad nodes.\n\nNow consider operation $$$3$$$. Apart from $$$O(\\\\log n)$$$ nodes on the two boundary paths, every internal node into which the recursion continues was bad before the operation. After the operation, its interval contains no negative value, so that bad node has disappeared. A recursion tree has at most one more terminal node than internal nodes. Hence one operation costs $$$O(\\\\log n+D)$$$, where $$$D$$$ is the number of bad nodes destroyed by this operation. Operation $$$3$$$ creates no bad nodes. The initial potential is $$$O(n)$$$, while all other operations increase it by only $$$O(q\\\\log n)$$$ in total. Therefore all operation-3 recursions take $$$O(n+q\\\\log n)$$$ time altogether.\n\nThe historical part still uses the four-value sequence summary from G1\\. After every operation, append one record-current-value event at the root. Since every clamp has already been materialized as assignment to zero, pending historical actions contain only assignment, negation, and recording.\n\nBefore the first assignment, an initial value $$$x$$$ can only become $$$x$$$ or $$$-x$$$. It is therefore enough to keep two initial states, negative and nonnegative, storing the generated coefficient-sequence summary and the final sign. After the first assignment, all later values are independent of the initial input, so keep a fixed numeric suffix and the final value. Two such tags still compose in $$$O(1)$$$ in chronological order.\n\nFor a query at $$$p$$$, push all tags on the root-to-leaf path and append the new history summary to the already materialized history of that position. Version $$$0$$$ and type-4 recording are handled in the same order as in Solution 1.\n\nThe total time complexity is $$$O(n+q\\\\log n)$$$ and the space complexity is $$$O(n)$$$.\n\n[2255F-Who Will Witness the End?](https://codeforces.com/contest/2255/problem/F) Idea: [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\")\n\n**Hint 1**\n\nView each factor $$$a\\_u+a\\_v$$$ as choosing one endpoint of edge $$$(u,v)$$$. Every exponent is $$$0$$$, $$$1$$$, or $$$2$$$. The numbers of vertices with exponents $$$0$$$ and $$$2$$$ are equal, and these two types alternate after all other vertices are removed.\n\n**Hint 2**\n\nThe coefficient of a monomial type depends only on the number of variables appearing twice. Express the answer using $$$e\\_re\\_{n-r}$$$, where $$$e\\_r$$$ is the $$$r$$$-th elementary symmetric polynomial.\n\n**Hint 3**\n\nCoefficient comparison gives a linear system that can be eliminated from high indices to low indices. Direct elimination is $$$O(n^2)$$$; use the ratio of adjacent $$$c\\_k$$$ and Pascal's identity to derive a short recurrence.\n\n**Hint 4**\n\nAdjacent equations give a second-order recurrence in $$$h\\_r,h\\_{r+1},h\\_{r+2}$$$. All $$$e\\_r$$$ are coefficients of $$$\\\\prod\\_{i=1}^n(1+a\\_ix)$$$ and can be computed by divide-and-conquer NTT.\n\n**Tutorial**\n\nExpand the weight of one cyclic ordering. Choosing one term from $$$a\\_u+a\\_v$$$ is equivalent to orienting edge $$$(u,v)$$$ toward the chosen endpoint.\n\nEvery vertex has degree two, so its exponent is $$$0$$$, $$$1$$$, or $$$2$$$. The exponent sum is $$$n$$$, hence the numbers of vertices with exponents $$$0$$$ and $$$2$$$ are equal; call both numbers $$$k$$$.\n\nFix these two sets. An exponent-0 vertex has both edges directed outward, an exponent-2 vertex has both directed inward, and every ordinary vertex has one edge in and one out. After ordinary vertices are removed, the two special types must alternate. Conversely, alternation uniquely determines the orientations of all paths between them.\n\nFor $$$k\\\\ge1$$$, the $$$2k$$$ special vertices have $$$(2k-1)!$$$ relative cyclic orders. Fix one exponent-0 vertex as the start. The exponent-2 vertices can be permuted arbitrarily, as can the remaining exponent-0 vertices, giving $$$k!(k-1)!$$$ alternating orders. Inserting the other $$$n-2k$$$ vertices does not change this ratio. Therefore the coefficient of this monomial type over all cyclic orders is\n\n$$$ c\\_k=\\\\frac{(n-1)!k!(k-1)!}{(2k-1)!}. $$$\n\nFor $$$k=0$$$, every vertex has one edge in and one out. The whole cycle must be consistently clockwise or counterclockwise, so\n\n$$$ c\\_0=2(n-1)!. $$$\n\nNow consider a symmetric-polynomial representation. Let $$$e\\_r$$$ be defined by\n\n$$$ \\\\prod\\_{i=1}^n(1+a\\_ix)=\\\\sum\\_{r=0}^n e\\_rx^r. $$$\n\nIn $$$e\\_re\\_{n-r}$$$, fix a monomial with $$$k$$$ variables appearing twice and $$$k$$$ absent. Repeated variables must be chosen from both factors and absent variables from neither. Choose $$$r-k$$$ of the remaining $$$n-2k$$$ variables for the first factor, so its coefficient is\n\n$$$ \\\\binom{n-2k}{r-k}. $$$\n\nLet $$$m=\\\\lfloor n/2\\\\rfloor$$$ and write the answer as\n\n$$$ \\\\sum\\_{r=0}^m h\\_re\\_re\\_{n-r}. $$$\n\nComparing each monomial type gives\n\n$$$ c\\_k=\\\\sum\\_{r=k}^m\\\\binom{n-2k}{r-k}h\\_r \\\\qquad(0\\\\le k\\\\le m). $$$\n\nEquation $$$k$$$ contains only $$$h\\_k,h\\_{k+1},\\\\ldots,h\\_m$$$, with coefficient $$$1$$$ on $$$h\\_k$$$. Thus the variables can be solved from large indices to small indices, but doing so directly takes $$$O(n^2)$$$.\n\nFor $$$k\\\\ge1$$$, divide two adjacent coefficients:\n\n$$$ \\\\begin{aligned} \\\\frac{c\\_{k+1}}{c\\_k} &=\\\\frac{(k+1)!k!}{(2k+1)!} \\\\frac{(2k-1)!}{k!(k-1)!}\\\\\\\\ &=\\\\frac{k(k+1)}{(2k)(2k+1)}\\\\\\\\ &=\\\\frac{k+1}{2(2k+1)}. \\\\end{aligned} $$$\n\nThe $$$k=0$$$ case follows directly from $$$c\\_0,c\\_1$$$. Hence\n\n$$$ (4k+2)c\\_{k+1}-(k+1)c\\_k=0\\. $$$\n\nDefine\n\n$$$ L\\_k=(k+1)c\\_k-(4k+2)c\\_{k+1}=0 \\\\qquad(0\\\\le k \\\\lt m). $$$\n\nFix $$$k$$$, set $$$N=n-2k$$$ and $$$j=r-k$$$, and let $$$A\\_t=\\\\binom{N-2}{t}$$$, with out-of-range binomial coefficients equal to zero.\n\nUsing Pascal's identity twice,\n\n$$$ \\\\binom Nj=A\\_j+2A\\_{j-1}+A\\_{j-2}, $$$\n\nNow use\n\n$$$ jA\\_j=(N-j-1)A\\_{j-1},\\\\qquad (N-j)A\\_{j-2}=(j-1)A\\_{j-1}, $$$\n\ntogether with $$$r=k+j$$$ and $$$n=N+2k$$$. The entire coefficient calculation can be displayed as\n\n$$$ \\\\begin{aligned} \\[h\\_r\\]L\\_k &=(k+1)\\\\binom Nj-(4k+2)\\\\binom{N-2}{j-1}\\\\\\\\ &=(k+1)(A\\_j+2A\\_{j-1}+A\\_{j-2})-(4k+2)A\\_{j-1}\\\\\\\\ &=(k+1)A\\_j-2kA\\_{j-1}+(k+1)A\\_{j-2}\\\\\\\\ &=(r+1)A\\_j-(n-2)A\\_{j-1}+(n-r+1)A\\_{j-2}. \\\\end{aligned} $$$\n\nSubstituting into $$$L\\_k$$$ and shifting the latter two indices gives\n\n$$$ 0=L\\_k=\\\\sum\\_{r=k}^m\\\\binom{n-2k-2}{r-k} \\\\left((r+1)h\\_r-(n-2)h\\_{r+1}+(n-r-1)h\\_{r+2}\\\\right), $$$\n\nwhere out-of-range $$$h$$$ values are zero. Define\n\n$$$ E\\_r=(r+1)h\\_r-(n-2)h\\_{r+1}+(n-r-1)h\\_{r+2}. $$$\n\nThen\n\n$$$ \\\\sum\\_{r=k}^m\\\\binom{n-2k-2}{r-k}E\\_r=0\\. $$$\n\nEquation $$$k$$$ contains only $$$E\\_k,E\\_{k+1},\\\\ldots,E\\_m$$$, with coefficient $$$1$$$ on $$$E\\_k$$$. Apart from the top boundary, backward elimination gives $$$E\\_r=0$$$, or\n\n$$$ (r+1)h\\_r=(n-2)h\\_{r+1}-(n-r-1)h\\_{r+2}. $$$\n\nThe highest two terms come directly from the original system:\n\n$$$ h\\_m=c\\_m,\\\\qquad h\\_{m-1}=c\\_{m-1}-(n-2m+2)h\\_m. $$$\n\nIf $$$n=2m+1$$$ is odd, these values satisfy $$$E\\_{m-1}=-E\\_m$$$. Their binomial coefficients in every sum are equal, so they cancel. The remaining equations give $$$E\\_{m-2},E\\_{m-3},\\\\ldots,E\\_0=0$$$. Apply the second-order recurrence starting from $$$m-2$$$.\n\nIf $$$n=2m$$$ is even, the first two values give $$$E\\_{m-1}=0$$$, but $$$E\\_m$$$ has no matching boundary term. Compute one more value from the original system:\n\n$$$ h\\_{m-2}=c\\_{m-2}-4h\\_{m-1}-6h\\_m. $$$\n\nIt satisfies $$$E\\_{m-2}=-E\\_m$$$. Again their coefficients are equal and they cancel, giving $$$E\\_{m-3},E\\_{m-4},\\\\ldots,E\\_0=0$$$. Apply the recurrence starting from $$$m-3$$$.\n\nFinally compute all $$$e\\_r$$$ by multiplying the $$$n$$$ linear polynomials $$$1+a\\_ix$$$ with divide and conquer, using NTT for every convolution. Substitute the resulting coefficients into $$$\\\\sum\\_{r=0}^m h\\_re\\_re\\_{n-r}$$$.\n\nThe time complexity is $$$O(n\\\\log^2 n)$$$ and the space complexity is $$$O(n\\\\log n)$$$.\n\n![](//codeforces.com/codeforces.org/s/70137/images/icons/paperclip-16x16.png) Tutorial of [Codeforces Round 1116 (Div. 1)](https://codeforces.com/contest/2255) \n\n![](//codeforces.com/codeforces.org/s/70137/images/icons/paperclip-16x16.png) Tutorial of [Codeforces Round 1116 (Div. 2)](https://codeforces.com/contest/2256) \n\n* [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/voteup-gray.png \"Vote: I like it\")](#)\n* +166\n* [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/votedown-gray.png \"Vote: I do not like it\")](#)\n\n* [![Author](//codeforces.com/codeforces.org/s/70137/images/blog/user_16x16.png \"Author\")](https://codeforces.com/profile/paulzrm) [paulzrm ](https://codeforces.com/profile/paulzrm)\n* ![Publication date](//codeforces.com/codeforces.org/s/70137/images/blog/date_16x16.png \"Publication date\") 6 weeks ago\n* [![Comments](//codeforces.com/codeforces.org/s/70137/images/blog/comments_16x16.png \"Comments\")](https://codeforces.com/blog/entry/155840#comments) [51 ](https://codeforces.com/blog/entry/155840#comments)\n\n  \n![Comments](//codeforces.com/codeforces.org/s/70137/images/icons/comments-48x48.png \"Comments\") Comments (51) \n\n[Write comment?](#) \n\n| » [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\") | 6 weeks ago, [show](#) # |\n| ------------------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/459910/avatar/1fe3f753b1f42bfe.jpg) ](https://codeforces.com/profile/paulzrm) [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155840?#comment-1384425 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +3 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) _Auto comment: topic has been updated by [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\") ([previous revision](https://codeforces.com/topic/156483/en1), [new revision](https://codeforces.com/topic/156483/en2), [compare](https://codeforces.com/topic/156483/diff/en1/en2))._ [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [esc\\_qwq](https://codeforces.com/profile/esc%5Fqwq \"Candidate Master esc_qwq\") | 6 weeks ago, [show](#) # |\n| --------------------------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/4963270/avatar/be164ba8d5d4a72e.jpg) ](https://codeforces.com/profile/esc%5Fqwq) [esc\\_qwq](https://codeforces.com/profile/esc%5Fqwq \"Candidate Master esc_qwq\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155840?#comment-1384426 \"Link to comment\") \\| [←](#)Rev. 3 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +12 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Lightning Fast Editorial, Thank you!!Great contest imo, at least the problems themselves :) [→](#) [Reply](#) |\n| ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [AksLolCoding](https://codeforces.com/profile/AksLolCoding \"Grandmaster AksLolCoding\") | 6 weeks ago, [show](#) # |\n| ---------------------------------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/2948677/avatar/f15474cd5b82d030.jpg) ](https://codeforces.com/profile/AksLolCoding) [AksLolCoding](https://codeforces.com/profile/AksLolCoding \"Grandmaster AksLolCoding\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155840?#comment-1384454 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +80 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) D1C was amazing [→](#) [Reply](#) |\n| -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [megahertz13](https://codeforces.com/profile/megahertz13 \"Pupil megahertz13\") | 6 weeks ago, [show](#) # |\n| ------------------------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/2827759/avatar/2d153e90dd0538bd.jpg) ](https://codeforces.com/profile/megahertz13) [megahertz13](https://codeforces.com/profile/megahertz13 \"Pupil megahertz13\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155840?#comment-1384463 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Great contest, I thought i did something wrong when i got that the answer for 2256B was <=4, lol [→](#) [Reply](#) |\n| ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [enaschair](https://codeforces.com/profile/enaschair \"Candidate Master enaschair\") | 6 weeks ago, [show](#) # |\n| ------------------------------------------------------------------------------------ | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/4788176/avatar/c6c7c5b744a37339.jpg) ](https://codeforces.com/profile/enaschair) [enaschair](https://codeforces.com/profile/enaschair \"Candidate Master enaschair\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155840?#comment-1384464 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) really fun contest to do! going to watch sukasuka because of this! [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [KiKoS](https://codeforces.com/profile/KiKoS \"Master KiKoS\") | 6 weeks ago, [show (+1)](#) # |\n| -------------------------------------------------------------- | ----------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/473219/avatar/a382879284586aa0.jpg) ](https://codeforces.com/profile/KiKoS) [KiKoS](https://codeforces.com/profile/KiKoS \"Master KiKoS\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155840?#comment-1384470 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I personally switched from C to D when I observed two things that i found hard to combine: 1\\. there is a case when we have one black tile — i assume, it should be moved to the secret location 2\\. there is a case where all tiles but one are black — i assumed we should swap empty one with the target. I think editorial accomplishes 1\\. but for 2\\. it does something else, right? Does it cut the \"opposite\" tile, whatever this means? [→](#) [Reply](#) |\n| ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\") | 6 weeks ago, [show (+1)](#) # |\n| --------------------------------------------------------------------------- | ----------------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/459910/avatar/1fe3f753b1f42bfe.jpg) ](https://codeforces.com/profile/paulzrm) [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155840?#comment-1384491 \"Link to comment\") [^](#comment-1384470 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +19 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I gave 1C and 1D the same score so that everyone could freely choose which problem to solve. I wonder how many contestants noticed this :-) [→](#) [Reply](#) |\n\n...89723 bytes truncated...\n\nThe tool call succeeded but the output was truncated. Full output saved to: /Users/zayd/.local/share/opencode/tool-output/tool_0b23a202d001O3cD0uaKiuyL2S\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_0b23a202d001O3cD0uaKiuyL2S"
            },
            "title": "https://codeforces.com/blog/entry/155840 (text/markdown; charset=utf-8)",
            "time": {
              "start": 1789696546303,
              "end": 1789696548911
            }
          },
          "metadata": {
            "openai": {
              "itemId": "fc_0aec25d7cde6a1c1016aac9a217a8487d0b2acb4d108deb8b7"
            }
          },
          "id": "prt_0b23a13520018GxOwhM9eeHmpb",
          "sessionID": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX",
          "messageID": "msg_0b23a0509001GLLmzALjR21VnY"
        },
        {
          "reason": "tool-calls",
          "type": "step-finish",
          "tokens": {
            "total": 10008,
            "input": 5368,
            "output": 32,
            "reasoning": 0,
            "cache": {
              "write": 0,
              "read": 4608
            }
          },
          "cost": 0,
          "id": "prt_0b23a2032001BiXq4LNbSqQWOY",
          "sessionID": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX",
          "messageID": "msg_0b23a0509001GLLmzALjR21VnY"
        }
      ]
    },
    {
      "info": {
        "parentID": "msg_0b23a048e001tw3mYpro3ozh19",
        "role": "assistant",
        "mode": "nudge-generation",
        "agent": "nudge-generation",
        "variant": "low",
        "path": {
          "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-rwrVSG",
          "root": "/"
        },
        "cost": 0,
        "tokens": {
          "total": 24098,
          "input": 13793,
          "output": 547,
          "reasoning": 30,
          "cache": {
            "write": 0,
            "read": 9728
          }
        },
        "modelID": "gpt-6-astra",
        "providerID": "openai",
        "time": {
          "created": 1789696548921,
          "completed": 1789696568612
        },
        "finish": "tool-calls",
        "id": "msg_0b23a2039001k4HoDprh0PBf2l",
        "sessionID": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX"
      },
      "parts": [
        {
          "type": "step-start",
          "id": "prt_0b23a2670001GHdUEJQj1Jx9ob",
          "sessionID": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX",
          "messageID": "msg_0b23a2039001k4HoDprh0PBf2l"
        },
        {
          "type": "reasoning",
          "text": "**Preparing exhaustive checks**",
          "time": {
            "start": 1789696550513,
            "end": 1789696554103
          },
          "metadata": {
            "openai": {
              "itemId": "rs_06ffecb01f137cd7016aac9a26740887d09e1811a241da7092",
              "reasoningEncryptedContent": "gAAAAABqrJopEfzZgaCF3eDhXaR4e19sDqHSAixO83W2dV6pw2SpDBwaFugLjjol2BCnRDxoj-1MIl8CNG_yyIwq1bZpp3xNkZY3TeEHP0pTPFdu0t2GCZa0fjeBA2jYIf73VPqnUzbRLuGvv36q0Giex9RRFvDgQdrglZ19eG3ffcG0Li0U3xXDW6l8eEQOxNrCyzW82M0xVXvJkJqdoMaFPAfUVarlwhAvyGPIhQTaAJfrBiSJ8QDa58EtZ8TDPybB3YczWuEjaljvtVEN3HMPSEmtQvHuzAfbpHsOvNT7Bk5NjwsEfdcaP2IcP2ZZA9Ay2XKy4yFdmaDF31qd3YsVYRC8d2E2CKu8T_i6s6B3j9qRyfHVnLhzTk6sx4B4u4kOCjvmfSyJNyuHx0bS0aNch5g26OF9S-74F6zBgsSI7ewb_hFJZBB1B77ezRZ69bNqx4DTh00hVaRFmqdPaLaHQS5lhL6R0FItw2jTb2Y71RV68JJ1DUzzrIwd6O10Z0fiGASuqrtvBya76rkNi2oqVaIBVsmdRCG-_V_NHWw_uBoKJGeWjqPLAC2QcHujMtZoRaxDTOhYTtl1M-4brS6c58eAFg3dp3acef1ZL6UjQgoS67J_Zo4AobYMg_xPJZvkm_sWRzR-6rH8PMiQ8rCxUxOLsKjjvBXrHM0v6AQ_Rm7coTruh2VgC0aRL-eEjI6KqIb3C7L4yNv_sujvjo_3AkyvMkJFfEbWNl9hUzaW7F-e6TGcZ87ZIpuOYH4RCNBWkbUjS8x_-8wCp2LfXKGPL_MGhe_008LDD3P_QUX0GF-GpTfQbsErkJof1kR9UwmNO29EHygPtO1b7nBv2MH3L4an00jvfj12-j5WKNmqBoFQLETeodYHFg1Ww06xSsGJViqy9DgovIvw3cpg-0ReUOHZAM0KcNW08p-8I4ZKM0EG93X190nwoPuuJ2PnJl9EEtdlBP-mZOiI_NNf7QkZeQOPW5QsLWTH3qJAwtvqpkfzRRewSX5MLTXeOWq-krU3wYbp6_xqvT6RNzDVXocp1II3MVl1ei1u1jvFEbuVuVpth9McOCgu7EWej58tKNIUgyQHLvXaaxG_DvI3HonmEFyjDP4GvIuK0VD8Krtlu4aBJ7pGA81Vjj6HgRpIs8AHtpm4ZxJUjsaYdD3Bpis77OJZLIWnO-wysd3YmDY5pO42kDd7u7i5d9CPef12HKSn7yFuIC23s1FY9n8BlgTuW7qYgq48c6NDii7H9OiZm3WQLn8EmBMxQvEzfFiSiBy491szReAffVMGi-t94-VIPGFhnSdbuOLMcrEQHL7wxIpfVbkjnXWM--SAnbo24oBvfQA3wH62UpUDk6R2dpsMO31KptbQJ4aSWM6kpzckaTwpdMq3idJSWcm6DSyf0OpifzUQTa0WDxZ8EtdcjN4IAtQUqRkPnLAjUMtD7hYnF3FoybryVSGSIitHA2jyius7CMcVbHkNDAJOWxkXVxw8g1ZX2jUDDA=="
            }
          },
          "id": "prt_0b23a2671001Aovvi4woKLqCzp",
          "sessionID": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX",
          "messageID": "msg_0b23a2039001k4HoDprh0PBf2l"
        },
        {
          "type": "tool",
          "tool": "bash",
          "callID": "call_B5KXSjMoY8XZ1RqX7UirPoIn",
          "state": {
            "status": "completed",
            "input": {
              "command": "python3 -c 'import math, random\nrng=random.Random(2255)\ncases=0\nfor n in range(2,13):\n boards=([[(mask>>i)&1 for i in range(n*n)] for mask in range(1<<(n*n))] if n<=3 else [[rng.randrange(2) for _ in range(n*n)] for _ in range(300)])\n for a in boards:\n  w=sum(a)\n  if math.gcd(w,n)!=1: continue\n  inv=pow(w,-1,n)\n  sr=sum(i//n for i in range(n*n) if a[i])%n\n  sc=sum(i%n for i in range(n*n) if a[i])%n\n  targets=range(n*n) if n<=3 else [rng.randrange(n*n) for _ in range(5)]\n  for x in targets:\n   r,c=divmod(x,n); dr=(w*r-sr)%n; dc=(w*c-sc)%n\n   b=a[:]\n   if dr or dc:\n    pair=next(((i,((i//n+dr)%n)*n+(i%n+dc)%n) for i in range(n*n) if a[i] and not a[((i//n+dr)%n)*n+(i%n+dc)%n]),None)\n    assert pair is not None\n    i,j=pair; b[i],b[j]=b[j],b[i]\n   for op in range(5):\n    out=[0]*(n*n)\n    def transform(r,c):\n     if op==0: return ((r+1)%n,(c+2)%n)\n     if op==1: return (c,n-1-r)\n     if op==2: return (r,n-1-c)\n     return (r,c)\n    for i in range(n*n):\n     rr,cc=transform(*divmod(i,n)); out[rr*n+cc]=b[i]^(op==3)\n    ww=sum(out); ii=pow(ww,-1,n)\n    center=(sum(i//n for i in range(n*n) if out[i])*ii%n,sum(i%n for i in range(n*n) if out[i])*ii%n)\n    assert center==transform(r,c)\n   cases+=1\nprint(\"Verified\",cases,\"encodings and all transformation generators, including composite sizes.\")'"
            },
            "output": "Verified 11205 encodings and all transformation generators, including composite sizes.\n",
            "metadata": {
              "output": "Verified 11205 encodings and all transformation generators, including composite sizes.\n",
              "exit": 0,
              "truncated": false
            },
            "title": "python3 -c 'import math, random\nrng=random.Random(2255)\ncases=0\nfor n in range(2,13):\n boards=([[(mask>>i)&1 for i in range(n*n)] for mask in range(1<<(n*n))] if n<=3 else [[rng.randrange(2) for _ in range(n*n)] for _ in range(300)])\n for a in boards:\n  w=sum(a)\n  if math.gcd(w,n)!=1: continue\n  inv=pow(w,-1,n)\n  sr=sum(i//n for i in range(n*n) if a[i])%n\n  sc=sum(i%n for i in range(n*n) if a[i])%n\n  targets=range(n*n) if n<=3 else [rng.randrange(n*n) for _ in range(5)]\n  for x in targets:\n   r,c=divmod(x,n); dr=(w*r-sr)%n; dc=(w*c-sc)%n\n   b=a[:]\n   if dr or dc:\n    pair=next(((i,((i//n+dr)%n)*n+(i%n+dc)%n) for i in range(n*n) if a[i] and not a[((i//n+dr)%n)*n+(i%n+dc)%n]),None)\n    assert pair is not None\n    i,j=pair; b[i],b[j]=b[j],b[i]\n   for op in range(5):\n    out=[0]*(n*n)\n    def transform(r,c):\n     if op==0: return ((r+1)%n,(c+2)%n)\n     if op==1: return (c,n-1-r)\n     if op==2: return (r,n-1-c)\n     return (r,c)\n    for i in range(n*n):\n     rr,cc=transform(*divmod(i,n)); out[rr*n+cc]=b[i]^(op==3)\n    ww=sum(out); ii=pow(ww,-1,n)\n    center=(sum(i//n for i in range(n*n) if out[i])*ii%n,sum(i%n for i in range(n*n) if out[i])*ii%n)\n    assert center==transform(r,c)\n   cases+=1\nprint(\"Verified\",cases,\"encodings and all transformation generators, including composite sizes.\")'",
            "time": {
              "start": 1789696567851,
              "end": 1789696568609
            }
          },
          "metadata": {
            "openai": {
              "itemId": "fc_06ffecb01f137cd7016aac9a29fbe487d095829ccdae754d27"
            }
          },
          "id": "prt_0b23a347d001MdV3h97cvL6xo3",
          "sessionID": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX",
          "messageID": "msg_0b23a2039001k4HoDprh0PBf2l"
        },
        {
          "reason": "tool-calls",
          "type": "step-finish",
          "tokens": {
            "total": 24098,
            "input": 13793,
            "output": 547,
            "reasoning": 30,
            "cache": {
              "write": 0,
              "read": 9728
            }
          },
          "cost": 0,
          "id": "prt_0b23a6d220015RX6qL3qOdeFBV",
          "sessionID": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX",
          "messageID": "msg_0b23a2039001k4HoDprh0PBf2l"
        }
      ]
    },
    {
      "info": {
        "parentID": "msg_0b23a048e001tw3mYpro3ozh19",
        "role": "assistant",
        "mode": "nudge-generation",
        "agent": "nudge-generation",
        "variant": "low",
        "path": {
          "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-rwrVSG",
          "root": "/"
        },
        "cost": 0,
        "tokens": {
          "total": 26782,
          "input": 828,
          "output": 2658,
          "reasoning": 0,
          "cache": {
            "write": 0,
            "read": 23296
          }
        },
        "modelID": "gpt-6-astra",
        "providerID": "openai",
        "time": {
          "created": 1789696568614,
          "completed": 1789696649953
        },
        "finish": "tool-calls",
        "structured": {
          "status": "success",
          "reason": null,
          "hints": [
            {
              "order": 1,
              "content": "Don't try to recover the picture's original orientation. Instead, look for a cell computed from the picture that moves exactly like an ordinary cell under every geometric transformation. Chtholly only needs to make that cell equal the target."
            },
            {
              "order": 2,
              "content": "Use zero-based coordinates modulo $n$. Let $S=(S_r,S_c)$ be the sum of all black-cell coordinates. A translation by $d$ changes $S$ to $S+wd$. Since $\\gcd(w,n)=1$, what happens if you define a modular center $C=w^{-1}S$?"
            },
            {
              "order": 3,
              "content": "Rotations and reflections have the form $p\\mapsto Mp+t$, so the modular center transforms to $MC+t$. Also, the coordinate sum of the entire board is $(0,0)$ modulo $n$. Color inversion therefore replaces $(w,S)$ by $(-w,-S)$ modulo $n$, leaving $C$ unchanged."
            },
            {
              "order": 4,
              "content": "To make the center equal the target $x$, the black-cell sum must become $wx$. Set $\\delta=wx-S$ modulo $n$. Swapping a black cell $p$ with a white cell $p+\\delta$ changes the sum by exactly $\\delta$. If $\\delta=(0,0)$, just swap a cell with itself."
            },
            {
              "order": 5,
              "content": "Why must the required black-to-white pair exist when $\\delta\\ne(0,0)$? Otherwise, translation by $\\delta$ would preserve the black set. Its cycles all have the same length $L>1$, and $L\\mid n$, so the black count would satisfy $L\\mid w$—contradicting $\\gcd(n,w)=1$. Scan all cells to find the pair; in the second run, simply output $w^{-1}S$."
            }
          ],
          "editorial": "The trick is to encode the target as a **modular center of mass** of the black cells. This center follows every shift, rotation, and reflection, and ignores color inversion. No orientation detective work required.\n\nUse zero-based coordinates throughout the math and the implementation. All coordinate arithmetic below is modulo $n$.\n\n## A center that follows the picture\n\nLet $B$ be the set of black cells, $w=|B|$, and\n\n$$\nS=\\sum_{p\\in B}p=(S_r,S_c).\n$$\n\nBecause $\\gcd(w,n)=1$, $w$ has a multiplicative inverse modulo $n$. Define\n\n$$\nC=w^{-1}S.\n$$\n\nThis is modular division, not an ordinary average. Both coordinates of $C$ are residues modulo $n$, so they identify exactly one cell.\n\nOur strategy is simple:\n\n- **Chtholly:** use the swap to make $C$ equal the target $x$.\n- **Willem:** compute $C$ from the picture he receives and report it.\n\nFirst, let's prove the jury cannot break this agreement.\n\n### Shifts, rotations, and reflections\n\nEach geometric operation can be written as\n\n$$\np\\mapsto Mp+t,\n$$\n\nwhere $M$ swaps and/or negates coordinates, and $t$ is a fixed translation vector. For example, a clockwise rotation sends $(r,c)$ to $(c,n-1-r)$.\n\nThere are still $w$ black cells after the operation, and their coordinate sum becomes\n\n$$\nS'=\\sum_{p\\in B}(Mp+t)=MS+wt.\n$$\n\nThus the new center is\n\n$$\nC'=w^{-1}S'=MC+t.\n$$\n\n**The center moves exactly like a cell.** In particular, once it equals the target, every geometric operation keeps them together.\n\n### Color inversion\n\nFor each coordinate, the sum over the entire board is\n\n$$\nn\\sum_{i=0}^{n-1}i\\equiv0\\pmod n,\n$$\n\nbecause every row index and every column index occurs $n$ times. This argument also works when $n$ is even; no division modulo $n$ is involved.\n\nAfter inversion, the new black cells are the old white cells. Therefore,\n\n$$\nS'=-S,\\qquad w'=n^2-w\\equiv-w\\pmod n.\n$$\n\nThe new count is still coprime to $n$, and\n\n$$\nC'=(-w)^{-1}(-S)=w^{-1}S=C.\n$$\n\nSo inversion changes neither the target nor the center. Since each allowed operation preserves their relationship, any sequence of operations does too.\n\n## Encoding the target with one swap\n\nChtholly wants the final black-cell sum to satisfy\n\n$$\nS_{\\text{new}}=wx.\n$$\n\nThe required change is\n\n$$\n\\delta=wx-S.\n$$\n\nIf $\\delta=(0,0)$, the target is already the center. Output the same cell twice.\n\nOtherwise, suppose we find a black cell $p$ such that\n\n$$\nq=p+\\delta\n$$\n\nis white, with both coordinates wrapped modulo $n$. Swapping these cells removes $p$ from the black set and adds $q$, so\n\n$$\nS_{\\text{new}}=S-p+q=S+\\delta=wx.\n$$\n\nThat is exactly what we need. We can find such a pair by scanning all black cells and checking the corresponding $q$.\n\n### Why does this pair always exist?\n\nThis is the key existence argument. A scan is only an algorithm if there is something to find.\n\nAssume $\\delta\\ne(0,0)$ but no valid pair exists. Then whenever $p$ is black, $p+\\delta$ is also black.\n\nConsider repeatedly applying the translation\n\n$$\nT(p)=p+\\delta.\n$$\n\nIt partitions the board into cycles. Every cycle has the same length $L$: the smallest positive integer such that\n\n$$\nL\\delta=(0,0)\\pmod n.\n$$\n\nWe have:\n\n- $L>1$, since $\\delta$ is nonzero;\n- $L\\mid n$, since $n\\delta=(0,0)$ and the smallest return time divides every return time.\n\nIf a cycle contains one black cell, our assumption forces every cell on that cycle to be black. Thus the black cells form a union of whole cycles, giving $L\\mid w$.\n\nNow $L>1$ divides both $n$ and $w$, contradicting $\\gcd(n,w)=1$.\n\nTherefore, **a suitable black-to-white pair always exists**.\n\n## Algorithm\n\nRead the run identifier, then process each test case independently.\n\n### First run\n\n1. Read the board and calculate $w,S_r,S_c$.\n2. Read the target and convert it to zero-based coordinates.\n3. Calculate\n   $$\n   \\delta_r=(wr_x-S_r)\\bmod n,\\qquad\n   \\delta_c=(wc_x-S_c)\\bmod n.\n   $$\n4. If both are zero, output `1 1 1 1`.\n5. Otherwise, scan black cells $(r,c)$. For each one, check whether\n   $$\n   ((r+\\delta_r)\\bmod n,\\ (c+\\delta_c)\\bmod n)\n   $$\n   is white. Output the first such pair, converting back to one-based coordinates.\n\n### Second run\n\n1. Read the resulting board and calculate its current $w,S_r,S_c$.\n2. Find $u$ such that $wu\\equiv1\\pmod n$.\n3. Output\n   $$\n   (uS_r\\bmod n)+1,\\qquad (uS_c\\bmod n)+1.\n   $$\n\nSince $n\\le800$, simply trying $u=1,2,\\ldots,n-1$ is fast enough. The modulus is not necessarily prime, so do not use Fermat's little theorem to compute the inverse.\n\n## Correctness\n\nFor every first-run test case, either the center already equals the target, or the existence argument guarantees a pair whose swap makes $S=wx$. Thus Chtholly always leaves a picture whose modular center is the target.\n\nEvery allowed geometric transformation moves the center exactly as it moves the target. Color inversion leaves the center and target unchanged. Consequently, the center of Willem's picture is the final target position.\n\nWillem computes and outputs that center, so the protocol is correct. Every test case is self-contained, so reordering cases between runs has no effect.\n\n## Complexity and implementation details\n\nEach test case takes $O(n^2)$ time in either run. The inverse search costs only $O(n)$, and the pair search checks at most $n^2$ cells. Storing the board uses $O(n^2)$ space.\n\nUse `long long` for coordinate sums and products, and normalize negative remainders when calculating $\\delta$. Only the first run reads a target. This is a run-twice protocol: print the required answer for the current run, with no queries or saved state between executions.",
          "solution": "#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n\n    string mode;\n    cin >> mode;\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        vector<string> a(n);\n        ll w = 0, sr = 0, sc = 0;\n        for (int r = 0; r < n; ++r) {\n            cin >> a[r];\n            for (int c = 0; c < n; ++c) {\n                if (a[r][c] == '#') {\n                    ++w;\n                    sr += r;\n                    sc += c;\n                }\n            }\n        }\n        sr %= n;\n        sc %= n;\n\n        if (mode == \"first\") {\n            int rx, cx;\n            cin >> rx >> cx;\n            --rx;\n            --cx;\n            int dr = ((w * rx - sr) % n + n) % n;\n            int dc = ((w * cx - sc) % n + n) % n;\n\n            if (dr == 0 && dc == 0) {\n                cout << \"1 1 1 1\\n\";\n                continue;\n            }\n\n            bool found = false;\n            for (int r = 0; r < n && !found; ++r) {\n                for (int c = 0; c < n; ++c) {\n                    int rr = (r + dr) % n;\n                    int cc = (c + dc) % n;\n                    if (a[r][c] == '#' && a[rr][cc] == '.') {\n                        cout << r + 1 << ' ' << c + 1 << ' '\n                             << rr + 1 << ' ' << cc + 1 << '\\n';\n                        found = true;\n                        break;\n                    }\n                }\n            }\n        } else {\n            int inv = 1;\n            while (w * inv % n != 1) ++inv;\n            cout << sr * inv % n + 1 << ' '\n                 << sc * inv % n + 1 << '\\n';\n        }\n    }\n}\n"
        },
        "id": "msg_0b23a6d2600143yDDfzRuQZJKs",
        "sessionID": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX"
      },
      "parts": [
        {
          "type": "step-start",
          "id": "prt_0b23a741f001o69A4Kh2TMfCtG",
          "sessionID": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX",
          "messageID": "msg_0b23a6d2600143yDDfzRuQZJKs"
        },
        {
          "type": "tool",
          "tool": "StructuredOutput",
          "callID": "call_Y1A5qqjmKGv6boothcbumt9i",
          "state": {
            "status": "completed",
            "input": {
              "status": "success",
              "reason": null,
              "hints": [
                {
                  "order": 1,
                  "content": "Don't try to recover the picture's original orientation. Instead, look for a cell computed from the picture that moves exactly like an ordinary cell under every geometric transformation. Chtholly only needs to make that cell equal the target."
                },
                {
                  "order": 2,
                  "content": "Use zero-based coordinates modulo $n$. Let $S=(S_r,S_c)$ be the sum of all black-cell coordinates. A translation by $d$ changes $S$ to $S+wd$. Since $\\gcd(w,n)=1$, what happens if you define a modular center $C=w^{-1}S$?"
                },
                {
                  "order": 3,
                  "content": "Rotations and reflections have the form $p\\mapsto Mp+t$, so the modular center transforms to $MC+t$. Also, the coordinate sum of the entire board is $(0,0)$ modulo $n$. Color inversion therefore replaces $(w,S)$ by $(-w,-S)$ modulo $n$, leaving $C$ unchanged."
                },
                {
                  "order": 4,
                  "content": "To make the center equal the target $x$, the black-cell sum must become $wx$. Set $\\delta=wx-S$ modulo $n$. Swapping a black cell $p$ with a white cell $p+\\delta$ changes the sum by exactly $\\delta$. If $\\delta=(0,0)$, just swap a cell with itself."
                },
                {
                  "order": 5,
                  "content": "Why must the required black-to-white pair exist when $\\delta\\ne(0,0)$? Otherwise, translation by $\\delta$ would preserve the black set. Its cycles all have the same length $L>1$, and $L\\mid n$, so the black count would satisfy $L\\mid w$—contradicting $\\gcd(n,w)=1$. Scan all cells to find the pair; in the second run, simply output $w^{-1}S$."
                }
              ],
              "editorial": "The trick is to encode the target as a **modular center of mass** of the black cells. This center follows every shift, rotation, and reflection, and ignores color inversion. No orientation detective work required.\n\nUse zero-based coordinates throughout the math and the implementation. All coordinate arithmetic below is modulo $n$.\n\n## A center that follows the picture\n\nLet $B$ be the set of black cells, $w=|B|$, and\n\n$$\nS=\\sum_{p\\in B}p=(S_r,S_c).\n$$\n\nBecause $\\gcd(w,n)=1$, $w$ has a multiplicative inverse modulo $n$. Define\n\n$$\nC=w^{-1}S.\n$$\n\nThis is modular division, not an ordinary average. Both coordinates of $C$ are residues modulo $n$, so they identify exactly one cell.\n\nOur strategy is simple:\n\n- **Chtholly:** use the swap to make $C$ equal the target $x$.\n- **Willem:** compute $C$ from the picture he receives and report it.\n\nFirst, let's prove the jury cannot break this agreement.\n\n### Shifts, rotations, and reflections\n\nEach geometric operation can be written as\n\n$$\np\\mapsto Mp+t,\n$$\n\nwhere $M$ swaps and/or negates coordinates, and $t$ is a fixed translation vector. For example, a clockwise rotation sends $(r,c)$ to $(c,n-1-r)$.\n\nThere are still $w$ black cells after the operation, and their coordinate sum becomes\n\n$$\nS'=\\sum_{p\\in B}(Mp+t)=MS+wt.\n$$\n\nThus the new center is\n\n$$\nC'=w^{-1}S'=MC+t.\n$$\n\n**The center moves exactly like a cell.** In particular, once it equals the target, every geometric operation keeps them together.\n\n### Color inversion\n\nFor each coordinate, the sum over the entire board is\n\n$$\nn\\sum_{i=0}^{n-1}i\\equiv0\\pmod n,\n$$\n\nbecause every row index and every column index occurs $n$ times. This argument also works when $n$ is even; no division modulo $n$ is involved.\n\nAfter inversion, the new black cells are the old white cells. Therefore,\n\n$$\nS'=-S,\\qquad w'=n^2-w\\equiv-w\\pmod n.\n$$\n\nThe new count is still coprime to $n$, and\n\n$$\nC'=(-w)^{-1}(-S)=w^{-1}S=C.\n$$\n\nSo inversion changes neither the target nor the center. Since each allowed operation preserves their relationship, any sequence of operations does too.\n\n## Encoding the target with one swap\n\nChtholly wants the final black-cell sum to satisfy\n\n$$\nS_{\\text{new}}=wx.\n$$\n\nThe required change is\n\n$$\n\\delta=wx-S.\n$$\n\nIf $\\delta=(0,0)$, the target is already the center. Output the same cell twice.\n\nOtherwise, suppose we find a black cell $p$ such that\n\n$$\nq=p+\\delta\n$$\n\nis white, with both coordinates wrapped modulo $n$. Swapping these cells removes $p$ from the black set and adds $q$, so\n\n$$\nS_{\\text{new}}=S-p+q=S+\\delta=wx.\n$$\n\nThat is exactly what we need. We can find such a pair by scanning all black cells and checking the corresponding $q$.\n\n### Why does this pair always exist?\n\nThis is the key existence argument. A scan is only an algorithm if there is something to find.\n\nAssume $\\delta\\ne(0,0)$ but no valid pair exists. Then whenever $p$ is black, $p+\\delta$ is also black.\n\nConsider repeatedly applying the translation\n\n$$\nT(p)=p+\\delta.\n$$\n\nIt partitions the board into cycles. Every cycle has the same length $L$: the smallest positive integer such that\n\n$$\nL\\delta=(0,0)\\pmod n.\n$$\n\nWe have:\n\n- $L>1$, since $\\delta$ is nonzero;\n- $L\\mid n$, since $n\\delta=(0,0)$ and the smallest return time divides every return time.\n\nIf a cycle contains one black cell, our assumption forces every cell on that cycle to be black. Thus the black cells form a union of whole cycles, giving $L\\mid w$.\n\nNow $L>1$ divides both $n$ and $w$, contradicting $\\gcd(n,w)=1$.\n\nTherefore, **a suitable black-to-white pair always exists**.\n\n## Algorithm\n\nRead the run identifier, then process each test case independently.\n\n### First run\n\n1. Read the board and calculate $w,S_r,S_c$.\n2. Read the target and convert it to zero-based coordinates.\n3. Calculate\n   $$\n   \\delta_r=(wr_x-S_r)\\bmod n,\\qquad\n   \\delta_c=(wc_x-S_c)\\bmod n.\n   $$\n4. If both are zero, output `1 1 1 1`.\n5. Otherwise, scan black cells $(r,c)$. For each one, check whether\n   $$\n   ((r+\\delta_r)\\bmod n,\\ (c+\\delta_c)\\bmod n)\n   $$\n   is white. Output the first such pair, converting back to one-based coordinates.\n\n### Second run\n\n1. Read the resulting board and calculate its current $w,S_r,S_c$.\n2. Find $u$ such that $wu\\equiv1\\pmod n$.\n3. Output\n   $$\n   (uS_r\\bmod n)+1,\\qquad (uS_c\\bmod n)+1.\n   $$\n\nSince $n\\le800$, simply trying $u=1,2,\\ldots,n-1$ is fast enough. The modulus is not necessarily prime, so do not use Fermat's little theorem to compute the inverse.\n\n## Correctness\n\nFor every first-run test case, either the center already equals the target, or the existence argument guarantees a pair whose swap makes $S=wx$. Thus Chtholly always leaves a picture whose modular center is the target.\n\nEvery allowed geometric transformation moves the center exactly as it moves the target. Color inversion leaves the center and target unchanged. Consequently, the center of Willem's picture is the final target position.\n\nWillem computes and outputs that center, so the protocol is correct. Every test case is self-contained, so reordering cases between runs has no effect.\n\n## Complexity and implementation details\n\nEach test case takes $O(n^2)$ time in either run. The inverse search costs only $O(n)$, and the pair search checks at most $n^2$ cells. Storing the board uses $O(n^2)$ space.\n\nUse `long long` for coordinate sums and products, and normalize negative remainders when calculating $\\delta$. Only the first run reads a target. This is a run-twice protocol: print the required answer for the current run, with no queries or saved state between executions.",
              "solution": "#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n\n    string mode;\n    cin >> mode;\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        vector<string> a(n);\n        ll w = 0, sr = 0, sc = 0;\n        for (int r = 0; r < n; ++r) {\n            cin >> a[r];\n            for (int c = 0; c < n; ++c) {\n                if (a[r][c] == '#') {\n                    ++w;\n                    sr += r;\n                    sc += c;\n                }\n            }\n        }\n        sr %= n;\n        sc %= n;\n\n        if (mode == \"first\") {\n            int rx, cx;\n            cin >> rx >> cx;\n            --rx;\n            --cx;\n            int dr = ((w * rx - sr) % n + n) % n;\n            int dc = ((w * cx - sc) % n + n) % n;\n\n            if (dr == 0 && dc == 0) {\n                cout << \"1 1 1 1\\n\";\n                continue;\n            }\n\n            bool found = false;\n            for (int r = 0; r < n && !found; ++r) {\n                for (int c = 0; c < n; ++c) {\n                    int rr = (r + dr) % n;\n                    int cc = (c + dc) % n;\n                    if (a[r][c] == '#' && a[rr][cc] == '.') {\n                        cout << r + 1 << ' ' << c + 1 << ' '\n                             << rr + 1 << ' ' << cc + 1 << '\\n';\n                        found = true;\n                        break;\n                    }\n                }\n            }\n        } else {\n            int inv = 1;\n            while (w * inv % n != 1) ++inv;\n            cout << sr * inv % n + 1 << ' '\n                 << sc * inv % n + 1 << '\\n';\n        }\n    }\n}\n"
            },
            "output": "Structured output captured successfully.",
            "metadata": {
              "valid": true
            },
            "title": "Structured Output",
            "time": {
              "start": 1789696649937,
              "end": 1789696649939
            }
          },
          "metadata": {
            "openai": {
              "itemId": "fc_0572c5f4b87c993c016aac9a3a446c87d0b5014fc3ad6dd39e"
            }
          },
          "id": "prt_0b23a74210012rjZzCB7hs3IlF",
          "sessionID": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX",
          "messageID": "msg_0b23a6d2600143yDDfzRuQZJKs"
        },
        {
          "reason": "tool-calls",
          "type": "step-finish",
          "tokens": {
            "total": 26782,
            "input": 828,
            "output": 2658,
            "reasoning": 0,
            "cache": {
              "write": 0,
              "read": 23296
            }
          },
          "cost": 0,
          "id": "prt_0b23baadf001Mi3pI8u3xPWMx7",
          "sessionID": "ses_f4dc5fb84ffemMM0l0Vx6wFhGX",
          "messageID": "msg_0b23a6d2600143yDDfzRuQZJKs"
        }
      ]
    }
  ]
}
