{
  "info": {
    "id": "ses_f4dad6b4effeSvHMDGQZu8C5ho",
    "slug": "quiet-nebula",
    "projectID": "global",
    "directory": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-0Ax43G",
    "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-0Ax43G",
    "title": "Nudge content generation",
    "agent": "nudge-generation",
    "model": {
      "id": "gpt-6-astra",
      "providerID": "openai",
      "variant": "low"
    },
    "version": "1.18.30",
    "summary": {
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    "tokens": {
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      "reasoning": 271,
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  "messages": [
    {
      "info": {
        "role": "user",
        "time": {
          "created": 1789698151619
        },
        "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": {
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          "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_0b25294c30019aFC9kQJEj0CGq",
        "sessionID": "ses_f4dad6b4effeSvHMDGQZu8C5ho"
      },
      "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 2249B: \"Permutation Cuts\" (rated 1700).\n\nKnown problem metadata:\n- Contest ID: 2249\n- Index: B\n- Name: Permutation Cuts\n- Rating: 1700 (Codeforces difficulty rating)\n- Tags: combinatorics, implementation, math\n- Problem URL: https://codeforces.com/contest/2249/problem/B\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\">B. Permutation Cuts</div><div class=\"time-limit\"><div class=\"property-title\">time limit per test</div>4 seconds</div><div class=\"memory-limit\"><div class=\"property-title\">memory limit per test</div>256 megabytes</div><div class=\"input-file input-standard\"><div class=\"property-title\">input</div>standard input</div><div class=\"output-file output-standard\"><div class=\"property-title\">output</div>standard output</div></div><div><p>  </p><p>You are given an integer $$$n$$$ and an array $$$a$$$ of length $$$n-1$$$.</p><p>For a permutation$$$^{\\text{∗}}$$$ $$$p$$$ of length $$$n$$$, define</p><p>$$$$$$ v_i=\\min\\left(\\max_{1\\le j\\le i}p_j, \\max_{i+1\\le j\\le n}p_j\\right), $$$$$$</p><p>where $$$1\\le i\\le n-1$$$. </p><p>In other words, cut the permutation between positions $$$i$$$ and $$$i+1$$$, take the maximum element on each side, and let $$$v_i$$$ be the smaller of these two values.</p><p>Your task is to count the number of permutations $$$p$$$ of length $$$n$$$ such that $$$v_i=a_i$$$ for each $$$1\\le i\\le n-1$$$.</p><p>Output the answer modulo $$$998\\,244\\,353$$$.</p><div class=\"statement-footnote\"><p>$$$^{\\text{∗}}$$$A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).</p></div></div><div class=\"input-specification\"><div class=\"section-title\">Input</div><p>Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 10^4$$$). The description of the test cases follows.</p><p>The first line of each test case contains one integer $$$n$$$ ($$$2\\le n\\le 10^6$$$) — the length of $$$p$$$.</p><p>The second line contains $$$n-1$$$ integers $$$a_1,a_2,\\ldots,a_{n-1}$$$ ($$$1\\le a_i\\le n$$$) — the elements of $$$a$$$.</p><p>It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.</p></div><div class=\"output-specification\"><div class=\"section-title\">Output</div><p>For each test case, output one integer — the number of suitable permutations, modulo $$$998\\,244\\,353$$$.</p></div><div class=\"sample-tests\"><div class=\"section-title\">Example</div><div class=\"sample-test\"><div class=\"input\"><div class=\"title\">Input</div><pre><div class=\"test-example-line test-example-line-even test-example-line-0\">11</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">2</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">1</div><div class=\"test-example-line test-example-line-even test-example-line-2\">3</div><div class=\"test-example-line test-example-line-even test-example-line-2\">2 2</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">3</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">1 1</div><div class=\"test-example-line test-example-line-even test-example-line-4\">4</div><div class=\"test-example-line test-example-line-even test-example-line-4\">2 3 2</div><div class=\"test-example-line test-example-line-odd test-example-line-5\">5</div><div class=\"test-example-line test-example-line-odd test-example-line-5\">3 3 4 2</div><div class=\"test-example-line test-example-line-even test-example-line-6\">2</div><div class=\"test-example-line test-example-line-even test-example-line-6\">2</div><div class=\"test-example-line test-example-line-odd test-example-line-7\">3</div><div class=\"test-example-line test-example-line-odd test-example-line-7\">1 2</div><div class=\"test-example-line test-example-line-even test-example-line-8\">4</div><div class=\"test-example-line test-example-line-even test-example-line-8\">3 3 3</div><div class=\"test-example-line test-example-line-odd test-example-line-9\">5</div><div class=\"test-example-line test-example-line-odd test-example-line-9\">4 4 4 4</div><div class=\"test-example-line test-example-line-even test-example-line-10\">4</div><div class=\"test-example-line test-example-line-even test-example-line-10\">2 1 2</div><div class=\"test-example-line test-example-line-odd test-example-line-11\">6</div><div class=\"test-example-line test-example-line-odd test-example-line-11\">3 3 5 5 5</div></pre></div><div class=\"output\"><div class=\"title\">Output</div><pre><div class=\"test-example-line test-example-line-odd test-example-line-1\">2</div><div class=\"test-example-line test-example-line-even test-example-line-2\">2</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">0</div><div class=\"test-example-line test-example-line-even test-example-line-4\">0</div><div class=\"test-example-line test-example-line-odd test-example-line-5\">2</div><div class=\"test-example-line test-example-line-even test-example-line-6\">0</div><div class=\"test-example-line test-example-line-odd test-example-line-7\">2</div><div class=\"test-example-line test-example-line-even test-example-line-8\">4</div><div class=\"test-example-line test-example-line-odd test-example-line-9\">12</div><div class=\"test-example-line test-example-line-even test-example-line-10\">0</div><div class=\"test-example-line test-example-line-odd test-example-line-11\">8</div></pre></div></div></div><div class=\"note\"><div class=\"section-title\">Note</div><p>In the first test case, both permutations $$$[1,2]$$$ and $$$[2,1]$$$ satisfy $$$v_1=1$$$.</p><p>In the second test case, the only suitable permutations are $$$[2,1,3]$$$ and $$$[3,1,2]$$$.</p><p>In the third test case, no suitable permutation exists.</p></div>\n</problem-statement>\n\n\nSource lookup status:\n<source-lookup-status>\n- Tutorial (en): https://codeforces.com/blog/entry/155516 loaded (200 OK; title: \"Codeforces Round 1112 (Div. 1, Div. 2) Editorial - Codeforces\"; mentions 2249B)\n</source-lookup-status>\n\nGenerate:\n1. Five progressive hints, from a gentle nudge to the key insight.\n2. A self-contained editorial that teaches the solution clearly, including the key insight, algorithm, correctness argument, complexity, and necessary implementation details. Explain non-obvious steps without padding or a discovery-process transcript.\n3. A complete C++26 solution that gets AC on Codeforces.\n\nFormatting & Style Rules:\n- Hints and editorial: valid Markdown with LaTeX as needed (e.g., $dp[i]$, $$\\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```",
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        "role": "assistant",
        "mode": "nudge-generation",
        "agent": "nudge-generation",
        "variant": "low",
        "path": {
          "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-0Ax43G",
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        "providerID": "openai",
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          "completed": 1789698157113
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        "finish": "tool-calls",
        "id": "msg_0b2529553001iaT26iq4sCWMgK",
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      "parts": [
        {
          "type": "step-start",
          "id": "prt_0b2529e4c001BFoGawFp1PGHmY",
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            "output": "---\nimage: https://codeforces.com/codeforces.org/s/64117/images/codeforces-telegram-square-1024x1024.png\ntitle: Codeforces Round 1112 (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/64117/images/codeforces-logo-with-telegram-555x130.png \"Codeforces\")](https://codeforces.com/) \n\n[![In English](//codeforces.com/codeforces.org/s/64117/images/flags/24/gb.png \"In English\")](https://codeforces.com/blog/entry/155516?locale=en) [![По-русски](//codeforces.com/codeforces.org/s/64117/images/flags/24/ru.png \"По-русски\")](https://codeforces.com/blog/entry/155516?locale=ru) \n\n[Enter](https://codeforces.com/enter?back=%2Fblog%2Fentry%2F155516)| [Register](https://codeforces.com/register) \n\n  \n* [Home](https://codeforces.com/)\n* [Top](https://codeforces.com/top)\n* [Catalog](https://codeforces.com/catalog)\n* [Contests](https://codeforces.com/contests)\n* [Gym](https://codeforces.com/gyms)\n* [Problemset](https://codeforces.com/problemset)\n* [Groups](https://codeforces.com/groups)\n* 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[Contests](https://codeforces.com/contests/with/%5FMCYYDS%5F)\n* [Problemsetting](https://codeforces.com/contests/writer/%5FMCYYDS%5F)\n\n### [\\_MCYYDS\\_'s blog](https://codeforces.com/blog/%5FMCYYDS%5F)\n\n[ Codeforces Round 1112 (Div. 1, Div. 2) Editorial ](https://codeforces.com/blog/entry/155516) \n\nBy [\\_MCYYDS\\_](https://codeforces.com/profile/%5FMCYYDS%5F \"Grandmaster _MCYYDS_\"), [history](https://codeforces.com/topic/156158/en3), 2 months ago, ![In English](//codeforces.com/codeforces.org/s/64117/images/flags/24/gb.png \"In English\") \n\n## [2250A - Threshold Movement](https://codeforces.com/contest/2250/problem/A \"Codeforces Round 1112 (Div. 2)\")\n\n* Idea:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n* Solution:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n\n**Solution**\n\nLet us determine what is required for every position to contain exactly one element after the simultaneous move.\n\nThe element initially at position $$$1$$$ cannot move to the left, so it must move to the right. Therefore, $$$w\\_1 \\\\gt k$$$. Once it leaves, only the element initially at position $$$2$$$ can fill position $$$1$$$, so $$$w\\_2 \\\\lt k$$$.\n\nThus, the elements at positions $$$1$$$ and $$$2$$$ must swap. Repeating the same argument for the remaining positions, the only possible movement pattern is\n\n$$$ 1\\\\leftrightarrow 2,\\\\quad 3\\\\leftrightarrow 4,\\\\quad 5\\\\leftrightarrow 6,\\\\ldots $$$\n\nConsequently, $$$n$$$ must be even, every element at an odd position must move to the right, and every element at an even position must move to the left.\n\nLet\n\n$$$ L=\\\\max\\_{i\\\\bmod 2=0}w\\_i,\\\\qquad R=\\\\min\\_{i\\\\bmod 2=1}w\\_i. $$$\n\nWe need an integer $$$k$$$ such that $$$L \\\\lt k \\\\lt R$$$. Such an integer exists if and only if $$$L+2\\\\le R$$$.\n\nThe time complexity is $$$O(n)$$$ per test case, and the extra space complexity is $$$O(1)$$$.\n\n**Code**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nvoid work() {\n\tint n; cin >> n;\n\tlong long L = 0, R = 1000000001LL;\n\n\tfor (int i = 1; i <= n; ++i) {\n\t\tlong long w; cin >> w;\n\t\tif (i & 1) R = min(R, w);\n\t\telse L = max(L, w);\n\t}\n\n\tcout << (n % 2 == 0 && L + 2 <= R ? \"YES\" : \"NO\") << '\\n';\n}\n\nint main() {\n\tios::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n\n\tint T; cin >> T;\n\twhile (T--) work();\n\n\treturn 0;\n}\n\n```\n\n## [2250B - String Construction](https://codeforces.com/contest/2250/problem/B \"Codeforces Round 1112 (Div. 2)\")\n\n* Idea:[\\_MCYYDS\\_](https://codeforces.com/profile/%5FMCYYDS%5F \"Grandmaster _MCYYDS_\")\n* Solution:[\\_MCYYDS\\_](https://codeforces.com/profile/%5FMCYYDS%5F \"Grandmaster _MCYYDS_\")\n\n**Hint 1**\n\nPartition the string into maximal blocks of equal characters. How does the number of blocks determine the number of adjacent equal pairs?\n\n**Solution**\n\nDefine a block as a maximal consecutive segment of equal characters. Suppose the string has $$$r$$$ blocks with lengths $$$l\\_1,l\\_2,\\\\ldots,l\\_r$$$. A block of length $$$l\\_i$$$ contributes exactly $$$l\\_i-1$$$ adjacent equal pairs, so the total number is\n\n$$$ \\\\sum\\_{i=1}^{r}(l\\_i-1)=n-r. $$$\n\nTherefore, we need to construct a string with exactly $$$r=n-k$$$ blocks.\n\nBecause $$$k \\\\lt n$$$, we have $$$r\\\\ge1$$$. Since $$$n\\\\ge2$$$ and the counts of $$$\\\\mathtt{0}$$$ and $$$\\\\mathtt{1}$$$ may differ by at most $$$1$$$, both characters must appear, so the string must contain at least two blocks. Hence, when $$$r=1$$$, or equivalently $$$k=n-1$$$, no solution exists.\n\nNow assume $$$r\\\\ge2$$$. Let the block characters alternate, starting with $$$\\\\mathtt{0}$$$. The numbers of $$$\\\\mathtt{0}$$$-blocks and $$$\\\\mathtt{1}$$$-blocks are\n\n$$$ r\\_0=\\\\left\\\\lceil\\\\frac r2\\\\right\\\\rceil,\\\\qquad r\\_1=\\\\left\\\\lfloor\\\\frac r2\\\\right\\\\rfloor. $$$\n\nWe want the total numbers of the two characters to be\n\n$$$ c\\_0=\\\\left\\\\lceil\\\\frac n2\\\\right\\\\rceil,\\\\qquad c\\_1=\\\\left\\\\lfloor\\\\frac n2\\\\right\\\\rfloor. $$$\n\nFirst assign one character to each block. This uses $$$r\\_0$$$ zeros and $$$r\\_1$$$ ones. Since $$$r\\\\le n$$$, we have $$$r\\_0\\\\le c\\_0$$$ and $$$r\\_1\\\\le c\\_1$$$.\n\nAppend all remaining zeros to any $$$\\\\mathtt{0}$$$-block and all remaining ones to any $$$\\\\mathtt{1}$$$-block, then concatenate the blocks in order. The two character counts differ by at most $$$1$$$, and the string has exactly $$$r$$$ blocks, so it contains exactly $$$n-r=k$$$ adjacent equal pairs.\n\nThe time complexity is $$$O(n)$$$ per test case and $$$O(\\\\sum n)$$$ in total. The space complexity is $$$O(n)$$$.\n\n**Code**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nvoid work() {\n\tint n, k; cin >> n >> k;\n\n\tif (n > 1 && k == n - 1) {\n\t\tcout << -1 << '\\n';\n\t\treturn;\n\t}\n\n\tk = n - k;\n\tint c0 = (n + 1) / 2, c1 = n / 2;\n\n\tfor (int i = 1; i <= k; ++i) {\n\t\tif (i & 1) {\n\t\t\tif (i + 2 > k) {\n\t\t\t\twhile (c0--) cout << 0;\n\t\t\t}\n\t\t\telse {\n\t\t\t\t--c0;\n\t\t\t\tcout << 0;\n\t\t\t}\n\t\t}\n\t\telse {\n\t\t\tif (i + 2 > k) {\n\t\t\t\twhile (c1--) cout << 1;\n\t\t\t}\n\t\t\telse {\n\t\t\t\t--c1;\n\t\t\t\tcout << 1;\n\t\t\t}\n\t\t}\n\t}\n\tcout << '\\n';\n}\n\nint main() {\n\tios::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n\n\tint T; cin >> T;\n\twhile (T--) work();\n\n\treturn 0;\n}\n\n```\n\n## [2249A - Rank Subsequence](https://codeforces.com/contest/2249/problem/A \"Codeforces Round 1112 (Div. 1)\")\n\n* Idea:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n* Solution:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n\n**Hint 1**\n\nAn element's rank from the right depends on both its position in the retained subsequence and the subsequence's final length.\n\n**Hint 2**\n\nEnumerate the final retained length $$$m$$$. Once $$$m$$$ is fixed, the condition for every position in the subsequence is fixed.\n\n**Solution**\n\nWhether an element can occupy the $$$j$$$-th position of the retained subsequence depends on the final length $$$m$$$, because its rank from the right is $$$m-j+1$$$.\n\nTherefore, we enumerate the final length $$$m$$$.\n\nFor a fixed $$$m$$$, scan the original sequence from left to right and fill positions $$$1,2,\\\\ldots,m$$$ of the subsequence in order. When filling position $$$j$$$, element $$$i$$$ can be chosen if and only if\n\n$$$ j\\\\notin\\[l\\_i,r\\_i\\] $$$\n\nand\n\n$$$ m-j+1\\\\notin\\[u\\_i,v\\_i\\]. $$$\n\nWhenever the current element satisfies both conditions, choose it immediately.\n\nTo prove this greedy choice, suppose a feasible solution chooses an element $$$x$$$ for position $$$j$$$, while the greedy algorithm chooses an earlier valid element $$$y$$$. Replacing $$$x$$$ with $$$y$$$ leaves every later element of the original solution available and does not reduce the remaining suffix in which subsequent choices must be made. Thus, whenever a feasible solution exists, the greedy algorithm also succeeds.\n\nEnumerate $$$m$$$ from $$$n$$$ down to $$$1$$$. The first feasible value is the maximum possible answer. If no positive value is feasible, the answer is $$$0$$$.\n\nThe time complexity is $$$O(n^2)$$$ per test case, and the space complexity is $$$O(n)$$$.\n\n**Code**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nconst int N = 5000;\n\nint n;\nint l[N + 10], r[N + 10], u[N + 10], v[N + 10];\n\nbool check(int m) {\n\tint j = 1;\n\n\tfor (int i = 1; i <= n && j <= m; ++i) {\n\t\tint x = m - j + 1;\n\t\tif ((j < l[i] || j > r[i]) && (x < u[i] || x > v[i])) ++j;\n\t}\n\n\treturn j == m + 1;\n}\n\nvoid work() {\n\tcin >> n;\n\tfor (int i = 1; i <= n; ++i) cin >> l[i] >> r[i] >> u[i] >> v[i];\n\n\tfor (int m = n; m >= 1; --m) {\n\t\tif (!check(m)) continue;\n\t\tcout << m << '\\n';\n\t\treturn;\n\t}\n\n\tcout << 0 << '\\n';\n}\n\nint main() {\n\tios::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n\n\tint T; cin >> T;\n\twhile (T--) work();\n\n\treturn 0;\n}\n\n```\n\n## [2249B - Permutation Cuts](https://codeforces.com/contest/2249/problem/B \"Codeforces Round 1112 (Div. 1)\")\n\n* Idea:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n* Solution:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n\n**Hint 1**\n\nThe value $$$n$$$ lies on exactly one side of every cut. Therefore, the value of the cut must be the maximum on the opposite side.\n\n**Hint 2**\n\nEnumerate the dividing point determined by the position of $$$n$$$. The cut values to its left must be non-decreasing, while those to its right must be non-increasing.\n\n**Hint 3**\n\nMerge the left sequence with the reversed right sequence. The first occurrence of a value forces its position, while every later occurrence contributes a number of choices.\n\n**Solution**\n\nFirst observe that no cut value can equal $$$n$$$. The value $$$n$$$ lies on exactly one side of a cut, while the maximum on the opposite side is at most $$$n-1$$$. Therefore, if some $$$a\\_i=n$$$, the answer is $$$0$$$.\n\nNow consider the position of $$$n$$$. For cut $$$i$$$:\n\n* If $$$n$$$ lies on the right, then $$$a\\_i$$$ is the maximum of the prefix on the left.\n* If $$$n$$$ lies on the left, then $$$a\\_i$$$ is the maximum of the suffix on the right.\n\nHence, there must be a dividing point $$$c$$$ such that $$$a\\_1,\\\\ldots,a\\_c$$$ is non-decreasing and $$$a\\_{c+1},\\\\ldots,a\\_{n-1}$$$ is non-increasing. If both parts are non-empty, we must also have $$$a\\_c\\\\ne a\\_{c+1}$$$. These two maxima are attained at positions on opposite sides of the dividing point, and a permutation cannot contain the same value at both positions.\n\nWe enumerate all values of $$$c$$$ satisfying these conditions. There may seem to be many candidates, but the indices belonging to both the non-decreasing prefix and the non-increasing suffix can only form a segment of equal values. Two adjacent equal values cannot form an internal boundary, so only a constant number of candidates need to be considered.\n\nAfter fixing $$$c$$$, merge\n\n$$$ a\\_1,a\\_2,\\\\ldots,a\\_c $$$\n\nwith the reversed right part\n\n$$$ a\\_{n-1},a\\_{n-2},\\\\ldots,a\\_{c+1} $$$\n\nto obtain one non-decreasing sequence. Process its $$$i$$$-th value $$$x$$$:\n\n* If this is the first occurrence of $$$x$$$, then the maximum on that side must become $$$x$$$ at this position, so the assigned permutation value is forced to be $$$x$$$.\n* Otherwise, we may assign any unused value smaller than $$$x$$$. Since $$$i-1$$$ values have already been used, there are $$$x-i+1$$$ choices.\n\nFor each repeated occurrence, multiply the answer by $$$x-i+1$$$. This gives the number of permutations corresponding to the fixed dividing point. Finally, sum these values over all valid dividing points.\n\nEach merge takes $$$O(n)$$$ time. Since only a constant number of dividing points are valid, the time complexity is $$$O(n)$$$ per test case, and the space complexity is $$$O(n)$$$.\n\n**Code**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nconst int N = 1000000;\nconst int mod = 998244353;\n\nint n;\nint a[N + 10], vis[N + 10];\nbool pre[N + 10], suf[N + 10];\n\nvoid work() {\n\tcin >> n;\n\tfor (int i = 1; i < n; ++i) cin >> a[i];\n\ta[n] = 0;\n\n\tfor (int i = 1; i < n; ++i) {\n\t\tif (a[i] != n) continue;\n\t\tcout << 0 << '\\n';\n\t\treturn;\n\t}\n\n\tpre[0] = 1;\n\tfor (int i = 1; i < n; ++i) pre[i] = pre[i - 1] && a[i] >= a[i - 1];\n\tsuf[n] = 1;\n\tfor (int i = n - 1; i >= 1; --i) suf[i] = suf[i + 1] && a[i] >= a[i + 1];\n\n\tlong long ans = 0;\n\tfor (int cut = 0; cut < n; ++cut) {\n\t\tif (!pre || !suf[cut + 1]) continue;\n                if (cut > 0 && cut < n - 1 && a[cut] == a[cut + 1]) continue;\n\n\t        vector<int> l, r;\n\t        for (int i = 1; i <= cut; ++i) l.push_back(a[i]);\n\t        for (int i = cut + 1; i < n; ++i) r.push_back(a[i]);\n\t        reverse(r.begin(), r.end());\n\n\t\tmemset(vis, 0, sizeof(int) * (n + 1));\n\t\tint x = 0, y = 0, used = 0;\n\t\tlong long ways = 1;\n\n\t\twhile (x < (int)l.size() || y < (int)r.size()) {\n\t\t\tint w;\n\t\t\tif (x == (int)l.size()) w = r[y++];\n\t\t\telse if (y == (int)r.size()) w = l[x++];\n\t\t\telse if (l[x] < r[y]) w = l[x++];\n\t\t\telse if (l[x] > r[y]) w = r[y++];\n\t\t\telse {\n\t\t\t\tcout << 0 << '\\n';\n\t\t\t\treturn;\n\t\t\t}\n\n\t\t\tif (!vis[w]) vis[w] = 1;\n\t\t\telse if (w < used) ways = 0;\n\t\t\telse ways = ways * (w - used) % mod;\n\t\t\t++used;\n\t\t}\n\n\t\tans += ways;\n\t}\n\n\tcout << ans % mod << '\\n';\n}\n\nint main() {\n\tios::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n\n\tint T; cin >> T;\n\twhile (T--) work();\n\n\treturn 0;\n}\n\n```\n\n## [2249C - Double-Rift Dial](https://codeforces.com/contest/2249/problem/C \"Codeforces Round 1112 (Div. 1)\")\n\n* Idea:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n* Solution:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n\n**Hint 1**\n\nDuplicate the permutation so that the cyclic sequence starting at any position becomes an ordinary contiguous segment of the duplicated array.\n\n**Hint 2**\n\nAfter inserting a new value $$$x$$$, the change in the number of consecutive-value blocks depends only on whether $$$x-1$$$ and $$$x+1$$$ have already appeared.\n\n**Hint 3**\n\nMaintain the number of blocks for all starting positions simultaneously. Once a starting position has produced more than two blocks, it can be discarded permanently.\n\n**Solution**\n\nChecking every prefix separately for every starting position would introduce an additional factor of $$$n$$$. Instead, duplicate the permutation and define\n\n$$$ q\\_i=p\\_{((i-1)\\\\bmod n)+1}. $$$\n\nFor a starting position $$$l$$$, the successive cyclic prefixes become the ordinary segments $$$\\[l,l\\],\\[l,l+1\\],\\\\ldots,\\[l,l+n-1\\]$$$.\n\nEnumerate the right endpoint $$$r$$$ from left to right. When $$$q\\_r$$$ is inserted, the starting positions whose current segment ends at $$$r$$$ form the interval\n\n$$$ L=\\\\max(1,r-n+1),\\\\qquad R=\\\\min(r,n). $$$\n\nLet the new value be $$$x$$$. Because the current segment has length at most $$$n$$$, no permutation value appears twice. The change in the number of consecutive-value blocks is\n\n$$$ \\\\Delta=1-\\[x-1\\\\text{ has already appeared}\\]-\\[x+1\\\\text{ has already appeared}\\]. $$$\n\nWe first regard $$$x$$$ as a new block, then merge it with the block on either side whenever the corresponding neighboring value has already appeared.\n\nTo update all starting positions simultaneously, let $$$\\\\operatorname{last}\\_v$$$ be the latest position before $$$r$$$ at which value $$$v$$$ appeared. For a starting position $$$l$$$, value $$$v$$$ appears in $$$\\[l,r-1\\]$$$ if and only if $$$\\\\operatorname{last}\\_v\\\\ge l$$$.\n\nTherefore, inserting $$$x$$$ requires only three range additions:\n\n1. Add $$$1$$$ to $$$\\[L,R\\]$$$.\n2. If $$$x-1$$$ exists, subtract $$$1$$$ from $$$\\[L,\\\\min(R,\\\\operatorname{last}\\_{x-1})\\]$$$.\n3. If $$$x+1$$$ exists, subtract $$$1$$$ from $$$\\[L,\\\\min(R,\\\\operatorname{last}\\_{x+1})\\]$$$.\n\nUse a lazy segment tree to maintain the current number of blocks for every starting position, with range addition and a global maximum query. Once a starting position has more than two blocks, it has already produced an invalid prefix. A later decrease cannot erase that earlier invalid prefix, so the starting position can be removed permanently.\n\nEach starting position is removed at most once. We process $$$2n-1$$$ right endpoints, performing only a constant number of range operations for each one. Thus, the time complexity is $$$O(n\\\\log n)$$$, and the space complexity is $$$O(n)$$$.\n\n**Code**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nconst int N = 200000;\nconst int inf = 1000000000;\n\nint n;\nint p[N + 10], last[N + 10];\nint mx[4 * N + 10], lazy[4 * N + 10];\n\nvoid apply(int u, int v) {\n\tmx[u] += v;\n\tlazy[u] += v;\n}\n\nvoid push(int u) {\n\tif (!lazy[u]) return;\n\tapply(u << 1, lazy[u]);\n\tapply(u << 1 | 1, lazy[u]);\n\tlazy[u] = 0;\n}\n\nvoid add(int u, int l, int r, int ql, int qr, int v) {\n\tif (ql <= l && r <= qr) {\n\t\tapply(u, v);\n\t\treturn;\n\t}\n\n\tpush(u);\n\tint mid = (l + r) >> 1;\n\tif (ql <= mid) add(u << 1, l, mid, ql, qr, v);\n\tif (qr > mid) add(u << 1 | 1, mid + 1, r, ql, qr, v);\n\tmx[u] = max(mx[u << 1], mx[u << 1 | 1]);\n}\n\nvoid add(int l, int r, int v) {\n\tif (l <= r) add(1, 1, n, l, r, v);\n}\n\nint find_bad(int u, int l, int r) {\n\tif (l == r) return l;\n\n\tpush(u);\n\tint mid = (l + r) >> 1;\n\tif (mx[u << 1] > 2) return find_bad(u << 1, l, mid);\n\treturn find_bad(u << 1 | 1, mid + 1, r);\n}\n\nvoid del(int u, int l, int r, int x) {\n\tif (l == r) {\n\t\tmx[u] = -inf;\n\t\tlazy[u] = 0;\n\t\treturn;\n\t}\n\n\tpush(u);\n\tint mid = (l + r) >> 1;\n\tif (x <= mid) del(u << 1, l, mid, x);\n\telse del(u << 1 | 1, mid + 1, r, x);\n\tmx[u] = max(mx[u << 1], mx[u << 1 | 1]);\n}\n\nvoid work() {\n\tcin >> n;\n\tfor (int i = 1; i <= n; ++i) cin >> p[i];\n\n\tmemset(last, 0, sizeof(int) * (n + 2));\n\tmemset(mx, 0, sizeof(int) * (4 * n + 5));\n\tmemset(lazy, 0, sizeof(int) * (4 * n + 5));\n\n\tint ans = n;\n\tfor (int r = 1; r < 2 * n; ++r) {\n\t\tint L = max(1, r - n + 1), R = min(r, n);\n\t\tint x = p[(r - 1) % n + 1];\n\n\t\tadd(L, R, 1);\n\t\tif (x > 1 && last[x - 1] >= L) add(L, min(R, last[x - 1]), -1);\n\t\tif (x < n && last[x + 1] >= L) add(L, min(R, last[x + 1]), -1);\n\t\tlast[x] = r;\n\n\t\twhile (mx[1] > 2) {\n\t\t\tint x = find_bad(1, 1, n);\n\t\t\tdel(1, 1, n, x);\n\t\t\t--ans;\n\t\t}\n\t}\n\n\tcout << ans << '\\n';\n}\n\nint main() {\n\tios::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n\n\tint T; cin >> T;\n\twhile (T--) work();\n\n\treturn 0;\n}\n\n```\n\n## [2249D - Xor Permutation Matrix](https://codeforces.com/contest/2249/problem/D \"Codeforces Round 1112 (Div. 1)\")\n\n* Idea:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n* Solution:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n\n**Hint 1**\n\nCompare two adjacent rows. For a fixed pair of rows, the $$$2\\\\times2$$$ condition forces the XOR difference between corresponding entries to alternate with the parity of the column.\n\n**Hint 2**\n\nWe need many values $$$h$$$ satisfying\n\n$$$ \\\\{0,1,\\\\ldots,n-1\\\\}\\\\mathbin{\\\\operatorname{xor}}h=\\\\{0,1,\\\\ldots,n-1\\\\}. $$$\n\nHow large is the XOR stabilizer of this set?\n\n**Hint 3**\n\nWhen $$$n$$$ is a power of two, pair the values as $$$y\\\\leftrightarrow y\\\\oplus x$$$, and place several complete pairs at odd positions.\n\n**Solution**\n\nWe first determine when a solution can exist. Let $$$S={0,1,\\\\ldots,n-1}$$$, and define the XOR difference between adjacent rows in column $$$j$$$ by\n\n$$$ d\\_i(j)=A\\_{i,j}\\\\oplus A\\_{i+1,j}. $$$\n\nThe condition on every adjacent $$$2\\\\times2$$$ submatrix implies\n\n$$$ d\\_i(j+1)=d\\_i(j)\\\\oplus x. $$$\n\nThus, once the parity of the column is fixed, the XOR difference between the two rows is fixed. At least $$$\\\\lceil n/2\\\\rceil$$$ rows have the same parity, and every column must be a permutation. Therefore, we need at least $$$\\\\lceil n/2\\\\rceil$$$ distinct values $$$h$$$ such that $$$S\\\\oplus h=S$$$.\n\nLet $$$L=\\\\operatorname{lowbit}(n)$$$. The XOR stabilizer of $$$S$$$ is exactly\n\n$$$ \\\\{0,1,\\\\ldots,L-1\\\\}. $$$\n\nEvery value smaller than $$$L$$$ merely permutes elements within blocks of length $$$L$$$. Conversely, write $$$h=gL+r$$$, where $$$0\\\\le r \\\\lt L$$$. If $$$g\\\\ne0$$$, XOR by $$$h$$$ induces a nontrivial pairing of the block indices. Since the number of blocks, $$$n/L$$$, is odd, the whole set of block indices cannot remain invariant. Hence, $$$g=0$$$.\n\nThe necessary condition is therefore\n\n$$$ \\\\operatorname{lowbit}(n)\\\\ge\\\\left\\\\lceil\\\\frac n2\\\\right\\\\rceil. $$$\n\nThis holds only when $$$n$$$ is a power of two. In addition, when $$$n=2$$$, each value appears twice in any $$$2\\\\times2$$$ Latin square, so the XOR of all four entries is always $$$0$$$. Thus, no solution exists for $$$n=2$$$ and $$$x \\\\gt 0$$$.\n\nWe now describe the construction.\n\nIf $$$x=0$$$, set\n\n$$$ A\\_{i,j}=i\\\\oplus j. $$$\n\nBecause $$$n$$$ is a power of two, XOR with a fixed value permutes $$$S$$$, so every row and every column is a permutation. In each adjacent $$$2\\\\times2$$$ submatrix, every index term appears twice, and the total XOR is $$$0$$$.\n\nNow assume $$$x \\\\gt 0$$$, $$$n$$$ is a power of two, and $$$n\\\\ge4$$$. The map $$$y\\\\mapsto y\\\\oplus x$$$ partitions $$$S$$$ into $$$n/2$$$ pairs. Choose $$$n/4$$$ of these pairs to form a set $$$O$$$, and let the remaining values form a set $$$E$$$. Then\n\n$$$ |E|=|O|=\\\\frac n2,\\\\qquad E\\\\oplus x=E,\\\\qquad O\\\\oplus x=O. $$$\n\nConstruct a permutation $$$p$$$ by placing the elements of $$$E$$$ at even positions and the elements of $$$O$$$ at odd positions. Then define\n\n$$$ A\\_{i,j}=p\\_i\\\\oplus p\\_j\\\\oplus\\\\bigl((i\\\\bmod2)(j\\\\bmod2)x\\\\bigr). $$$\n\nEvery even-indexed row is a fixed XOR translation of $$$p$$$. In an odd-indexed row, the even columns contain $$$p\\_i\\\\oplus E$$$, while the odd columns contain $$$p\\_i\\\\oplus(O\\\\oplus x)=p\\_i\\\\oplus O$$$. Together, they still form $$$S$$$. The formula is symmetric in $$$i$$$ and $$$j$$$, so the same argument applies to every column.\n\nFinally, consider any adjacent $$$2\\\\times2$$$ submatrix. The four terms involving $$$p$$$ cancel in pairs. Among the four parity pairs, exactly one is $$$(1,1)$$$, so the remaining XOR is exactly $$$x$$$.\n\nThe output contains $$$n^2$$$ numbers, so the time complexity is $$$O(n^2)$$$, and the extra space complexity is $$$O(n)$$$.\n\n**Code**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nconst int N = 2500;\n\nint n, x;\nint vis[N + 10], in_o[N + 10], e[N + 10], o[N + 10], a[N + 10];\npair<int, int> p[N + 10];\n\nbool check(int v) {\n\treturn (v & (v - 1)) == 0;\n}\n\nvoid work() {\n\tcin >> n >> x;\n\n\tif (!check(n) || (x && n < 4)) {\n\t\tcout << -1 << '\\n';\n\t\treturn;\n\t}\n\n\tif (!x) {\n\t\tfor (int i = 0; i < n; ++i) {\n\t\t\tfor (int j = 0; j < n; ++j) {\n\t\t\t\tcout << (i ^ j) << (j + 1 == n ? '\\n' : ' ');\n\t\t\t}\n\t\t}\n\t\treturn;\n\t}\n\n\tfor (int i = 0; i <= n; ++i) vis[i] = in_o[i] = e[i] = o[i] = a[i] = 0;\n\n\tint cnt = 0;\n\tfor (int i = 0; i < n; ++i) {\n\t\tif (vis[i]) continue;\n\t\tint j = i ^ x;\n\t\tvis[i] = vis[j] = 1;\n\t\tp[++cnt] = {i, j};\n\t}\n\n\tfor (int i = 1; i <= n / 4; ++i) in_o[p[i].first] = in_o[p[i].second] = 1;\n\n\tint ec = 0, oc = 0;\n\tfor (int i = 0; i < n; ++i) {\n\t\tif (in_o[i]) o[++oc] = i;\n\t\telse e[++ec] = i;\n\t}\n\n\tec = oc = 0;\n\tfor (int i = 0; i < n; ++i) {\n\t\tif (i & 1) a[i] = o[++oc];\n\t\telse a[i] = e[++ec];\n\t}\n\n\tfor (int i = 0; i < n; ++i) {\n\t\tfor (int j = 0; j < n; ++j) {\n\t\t\tint v = a[i] ^ a[j];\n\t\t\tif ((i & 1) && (j & 1)) v ^= x;\n\t\t\tcout << v << (j + 1 == n ? '\\n' : ' ');\n\t\t}\n\t}\n}\n\nint main() {\n\tios::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n\n\tint T; cin >> T;\n\twhile (T--) work();\n\n\treturn 0;\n}\n\n```\n\n## [2249E1 - String (Easy Version)](https://codeforces.com/contest/2249/problem/E1 \"Codeforces Round 1112 (Div. 1)\")\n\n* Idea:[131241](https://codeforces.com/profile/131241 \"Specialist 131241\")\n* Solution:[131241](https://codeforces.com/profile/131241 \"Specialist 131241\")\n\n**Hint 1**\n\nTreat a prefix of length $$$k^i$$$ as one block. How can it be decomposed into smaller blocks?\n\n**Hint 2**\n\nAdding the same value to every character modulo $$$k$$$ produces only finitely many translated versions of the pattern-matching problem.\n\n**Hint 3**\n\nFor each block, store its internal occurrence count and its prefix and suffix of length at most $$$n-1$$$. When two blocks are merged, only occurrences crossing their boundary need additional processing.\n\n**Solution**\n\nConsider the self-similar structure of the infinite string. Let $$$B\\_{i,j}$$$ denote the standard block of length $$$k^i$$$ whose characters are all shifted by $$$j$$$ modulo $$$k$$$. Then\n\n$$$ B\\_{i,j}=B\\_{i-1,j}B\\_{i-1,j+1}\\\\cdots B\\_{i-1,j+k-1}, $$$\n\nwhere all second indices are taken modulo $$$k$$$. This follows directly from how the base-$$$k$$$ digit sum changes when the highest digit is increased.\n\nFor the current pattern $$$t$$$, maintain the following information for every block:\n\n* The number of occurrences of $$$t$$$ entirely inside the block.\n* A prefix of the block of length at most $$$n-1$$$.\n* A suffix of the block of length at most $$$n-1$$$.\n\nWhen two adjacent blocks are concatenated, their internal occurrence counts are added. Every newly created occurrence must cross the boundary, so it is sufficient to run KMP on the suffix of the left block followed by the prefix of the right block. The new prefix and suffix are again truncated to at most $$$n-1$$$ characters.\n\nStoring these boundary strings explicitly at every level would still be too slow. Let $$$x$$$ be the smallest integer such that $$$n\\\\le k^x$$$. Once the block length reaches this scale, every boundary string is a uniform translation of the boundary of a standard block. Therefore, the contribution across the boundary of two large blocks depends only on their two translation values. There are only $$$k^2$$$ such pairs, so all of them can be precomputed.\n\nFor a query interval $$$\\[l,r\\]$$$, recursively decompose it in base $$$k$$$ into $$$O(k\\\\log\\_k V)$$$ standard blocks and concatenate them in order. The internal occurrence count of each block is obtained from preprocessing, while occurrences crossing block boundaries are counted using either the stored short boundaries or the precomputed translation transitions.\n\nAt levels where the block length is smaller than $$$n$$$, the pattern cannot occur entirely inside a block. The first critical level can be processed directly; at all larger levels, only the finite set of translation transitions is needed. The total time complexity is\n\n$$$ O\\\\left(\\\\sum n\\\\cdot k^2+qk^2\\\\log\\_k V\\\\right), $$$\n\nwhere $$$V=10^{17}$$$. The space complexity is $$$O(n+k^2\\\\log\\_k V)$$$.\n\n**Code**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nstruct node {\n\tlong long v = 0;\n\tbool b = 0;\n\tstring l = \"\", r = \"\";\n\tint l1, r1;\n};\n\nint n, k;\nint fai[2000005];\nlong long T, ql, qr, trans[10][10];\nlong long ma, L, R;\nnode a[66][10], ans;\nstring pl[10], pr[10], s;\n\nvoid make_fai() {\n\tfor (int i = 1; i <= n; ++i) s[i] ^= 48;\n\n\tint l = 0;\n\tfor (int i = 2; i <= n; ++i) {\n\t\twhile (l && s[l + 1] != s[i]) l = fai[l];\n\t\tl += s[l + 1] == s[i];\n\t\tfai[i] = l;\n\t}\n}\n\nint val(string x) {\n\tint l = 0, res = 0;\n\n\tfor (int i = 0; i < (int)x.size(); ++i) {\n\t\twhile (l && s[l + 1] != x[i]) l = fai[l];\n\t\tl += s[l + 1] == x[i];\n\t\tif (l == n) ++res;\n\t}\n\n\treturn res;\n}\n\nnode add(node &x, node &y) {\n\tnode res;\n\n\tif (x.b && y.b) {\n\t\tres.b = 1;\n\t\tres.l1 = x.l1;\n\t\tres.r1 = y.r1;\n\t\tres.v = x.v + y.v + trans[x.r1][y.l1];\n\t\treturn res;\n\t}\n\n\tif (x.b) x.l = pl[x.l1], x.r = pr[x.r1];\n\tif (y.b) y.l = pl[y.l1], y.r = pr[y.r1];\n\tres.v = x.v + y.v + val(x.r + y.l);\n\tres.l = x.l;\n\n\tif ((int)res.l.size() < n - 1) {\n\t\tint t = min(n - 1 - (int)res.l.size(), (int)y.l.size());\n\t\tres.l += y.l.substr(0, t);\n\t}\n\n\tres.r = y.r;\n\tif ((int)res.r.size() < n - 1) {\n\t\tint t = min(n - 1 - (int)res.r.size(), (int)x.r.size());\n\t\tres.r = x.r.substr(x.r.size() - t, t) + res.r;\n\t}\n\n\treturn res;\n}\n\nvoid init(long long x) {\n\tbool done = 0;\n\n\tfor (long long cur = 1, i = 0; cur <= x; cur *= k, ++i) {\n\t\tma = i;\n\t\tL = 0, R = cur;\n\n\t\tif (i == 0) {\n\t\t\tfor (int j = 0; j < k; ++j) {\n\t\t\t\tif (n == 1) {\n\t\t\t\t\ta[0][j].v = j == s[1];\n\t\t\t\t\ta[0][j].b = 0;\n\t\t\t\t\ta[0][j].l = a[0][j].r = \"\";\n\t\t\t\t}\n\t\t\t\telse {\n\t\t\t\t\ta[0][j].v = 0;\n\t\t\t\t\ta[0][j].b = 0;\n\t\t\t\t\ta[0][j].l = a[0][j].r = \"\";\n\t\t\t\t\ta[0][j].l += (char)j;\n\t\t\t\t\ta[0][j].r = a[0][j].l;\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t\telse {\n\t\t\tfor (int j = 0; j < k; ++j) {\n\t\t\t\tnode cur;\n\t\t\t\tfor (int t = 0; t < k; ++t) {\n\t\t\t\t\tif (!cur.v && !cur.b && cur.l.empty()) cur = a[i - 1][(j + t) % k];\n\t\t\t\t\telse cur = add(cur, a[i - 1][(j + t) % k]);\n\t\t\t\t}\n\t\t\t\ta[i][j] = cur;\n\t\t\t}\n\t\t}\n\n\t\tif (!done && cur >= n - 1) {\n\t\t\tdone = 1;\n\t\t\tfor (int j = 0; j < k; ++j) {\n\t\t\t\tpl[j] = a[i][j].l;\n\t\t\t\tpr[j] = a[i][j].r;\n\t\t\t\ta[i][j].b = 1;\n\t\t\t\ta[i][j].l1 = a[i][j].r1 = j;\n\t\t\t\ta[i][j].l = a[i][j].r = \"\";\n\t\t\t}\n\t\t\tfor (int j = 0; j < k; ++j) {\n\t\t\t\tfor (int t = 0; t < k; ++t) trans[j][t] = val(pr[j] + pl[t]);\n\t\t\t}\n\t\t}\n\t}\n}\n\nvoid calc(int dep, int st, long long l, long long r) {\n\tif (r <= ql || l > qr) return;\n\n\tif (ql <= l && r - 1 <= qr) {\n\t\tans = add(ans, a[dep][st]);\n\t\treturn;\n\t}\n\n\tnode cur;\n\tlong long len = r - l;\n\n\tfor (int i = 0; i < k; ++i) {\n\t\tlong long nl = l + len / k * i, nr = nl + len / k;\n\n\t\tif (ql <= nl && nr - 1 <= qr) {\n\t\t\tif (!cur.v && !cur.b && cur.l.empty()) cur = a[dep - 1][(st + i) % k];\n\t\t\telse cur = add(cur, a[dep - 1][(st + i) % k]);\n\t\t\tcontinue;\n\t\t}\n\n\t\tif (cur.v || cur.b || !cur.l.empty()) {\n\t\t\tans = add(ans, cur);\n\t\t\tcur.v = 0, cur.b = 0;\n\t\t\tcur.l = cur.r = \"\";\n\t\t}\n\t\tcalc(dep - 1, (st + i) % k, nl, nr);\n\t}\n\n\tif (cur.v || cur.b || !cur.l.empty()) ans = add(ans, cur);\n}\n\nint main() {\n\tios::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n\n\tcin >> k >> T;\n\twhile (T--) {\n\t\tcin >> ql >> qr >> n >> s;\n\t\ts = \" \" + s + \"!\";\n\n\t\tmake_fai();\n\t\tinit(qr);\n\t\t++ma, R *= k;\n\n\t\tans.v = 0, ans.b = 0;\n\t\tans.l = ans.r = \"\";\n\t\tcalc(ma, 0, L, R);\n\t\tcout << ans.v << '\\n';\n\t}\n\n\treturn 0;\n}\n\n```\n\n## [2249E2 - String (Hard Version)](https://codeforces.com/contest/2249/problem/E2 \"Codeforces Round 1112 (Div. 1)\")\n\n* Idea:[131241](https://codeforces.com/profile/131241 \"Specialist 131241\")\n* Solution:[131241](https://codeforces.com/profile/131241 \"Specialist 131241\")\n\n**Hint 1**\n\nPartition the infinite string into blocks of length $$$k$$$. Every block is a cyclic shift of $$$\\\\mathtt{012\\\\ldots(k-1)}$$$. If each block is replaced by its shift value, the resulting sequence is the original infinite string again.\n\n**Hint 2**\n\nLet\n\n$$$ d\\_i=(t\\_i-t\\_{i-1}+k)\\\\bmod k. $$$\n\nInside a complete block, $$$d\\_i$$$ is always $$$1$$$. Therefore, all positions with $$$d\\_i\\\\ne1$$$ must have the same residue modulo $$$k$$$.\n\n**Hint 3**\n\nIf the positions with $$$d\\_i\\\\ne1$$$ determine the block boundaries, compress both the pattern and the query interval by a factor of about $$$k$$$ and recurse. If no such position exists, an occurring pattern cannot be longer than $$$2k$$$.\n\n**Solution**\n\nFirst examine the block structure of the infinite string $$$s$$$. Every block of length $$$k$$$ is a cyclic shift of\n\n$$$ \\\\mathtt{012\\\\ldots(k-1)}. $$$\n\nIf each complete block is replaced by its shift value, the resulting sequence is again $$$s$$$. Therefore, once the pattern $$$t$$$ is aligned with these blocks, both the pattern and the query interval can be compressed by a factor of about $$$k$$$.\n\nTo determine the alignment, consider the adjacent differences of $$$t$$$ modulo $$$k$$$:\n\n$$$ d\\_i=(t\\_i-t\\_{i-1}+k)\\\\bmod k. $$$\n\nInside a complete cyclic block, every adjacent difference is $$$1$$$. Hence, every position with $$$d\\_i\\\\ne1$$$ must correspond to the same type of block boundary, so all such positions must have the same residue modulo $$$k$$$. If two different residues occur, the pattern cannot appear.\n\nIf exactly one residue class occurs, the block boundaries are uniquely determined. Split $$$t$$$ according to these boundaries, compress each complete block into its shift value, map $$$\\[l,r\\]$$$ to the corresponding compressed interval according to the two partial end blocks, and recurse.\n\nIt remains to handle the case in which every $$$d\\_i=1$$$, so the block boundaries are not determined. The string $$$s$$$ contains no substring longer than $$$2k$$$ whose adjacent differences are all $$$1$$$. Otherwise, there would be three corresponding terms satisfying $$$s\\_i=s\\_{i+k}=s\\_{i+2k}$$$. However, the equality $$$s\\_i=s\\_{i+k}$$$ requires a particular carry pattern when $$$i$$$ is increased by $$$k$$$, and the same pattern cannot hold again for $$$i+k$$$.\n\nTherefore, any pattern that occurs must satisfy $$$n\\\\le2k$$$. We can enumerate all possible block alignments, reducing the compressed pattern to length at most $$$2$$$. When $$$n\\\\le k$$$, we must also consider alignments that compress it to a single character. Different alignments correspond to different starting positions in the original string, so they are not counted more than once.\n\nWe are left with occurrence queries for patterns of length $$$1$$$ or $$$2$$$.\n\nFor a pattern of length $$$1$$$, every complete block of length $$$k$$$ contains each character exactly once. Only the final partial block of a prefix query needs to be checked directly.\n\nFor a pattern of length $$$2$$$, precompute\n\n$$$ f\\_{p,q,x,y}, $$$\n\nthe number of occurrences of the pair $$$xy$$$ in the first $$$qk^p$$$ characters. A block at the next level consists of $$$k$$$ shifted copies of a smaller block. Their internal contributions are accumulated with the appropriate shifts, and occurrences crossing block boundaries are counted by checking adjacent endpoints. A prefix can be decomposed in base $$$k$$$, and an interval answer is obtained by subtracting two prefix answers.\n\nThe total time complexity is\n\n$$$ O\\\\left(qk\\\\log\\_k V+\\\\sum n+k^3\\\\log\\_k V+k\\\\log\\_k^2V\\\\right), $$$\n\nwhere $$$V=10^{17}$$$.\n\n**Code**\n\n```\n#include<bits/stdc++.h>\n#define int long long\nusing namespace std;\nint lg;\nint q,L,R,n,k,d[10000005],t[100],b[100];\nchar c[10000005],s[10000005];\nstruct node{\n\tint f[65][65];\n}sum[65][65];\nnode merge(node a,node b,int v=1){\n\tfor(int i=0;i<k;i++){\n\t\tfor(int j=0;j<k;j++){\n\t\t\ta.f[(i+v)%k][(j+v)%k]+=b.f[i][j];\n\t\t}\n\t}\n\treturn a;\n}\nint calc(int x){\n\tint sum=0;\n\twhile(x){\n\t\tsum+=x%k;\n\t\tx=x/k;\n\t}\n\treturn sum%k;\n}\nint sp[70];\nint gs(int r,int x,int y){\n    // Count occurrences of the pattern \"xy\" in the range [0, r].\n\tr++;\n\tint cnt=0,xnow=0,ans=0,lowbit=0;\n\tfor(int i=0;i<=lg;i++){\n\t\tsp[i]=r%k;r/=k;\n\t}\n\tfor(int i=lg;i>=0;i--){\n\t\tint j=sp[i];\n\t\tif(!j) continue;\n\t\tans+=sum[i][j].f[(x+(k-1)*cnt)%k][(y+(k-1)*cnt)%k];\n\t\tif(xnow){\n\t\t\tint px=(cnt-1+lowbit*(k-1))%k,py=cnt%k;\n\t\t\tif(px==x&&py==y){\n\t\t\t\tans++;\n\t\t\t}\n\t\t}\n\t\txnow+=b[i]*j;\n\t\tcnt+=j;\n\t\tlowbit=i;\n\t}\n\treturn ans;\n}\nint gs(int r,int x){\n    // Count occurrences of the pattern \"x\" in the range [0, r].\n\tr++;\n\tint sum=r/k,pl=r/k*k,y=calc(pl);\n\twhile(pl<r){\n\t\tif(y==x){\n\t\t\tsum++;\n\t\t\tbreak;\n\t\t}\n\t\tpl++;\n\t\ty++;\n\t\ty>=k?y-=k:0;\n\t}\n\treturn sum;\n}\nvoid print(char c){\n\tif(c<10){\n\t\tcout<<char(c+'0');\n\t}else{\n\t\tcout<<char(c+'a'-10);\n\t}\n}\nint ans=0;\nvoid solve(int l,int r,int n){\n    // Compute the answer recursively.\n\tif(l>r){\n\t\treturn ;\n\t}\n\tfor(int i=0;i<k;i++){\n\t\tt[i]=0;\n\t}\n\tif(n<=2){\n        // For length at most 2, use prefix-count differences directly.\n\t\tif(n==2){\n\t\t\tint pl=c[0],pr=c[1];\n\t\t\tint sum=gs(r,pl,pr)-gs(l,pl,pr);\n\t\t\tans+=sum;\n\t\t\treturn ;\n\t\t}else{\n\t\t\tint sum=0;\n\t\t\tsum=gs(r,c[0])-gs(l-1,c[0]);\n\t\t\tans+=sum;\n\t\t\treturn ;\n\t\t}\n\t}\n\tint cnt=0;\n\tfor(int i=1;i<n;i++){\n\t\td[i]=(c[i]-c[i-1]+k)%k;\n\t\tif(d[i]!=1){\n\t\t\tif(!t[i%k]) cnt++;\n\t\t\tt[i%k]=1;\n\t\t}\n\t}\n\tif(cnt>1){\n        // If more than one residue occurs, the pattern cannot appear.\n\t\treturn ;\n\t}\n\tint m=0,pl=k,pr=k,p=0;\n\tif(cnt){\n        // At least one adjacent difference is not 1.\n\t\tfor(int i=1;i<n;i++){\n\t\t\tif(d[i]!=1) p=i%k;\n\t\t}\n\t\tif(p){\n\t\t\tpl=p;\n\t\t\ts[m++]=(c[p-1]+1)%k;\n\t\t}\n\t\tfor(int i=p;i<n;i+=k){\n\t\t\tif(i+k-1>=n) pr=n-i;\n\t\t\ts[m++]=c[i];\n\t\t}\n\n        // Handle the two partial boundary blocks separately.\n\t\tint bl=(r/k)*k,br=(l/k+1)*k-1;\n\t\tif(br-l+1<pl) l=l/k+1;\n\t\telse l=l/k;\n\t\tif(r-bl+1<pr) r=r/k-1;\n\t\telse r=r/k;\n\n\t\tfor(int i=0;i<m;i++) c[i]=s[i];\n\t\tsolve(l,r,m);\n\t}else{\n        // Every adjacent difference is 1.\n\t\tint szl=(l/k+1)*k-l,szr=r-(r/k)*k+1,L,R;\n\t\tif(n<=k){\n\t\t\tint x=c[0];\n\t\t\tfor(int i=0;i+n-1<k;i++){\n\t\t\t\tc[0]=(x-i+k)%k;\n                // Handle the two partial boundary blocks separately.\n\t\t\t\tpl=k-i,pr=i+n;\n\t\t\t\tif(szl<pl) L=l/k+1;\n\t\t\t\telse L=l/k;\n\t\t\t\tif(szr<pr) R=r/k-1;\n\t\t\t\telse R=r/k;\n\t\t\t\tsolve(L,R,1);\n\t\t\t}\n\t\t}\n\t\tfor(int i=0;i<n;i++){\n\t\t\ts[i]=c[i];\n\t\t}\n\t\tfor(int i=0;i<min(n-1,k);i++){\n            // Handle the two partial boundary blocks separately.\n\t\t\tpl=i+1,pr=n-1-i;\n\t\t\tif(pr>k) continue;\n\t\t\tif(szl<pl) L=l/k+1;\n\t\t\telse L=l/k;\n\t\t\tif(szr<pr) R=r/k-1;\n\t\t\telse R=r/k;\n\t\t\tc[0]=c[1]=s[i+1];\n\t\t\tsolve(L,R,2);\n\t\t}\n\t}\n}\nchar change(int x){\n    if(0<=x&&x<=9) return x+'0';\n    if(10<=x&&x<=35) return x-10+'A';\n    if(36<=x&&x<=61) return x-36+'a';\n}\nint change(char x){\n    if('0'<=x&&x<='9') return x-'0';\n    if('A'<=x&&x<='Z') return x+10-'A';\n    if('a'<=x&&x<='z') return x+36-'a';\n}\nsigned main(){\n\tios::sync_with_stdio(0);\n\tcin.tie(0);cout.tie(0);\n\tcin>>k>>q;\n\n    // sum[i][j] stores counts of length-2 substrings in the first j * k^i characters.\n\tb[0]=1;lg=0;\n\tfor(int i=1;i<=k;i++){\n\t\tfor(int j=1;j<i;j++){\n\t\t\tsum[0][i].f[j-1][j]=1;\n\t\t}\n\t}\n\tfor(int p=1;p;p++){\n\t\tb[p]=b[p-1]*k;\n\t\tlg++;\n\t\tif(b[p]>1e17){\n\t\t\tbreak;\n\t\t}\n\t\tsum[p][1]=sum[p-1][k];\n        // Concatenate the k shifted blocks.\n\t\tfor(int i=2;i<=k;i++){\n\t\t\tsum[p][i]=merge(sum[p][i-1],sum[p-1][k],i-1);\n\t\t\t// Count substrings that cross a block boundary directly.\n\t\t\tsum[p][i].f[calc(b[p]*(i-1)-1)][calc(b[p]*(i-1))]++;\n\t\t}\n\t}\n\n\twhile(q--){\n\t\tcin>>L>>R>>n;\n\t\tcin>>c;\n\t\tfor(int i=0;i<n;i++){\n\t\t\tc[i]=change(c[i]);\n\t\t}\n\t\tsolve(L,R,n);\n\t\tcout<<ans<<endl;\n\t\tans=0;\n\t}\n    return 0;\n}\n\n```\n\n## [2249F - Even Simple Path](https://codeforces.com/contest/2249/problem/F \"Codeforces Round 1112 (Div. 1)\")\n\n* Idea:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n* Solution:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n\n**Hint 1**\n\nIn a simple path, the two endpoints have degree $$$1$$$, every other used vertex has degree $$$2$$$, and every unused vertex has degree $$$0$$$. Can this degree condition be encoded by a matching?\n\n**Hint 2**\n\nSplit every vertex into left and right copies, but remove $$$R(1)$$$ and $$$L(n)$$$. Original edges connect copies within the same layer, while an unused internal vertex matches its two copies with a cross-layer edge.\n\n**Hint 3**\n\nGive every cross-layer edge weight $$$M$$$ and every same-layer edge corresponding to an original edge weight $$$M-1$$$. A maximum-weight matching eliminates extra cycles and minimizes the resulting path length.\n\n**Solution**\n\nThe main difficulty is the word \\`\\`simple''. A shortest path algorithm cannot directly handle the condition that no vertex may be visited twice, so we encode the path as a matching.\n\nFirst, consider the degree pattern of a simple path from vertex $$$1$$$ to vertex $$$n$$$:\n\n* vertices $$$1$$$ and $$$n$$$ have degree $$$1$$$ in the path;\n* every other used vertex has degree $$$2$$$;\n* every unused vertex has degree $$$0$$$.\n\nWe will build an auxiliary graph where:\n\n* an unused vertex corresponds to one internal matching edge;\n* a used internal vertex must match both of its copies to other vertices;\n* vertices $$$1$$$ and $$$n$$$ have only one copy each, so they contribute exactly one path edge.\n\nThen a perfect matching in the auxiliary graph will describe a path in the original graph.\n\n---\n\nLet the original graph be $$$G=(V,E)$$$.\n\nFor every vertex $$$v$$$, create two copies $$$L(v)$$$ and $$$R(v)$$$. Then delete $$$R(1)$$$ and $$$L(n)$$$. 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              "command": "python3 -c 'from itertools import permutations, product\nfrom collections import Counter\n\ndef solve(a):\n n=len(a)+1\n if max(a)!=n-1:return 0\n L=a.index(n-1); R=len(a)-1-a[::-1].index(n-1)\n ans=0\n for c in (L,R+1):\n  left=a[:c];right=a[c:][::-1]\n  if any(x>y for x,y in zip(left,left[1:])) or any(x>y for x,y in zip(right,right[1:])):continue\n  i=j=used=last=0;ways=1\n  while i<len(left) or j<len(right):\n   if i<len(left) and j<len(right) and left[i]==right[j]:ways=0;break\n   if j==len(right) or (i<len(left) and left[i]<right[j]):x=left[i];i+=1\n   else:x=right[j];j+=1\n   if x==last:ways*=max(0,x-used)\n   last=x;used+=1\n  ans+=ways\n return ans\nfor n in range(2,8):\n cnt=Counter()\n for p in permutations(range(1,n+1)):\n  a=tuple(min(max(p[:i]),max(p[i:])) for i in range(1,n));cnt[a]+=1\n cases=product(range(1,n+1),repeat=n-1) if n<=6 else cnt\n checked=0\n for a in cases:\n  assert solve(a)==cnt[a],(n,a,solve(a),cnt[a])\n  checked+=1\n print(n,checked,\"passed\")'",
              "timeout": 120000
            },
            "output": "2 2 passed\n3 9 passed\n4 64 passed\n5 625 passed\n6 7776 passed\n7 903 passed\n",
            "metadata": {
              "output": "2 2 passed\n3 9 passed\n4 64 passed\n5 625 passed\n6 7776 passed\n7 903 passed\n",
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            "title": "python3 -c 'from itertools import permutations, product\nfrom collections import Counter\n\ndef solve(a):\n n=len(a)+1\n if max(a)!=n-1:return 0\n L=a.index(n-1); R=len(a)-1-a[::-1].index(n-1)\n ans=0\n for c in (L,R+1):\n  left=a[:c];right=a[c:][::-1]\n  if any(x>y for x,y in zip(left,left[1:])) or any(x>y for x,y in zip(right,right[1:])):continue\n  i=j=used=last=0;ways=1\n  while i<len(left) or j<len(right):\n   if i<len(left) and j<len(right) and left[i]==right[j]:ways=0;break\n   if j==len(right) or (i<len(left) and left[i]<right[j]):x=left[i];i+=1\n   else:x=right[j];j+=1\n   if x==last:ways*=max(0,x-used)\n   last=x;used+=1\n  ans+=ways\n return ans\nfor n in range(2,8):\n cnt=Counter()\n for p in permutations(range(1,n+1)):\n  a=tuple(min(max(p[:i]),max(p[i:])) for i in range(1,n));cnt[a]+=1\n cases=product(range(1,n+1),repeat=n-1) if n<=6 else cnt\n checked=0\n for a in cases:\n  assert solve(a)==cnt[a],(n,a,solve(a),cnt[a])\n  checked+=1\n print(n,checked,\"passed\")'",
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      "info": {
        "parentID": "msg_0b25294c30019aFC9kQJEj0CGq",
        "role": "assistant",
        "mode": "nudge-generation",
        "agent": "nudge-generation",
        "variant": "low",
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          "hints": [
            {
              "order": 1,
              "content": "Every cut has $n$ on exactly one side. So the smaller maximum is simply the maximum on the side **without $n$**. What does that make the cut values to the left and right of $n$?"
            },
            {
              "order": 2,
              "content": "A cut has value $n-1$ exactly when it separates $n$ from $n-1$. Therefore, all occurrences of $n-1$ must form one nonempty consecutive block. If that block is $a_L,\\ldots,a_R$, the positions of $n$ and $n-1$ must be $L$ and $R+1$, in some order. Only two possible positions for $n$!"
            },
            {
              "order": 3,
              "content": "Fix $p_{c+1}=n$. Then $a_1,\\ldots,a_c$ must be non-decreasing, and $a_{c+1},\\ldots,a_{n-1}$ must be non-increasing. Read the left part forward and the right part backward: both are now sequences of required running maxima. A value appearing in both sequences makes this candidate impossible, since both sides would need their own copy of it."
            },
            {
              "order": 4,
              "content": "Merge those two non-decreasing sequences, keeping the order within each. When a required maximum $x$ appears for the first time, the corresponding permutation position is forced to contain $x$. When $x$ repeats, that position can contain any unused value smaller than $x$."
            },
            {
              "order": 5,
              "content": "Let $k$ be the number of positions already filled during the merge, excluding $n$. At a repeated maximum $x$, all $k$ used values are at most $x$, and exactly one is $x$. Thus there are $(x-1)-(k-1)=x-k$ choices. Multiply these factors, rejecting nonpositive factors or equal values at the two merge fronts. Add the counts for $c=L-1$ and $c=R$; each merge is linear."
            }
          ],
          "editorial": "The big simplification is that **one side of every cut contains $n$**. Its maximum is already settled. The cut value is just the maximum on the other side.\n\nThat turns the problem into counting permutations with prescribed running maxima—and we only need to try two positions for $n$.\n\n## 1. Locate the two largest values\n\nFor any cut,\n\n$$a_i=n-1 \\iff \\text{the cut separates }n\\text{ and }n-1.$$\n\nIf they are on opposite sides, the two maxima are $n$ and $n-1$. If they are on the same side, the other side has maximum at most $n-2$.\n\nSuppose their positions, in increasing order, are $u<v$. Exactly the cuts\n\n$$u,u+1,\\ldots,v-1$$\n\nseparate them. Therefore:\n\n- No cut value can be $n$.\n- The value $n-1$ must appear.\n- Its occurrences must form one consecutive block.\n\nLet $L$ and $R$ be the first and last indices with $a_i=n-1$. Then the positions of the two largest values must be\n\n$$\\{\\operatorname{pos}(n),\\operatorname{pos}(n-1)\\}=\\{L,R+1\\}.$$\n\nSo $n$ is either at position $L$ or at position $R+1$. No need to try all $n$ positions.\n\nWe reject immediately if $\\max(a)\\ne n-1$. The checks below also reject a broken, nonconsecutive block of $n-1$ values.\n\n## 2. Fix the position of $n$\n\nWrite that position as $c+1$, so there are $c$ elements to its left. We will try\n\n$$c=L-1\\quad\\text{and}\\quad c=R.$$\n\nFor cuts before $n$, the side without $n$ is the prefix:\n\n$$a_i=\\max(p_1,\\ldots,p_i),\\qquad i\\le c.$$\n\nFor cuts after $n$, it is the suffix:\n\n$$a_i=\\max(p_{i+1},\\ldots,p_n),\\qquad i>c.$$\n\nThus we require:\n\n- $a_1,\\ldots,a_c$ to be non-decreasing;\n- $a_{c+1},\\ldots,a_{n-1}$ to be non-increasing.\n\nReject the candidate if either condition fails. Empty parts are fine: $n$ may sit at either end.\n\nNow read the permutation inward from both ends. The required running maxima are\n\n$$A=(a_1,\\ldots,a_c),\\qquad B=(a_{n-1},\\ldots,a_{c+1}).$$\n\nBoth sequences are non-decreasing. Each entry corresponds to one actual permutation position that we need to fill.\n\n**A value cannot appear in both sequences.** If $x$ is a running maximum on both sides, both sides must contain $x$. A permutation gets one copy, not a spare.\n\n## 3. Merge and count\n\nMerge $A$ and $B$ in non-decreasing order with two pointers. This is only an order for assigning positions; we are not rearranging the permutation positions themselves.\n\nIf the two current entries are equal, reject the candidate. Since both sequences are sorted, this detects every value shared by the two sides.\n\nLet the next required maximum be $x$, and let $k$ be the number of positions already filled, excluding the position containing $n$.\n\n### First occurrence of $x$\n\nThe running maximum on that side has just increased to $x$. The corresponding position must contain $x$, so there is exactly one choice.\n\nAlso, $x$ is still unused: every earlier assignment was at most its required maximum, and all earlier required maxima were smaller than $x$.\n\n### Repeated occurrence of $x$\n\nThe running maximum stays $x$, so this position must contain an unused value smaller than $x$.\n\nThere are $x-1$ values smaller than $x$. All $k$ previously assigned values are at most $x$, and exactly one of them is $x$. Hence $k-1$ smaller values have already been used, leaving\n\n$$(x-1)-(k-1)=x-k$$\n\nchoices.\n\nMultiply the candidate's count by $x-k$ modulo $998244353$. If this number is nonpositive, the candidate contributes zero.\n\nAfter either case, increment $k$.\n\nThe important detail is that this number of choices does not depend on *which* smaller values were chosen earlier. Their count is all that matters, because we process required maxima in sorted order. That's why a simple product works.\n\n## 4. Example\n\nTake $n=6$ and\n\n$$a=[3,3,5,5,5].$$\n\nThe block of $n-1=5$ spans cuts $3$ through $5$, so $6$ can only be at position $3$ or $6$.\n\nFor position $3$, the two sequences are $[3,3]$ and $[5,5,5]$. Their merge is\n\n$$[3,3,5,5,5].$$\n\nThe first $3$ and first $5$ are forced. The repeated entries contribute\n\n$$3-1=2,\\qquad 5-3=2,\\qquad 5-4=1.$$\n\nThis candidate contributes $2\\cdot2\\cdot1=4$. Position $6$ contributes another $4$, giving $8$.\n\n## Correctness argument\n\nEvery valid permutation places $n$ at $L$ or $R+1$, because the cuts with value $n-1$ are exactly those between the two largest values. Thus the algorithm considers every possible position of $n$.\n\nFor a fixed candidate, the cut conditions are exactly the running-maximum conditions of $A$ and $B$. The monotonicity checks are necessary, and a value shared by both sides is impossible because permutation values are distinct.\n\nDuring the merge, the algorithm preserves the order of positions within each side. Whenever a required maximum increases to $x$, assigning $x$ is necessary and available. Whenever it repeats, assigning any unused value smaller than $x$ is necessary and sufficient, and there are exactly $x-k$ such values. Therefore the product counts precisely all assignments satisfying the two running-maximum sequences.\n\nThese assignments fill $n-1$ positions with distinct values from $1$ through $n-1$. Together with the fixed $n$, they form a permutation with exactly the requested cut values. Each assignment is counted once, and the two candidates have different positions of $n$, so their counts do not overlap.\n\nHence their sum is the required answer.\n\n## Complexity and implementation\n\nFinding $L,R$, validating a candidate, and merging each take $O(n)$ time. There are only two candidates, so the total is **$O(n)$ time per test case** and **$O(\\sum n)$ overall**.\n\nThe input array uses **$O(n)$ space**; the merge itself needs only two pointers and a few counters. Use a 64-bit integer for modular multiplication. No sorting, factorials, or modular inverses are needed.",
          "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    constexpr ll MOD = 998244353;\n\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        vector<int> a(n);\n        int mx = 0, L = -1, R = -1;\n        for (int i = 1; i < n; ++i) {\n            cin >> a[i];\n            mx = max(mx, a[i]);\n            if (a[i] == n - 1) {\n                if (L == -1) L = i;\n                R = i;\n            }\n        }\n\n        if (mx != n - 1) {\n            cout << 0 << '\\n';\n            continue;\n        }\n\n        auto countWays = [&](int c) -> ll {\n            for (int i = 2; i <= c; ++i)\n                if (a[i - 1] > a[i]) return 0;\n            for (int i = c + 2; i < n; ++i)\n                if (a[i - 1] < a[i]) return 0;\n\n            int l = 1, r = n - 1;\n            int used = 0, last = 0;\n            ll ways = 1;\n            while (l <= c || r > c) {\n                if (l <= c && r > c && a[l] == a[r])\n                    return 0;\n\n                int x;\n                if (r <= c || (l <= c && a[l] < a[r]))\n                    x = a[l++];\n                else\n                    x = a[r--];\n\n                if (x == last) {\n                    int choices = x - used;\n                    if (choices <= 0) return 0;\n                    ways = ways * choices % MOD;\n                }\n                last = x;\n                ++used;\n            }\n            return ways;\n        };\n\n        cout << (countWays(L - 1) + countWays(R)) % MOD << '\\n';\n    }\n}\n"
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              "hints": [
                {
                  "order": 1,
                  "content": "Every cut has $n$ on exactly one side. So the smaller maximum is simply the maximum on the side **without $n$**. What does that make the cut values to the left and right of $n$?"
                },
                {
                  "order": 2,
                  "content": "A cut has value $n-1$ exactly when it separates $n$ from $n-1$. Therefore, all occurrences of $n-1$ must form one nonempty consecutive block. If that block is $a_L,\\ldots,a_R$, the positions of $n$ and $n-1$ must be $L$ and $R+1$, in some order. Only two possible positions for $n$!"
                },
                {
                  "order": 3,
                  "content": "Fix $p_{c+1}=n$. Then $a_1,\\ldots,a_c$ must be non-decreasing, and $a_{c+1},\\ldots,a_{n-1}$ must be non-increasing. Read the left part forward and the right part backward: both are now sequences of required running maxima. A value appearing in both sequences makes this candidate impossible, since both sides would need their own copy of it."
                },
                {
                  "order": 4,
                  "content": "Merge those two non-decreasing sequences, keeping the order within each. When a required maximum $x$ appears for the first time, the corresponding permutation position is forced to contain $x$. When $x$ repeats, that position can contain any unused value smaller than $x$."
                },
                {
                  "order": 5,
                  "content": "Let $k$ be the number of positions already filled during the merge, excluding $n$. At a repeated maximum $x$, all $k$ used values are at most $x$, and exactly one is $x$. Thus there are $(x-1)-(k-1)=x-k$ choices. Multiply these factors, rejecting nonpositive factors or equal values at the two merge fronts. Add the counts for $c=L-1$ and $c=R$; each merge is linear."
                }
              ],
              "editorial": "The big simplification is that **one side of every cut contains $n$**. Its maximum is already settled. The cut value is just the maximum on the other side.\n\nThat turns the problem into counting permutations with prescribed running maxima—and we only need to try two positions for $n$.\n\n## 1. Locate the two largest values\n\nFor any cut,\n\n$$a_i=n-1 \\iff \\text{the cut separates }n\\text{ and }n-1.$$\n\nIf they are on opposite sides, the two maxima are $n$ and $n-1$. If they are on the same side, the other side has maximum at most $n-2$.\n\nSuppose their positions, in increasing order, are $u<v$. Exactly the cuts\n\n$$u,u+1,\\ldots,v-1$$\n\nseparate them. Therefore:\n\n- No cut value can be $n$.\n- The value $n-1$ must appear.\n- Its occurrences must form one consecutive block.\n\nLet $L$ and $R$ be the first and last indices with $a_i=n-1$. Then the positions of the two largest values must be\n\n$$\\{\\operatorname{pos}(n),\\operatorname{pos}(n-1)\\}=\\{L,R+1\\}.$$\n\nSo $n$ is either at position $L$ or at position $R+1$. No need to try all $n$ positions.\n\nWe reject immediately if $\\max(a)\\ne n-1$. The checks below also reject a broken, nonconsecutive block of $n-1$ values.\n\n## 2. Fix the position of $n$\n\nWrite that position as $c+1$, so there are $c$ elements to its left. We will try\n\n$$c=L-1\\quad\\text{and}\\quad c=R.$$\n\nFor cuts before $n$, the side without $n$ is the prefix:\n\n$$a_i=\\max(p_1,\\ldots,p_i),\\qquad i\\le c.$$\n\nFor cuts after $n$, it is the suffix:\n\n$$a_i=\\max(p_{i+1},\\ldots,p_n),\\qquad i>c.$$\n\nThus we require:\n\n- $a_1,\\ldots,a_c$ to be non-decreasing;\n- $a_{c+1},\\ldots,a_{n-1}$ to be non-increasing.\n\nReject the candidate if either condition fails. Empty parts are fine: $n$ may sit at either end.\n\nNow read the permutation inward from both ends. The required running maxima are\n\n$$A=(a_1,\\ldots,a_c),\\qquad B=(a_{n-1},\\ldots,a_{c+1}).$$\n\nBoth sequences are non-decreasing. Each entry corresponds to one actual permutation position that we need to fill.\n\n**A value cannot appear in both sequences.** If $x$ is a running maximum on both sides, both sides must contain $x$. A permutation gets one copy, not a spare.\n\n## 3. Merge and count\n\nMerge $A$ and $B$ in non-decreasing order with two pointers. This is only an order for assigning positions; we are not rearranging the permutation positions themselves.\n\nIf the two current entries are equal, reject the candidate. Since both sequences are sorted, this detects every value shared by the two sides.\n\nLet the next required maximum be $x$, and let $k$ be the number of positions already filled, excluding the position containing $n$.\n\n### First occurrence of $x$\n\nThe running maximum on that side has just increased to $x$. The corresponding position must contain $x$, so there is exactly one choice.\n\nAlso, $x$ is still unused: every earlier assignment was at most its required maximum, and all earlier required maxima were smaller than $x$.\n\n### Repeated occurrence of $x$\n\nThe running maximum stays $x$, so this position must contain an unused value smaller than $x$.\n\nThere are $x-1$ values smaller than $x$. All $k$ previously assigned values are at most $x$, and exactly one of them is $x$. Hence $k-1$ smaller values have already been used, leaving\n\n$$(x-1)-(k-1)=x-k$$\n\nchoices.\n\nMultiply the candidate's count by $x-k$ modulo $998244353$. If this number is nonpositive, the candidate contributes zero.\n\nAfter either case, increment $k$.\n\nThe important detail is that this number of choices does not depend on *which* smaller values were chosen earlier. Their count is all that matters, because we process required maxima in sorted order. That's why a simple product works.\n\n## 4. Example\n\nTake $n=6$ and\n\n$$a=[3,3,5,5,5].$$\n\nThe block of $n-1=5$ spans cuts $3$ through $5$, so $6$ can only be at position $3$ or $6$.\n\nFor position $3$, the two sequences are $[3,3]$ and $[5,5,5]$. Their merge is\n\n$$[3,3,5,5,5].$$\n\nThe first $3$ and first $5$ are forced. The repeated entries contribute\n\n$$3-1=2,\\qquad 5-3=2,\\qquad 5-4=1.$$\n\nThis candidate contributes $2\\cdot2\\cdot1=4$. Position $6$ contributes another $4$, giving $8$.\n\n## Correctness argument\n\nEvery valid permutation places $n$ at $L$ or $R+1$, because the cuts with value $n-1$ are exactly those between the two largest values. Thus the algorithm considers every possible position of $n$.\n\nFor a fixed candidate, the cut conditions are exactly the running-maximum conditions of $A$ and $B$. The monotonicity checks are necessary, and a value shared by both sides is impossible because permutation values are distinct.\n\nDuring the merge, the algorithm preserves the order of positions within each side. Whenever a required maximum increases to $x$, assigning $x$ is necessary and available. Whenever it repeats, assigning any unused value smaller than $x$ is necessary and sufficient, and there are exactly $x-k$ such values. Therefore the product counts precisely all assignments satisfying the two running-maximum sequences.\n\nThese assignments fill $n-1$ positions with distinct values from $1$ through $n-1$. Together with the fixed $n$, they form a permutation with exactly the requested cut values. Each assignment is counted once, and the two candidates have different positions of $n$, so their counts do not overlap.\n\nHence their sum is the required answer.\n\n## Complexity and implementation\n\nFinding $L,R$, validating a candidate, and merging each take $O(n)$ time. There are only two candidates, so the total is **$O(n)$ time per test case** and **$O(\\sum n)$ overall**.\n\nThe input array uses **$O(n)$ space**; the merge itself needs only two pointers and a few counters. Use a 64-bit integer for modular multiplication. No sorting, factorials, or modular inverses are needed.",
              "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    constexpr ll MOD = 998244353;\n\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        vector<int> a(n);\n        int mx = 0, L = -1, R = -1;\n        for (int i = 1; i < n; ++i) {\n            cin >> a[i];\n            mx = max(mx, a[i]);\n            if (a[i] == n - 1) {\n                if (L == -1) L = i;\n                R = i;\n            }\n        }\n\n        if (mx != n - 1) {\n            cout << 0 << '\\n';\n            continue;\n        }\n\n        auto countWays = [&](int c) -> ll {\n            for (int i = 2; i <= c; ++i)\n                if (a[i - 1] > a[i]) return 0;\n            for (int i = c + 2; i < n; ++i)\n                if (a[i - 1] < a[i]) return 0;\n\n            int l = 1, r = n - 1;\n            int used = 0, last = 0;\n            ll ways = 1;\n            while (l <= c || r > c) {\n                if (l <= c && r > c && a[l] == a[r])\n                    return 0;\n\n                int x;\n                if (r <= c || (l <= c && a[l] < a[r]))\n                    x = a[l++];\n                else\n                    x = a[r--];\n\n                if (x == last) {\n                    int choices = x - used;\n                    if (choices <= 0) return 0;\n                    ways = ways * choices % MOD;\n                }\n                last = x;\n                ++used;\n            }\n            return ways;\n        };\n\n        cout << (countWays(L - 1) + countWays(R)) % MOD << '\\n';\n    }\n}\n"
            },
            "output": "Structured output captured successfully.",
            "metadata": {
              "valid": true
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            "title": "Structured Output",
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