{ "info": { "id": "ses_f4dad64c3ffepbEdQ7a3zGKrNi", "slug": "crisp-forest", "projectID": "global", "directory": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-qxeA8o", "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-qxeA8o", "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": 26762, "output": 1825, "reasoning": 14, "cache": { "read": 34816, "write": 0 } }, "time": { "created": 1789698153276, "updated": 1789698218974 } }, "messages": [ { "info": { "role": "user", "time": { "created": 1789698153293 }, "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_0b2529b4d001pko9ghPIYKwNGC", "sessionID": "ses_f4dad64c3ffepbEdQ7a3zGKrNi" }, "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 2249A: \"Rank Subsequence\" (rated 1300).\n\nKnown problem metadata:\n- Contest ID: 2249\n- Index: A\n- Name: Rank Subsequence\n- Rating: 1300 (Codeforces difficulty rating)\n- Tags: brute force, greedy, implementation\n- Problem URL: https://codeforces.com/contest/2249/problem/A\n\nUse the rating and tags as weak signals only. The supplied statement is the source of truth.\n\nProblem Statement:\n\n
A. Rank Subsequence
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given $$$n$$$ elements arranged in a line, numbered $$$1, 2, \\ldots, n$$$ from left to right.

You may delete any number of elements (possibly none). The remaining elements form a subsequence and keep their relative order. Let the length of this subsequence be $$$m$$$. If the element with original index $$$i$$$ becomes the $$$j$$$-th element of the subsequence ($$$1 \\le j \\le m$$$), define

For each element, you are given two integer segments $$$[l_i, r_i]$$$ and $$$[u_i, v_i]$$$. The $$$i$$$-th element is valid in a subsequence of length $$$m$$$ if and only if, when it occupies position $$$j$$$ in that subsequence, both of the following hold:

A subsequence is valid if every element remaining in it is valid.

Find the maximum possible length of a valid subsequence. The answer may be $$$0$$$.

Input

Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5000$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1\\le n\\le 5000$$$) — the number of elements.

Then $$$n$$$ lines follow, the $$$i$$$-th line containing the four integers $$$l_i$$$, $$$r_i$$$, $$$u_i$$$, and $$$v_i$$$ ($$$1\\le i\\le n$$$, $$$1\\le l_i\\le r_i\\le n$$$, $$$1\\le u_i\\le v_i\\le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5000$$$.

Output

For each test case, print one integer — the maximum possible length of a valid subsequence.

Example
Input
5
1
1 1 1 1
4
2 2 2 2
1 1 1 1
1 1 1 1
1 1 2 2
5
1 1 3 3
2 2 2 2
3 3 1 1
1 5 1 5
5 5 5 5
2
2 2 1 1
1 1 2 2
6
1 1 6 6
1 2 5 6
2 3 4 5
3 4 3 4
4 5 2 3
5 6 1 2
Output
0
4
3
2
3
Note

In the first test case, the only element cannot form a valid subsequence of length $$$1$$$, so the answer is $$$0$$$.

In the second test case, all $$$4$$$ elements can be kept. Their pairs of left and right ranks are $$$(1, 4)$$$, $$$(2, 3)$$$, $$$(3, 2)$$$, and $$$(4, 1)$$$, respectively, and all of them are valid.

In the third test case, one optimal choice is to keep the elements with original indices $$$2$$$, $$$3$$$, and $$$5$$$.

In the fourth test case, keeping both elements yields a valid subsequence of length $$$2$$$. Note that no valid subsequence of length $$$1$$$ exists.

In the fifth test case, one optimal choice is to keep the elements with original indices $$$3$$$, $$$4$$$, and $$$5$$$.

\n\n\n\nSource lookup status:\n\n- Tutorial (en): https://codeforces.com/blog/entry/155516 loaded (200 OK; title: \"Codeforces Round 1112 (Div. 1, Div. 2) Editorial - Codeforces\"; mentions 2249A)\n\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 \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_0b2529b4e001qQ09sXP4O9lT3g", "sessionID": "ses_f4dad64c3ffepbEdQ7a3zGKrNi", "messageID": "msg_0b2529b4d001pko9ghPIYKwNGC" } ] }, { "info": { "parentID": "msg_0b2529b4d001pko9ghPIYKwNGC", "role": "assistant", "mode": "nudge-generation", "agent": "nudge-generation", "variant": "low", "path": { "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-qxeA8o", "root": "/" }, "cost": 0, "tokens": { "total": 9563, "input": 9531, "output": 32, "reasoning": 0, "cache": { "write": 0, "read": 0 } }, "modelID": "gpt-6-astra", "providerID": "openai", "time": { "created": 1789698153428, "completed": 1789698159854 }, "finish": "tool-calls", "id": "msg_0b2529bd40019qzmmopzxS2SZ5", "sessionID": "ses_f4dad64c3ffepbEdQ7a3zGKrNi" }, "parts": [ { "type": "step-start", "id": "prt_0b252a499001NbYposy0HKbV2p", "sessionID": "ses_f4dad64c3ffepbEdQ7a3zGKrNi", "messageID": "msg_0b2529bd40019qzmmopzxS2SZ5" }, { "type": "tool", "tool": "webfetch", "callID": "call_oDRh2yxXPqTcTZMYebl6adDF", "state": { "status": "completed", "input": { "url": "https://codeforces.com/blog/entry/155516", "format": "markdown" }, "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 \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* [Rating](https://codeforces.com/ratings)\n* 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[kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") → [Codeforces Round 1120 (Div 1, Div 2) Editorial](https://codeforces.com/blog/entry/156688) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\")\n* [borgar02](https://codeforces.com/profile/borgar02 \"Master borgar02\") → [IATI 2022 editorial](https://codeforces.com/blog/entry/109545) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/64117/images/icons/comment-12x12.png \"New comment(s)\") ![Necropost](//codeforces.com/codeforces.org/s/64117/images/icons/hourglass.png \"Necropost\")\n* [Kerolloz](https://codeforces.com/profile/Kerolloz \"Newbie Kerolloz\") → [Codeforces JSON-Problem-Scraper API](https://codeforces.com/blog/entry/75074) ![Text created or updated](//codeforces.com/codeforces.org/s/64117/images/icons/x-update-12x12.png \"Text created or updated\") ![Necropost](//codeforces.com/codeforces.org/s/64117/images/icons/hourglass.png \"Necropost\")\n\n| | [Detailed →](https://codeforces.com/recent-actions) |\n| | --------------------------------------------------- |\n\n* [\\_MCYYDS\\_](https://codeforces.com/profile/%5FMCYYDS%5F)\n* [Blog](https://codeforces.com/blog/%5FMCYYDS%5F)\n* [Teams](https://codeforces.com/teams/with/%5FMCYYDS%5F)\n* [Submissions](https://codeforces.com/submissions/%5FMCYYDS%5F)\n* [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)$$$.\n\n**Code**\n\n```\n#include \nusing namespace std;\n\nvoid work() {\n\tint n; cin >> n;\n\tlong long L = 0, R = 1000000001LL;\n\n\tfor (int i = 1; i <= n; ++i) {\n\t\tlong long w; cin >> w;\n\t\tif (i & 1) R = min(R, w);\n\t\telse L = max(L, w);\n\t}\n\n\tcout << (n % 2 == 0 && L + 2 <= R ? \"YES\" : \"NO\") << '\\n';\n}\n\nint main() {\n\tios::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n\n\tint T; cin >> T;\n\twhile (T--) work();\n\n\treturn 0;\n}\n\n```\n\n## [2250B - String Construction](https://codeforces.com/contest/2250/problem/B \"Codeforces Round 1112 (Div. 2)\")\n\n* Idea:[\\_MCYYDS\\_](https://codeforces.com/profile/%5FMCYYDS%5F \"Grandmaster _MCYYDS_\")\n* Solution:[\\_MCYYDS\\_](https://codeforces.com/profile/%5FMCYYDS%5F \"Grandmaster _MCYYDS_\")\n\n**Hint 1**\n\nPartition the string into maximal blocks of equal characters. How does the number of blocks determine the number of adjacent equal pairs?\n\n**Solution**\n\nDefine a block as a maximal consecutive segment of equal characters. Suppose the string has $$$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)$$$.\n\n**Code**\n\n```\n#include \nusing namespace std;\n\nvoid work() {\n\tint n, k; cin >> n >> k;\n\n\tif (n > 1 && k == n - 1) {\n\t\tcout << -1 << '\\n';\n\t\treturn;\n\t}\n\n\tk = n - k;\n\tint c0 = (n + 1) / 2, c1 = n / 2;\n\n\tfor (int i = 1; i <= k; ++i) {\n\t\tif (i & 1) {\n\t\t\tif (i + 2 > k) {\n\t\t\t\twhile (c0--) cout << 0;\n\t\t\t}\n\t\t\telse {\n\t\t\t\t--c0;\n\t\t\t\tcout << 0;\n\t\t\t}\n\t\t}\n\t\telse {\n\t\t\tif (i + 2 > k) {\n\t\t\t\twhile (c1--) cout << 1;\n\t\t\t}\n\t\t\telse {\n\t\t\t\t--c1;\n\t\t\t\tcout << 1;\n\t\t\t}\n\t\t}\n\t}\n\tcout << '\\n';\n}\n\nint main() {\n\tios::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n\n\tint T; cin >> T;\n\twhile (T--) work();\n\n\treturn 0;\n}\n\n```\n\n## [2249A - Rank Subsequence](https://codeforces.com/contest/2249/problem/A \"Codeforces Round 1112 (Div. 1)\")\n\n* Idea:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n* Solution:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n\n**Hint 1**\n\nAn element's rank from the right depends on both its position in the retained subsequence and the subsequence's final length.\n\n**Hint 2**\n\nEnumerate the final retained length $$$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)$$$.\n\n**Code**\n\n```\n#include \nusing namespace std;\n\nconst int N = 5000;\n\nint n;\nint l[N + 10], r[N + 10], u[N + 10], v[N + 10];\n\nbool check(int m) {\n\tint j = 1;\n\n\tfor (int i = 1; i <= n && j <= m; ++i) {\n\t\tint x = m - j + 1;\n\t\tif ((j < l[i] || j > r[i]) && (x < u[i] || x > v[i])) ++j;\n\t}\n\n\treturn j == m + 1;\n}\n\nvoid work() {\n\tcin >> n;\n\tfor (int i = 1; i <= n; ++i) cin >> l[i] >> r[i] >> u[i] >> v[i];\n\n\tfor (int m = n; m >= 1; --m) {\n\t\tif (!check(m)) continue;\n\t\tcout << m << '\\n';\n\t\treturn;\n\t}\n\n\tcout << 0 << '\\n';\n}\n\nint main() {\n\tios::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n\n\tint T; cin >> T;\n\twhile (T--) work();\n\n\treturn 0;\n}\n\n```\n\n## [2249B - Permutation Cuts](https://codeforces.com/contest/2249/problem/B \"Codeforces Round 1112 (Div. 1)\")\n\n* Idea:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n* Solution:[aaa\\_Pigeon2](https://codeforces.com/profile/aaa%5FPigeon2 \"International Grandmaster aaa_Pigeon2\")\n\n**Hint 1**\n\nThe value $$$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 \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 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 \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)$$$.\n\n**Code**\n\n```\n#include \nusing namespace std;\n\nconst int N = 2500;\n\nint n, x;\nint vis[N + 10], in_o[N + 10], e[N + 10], o[N + 10], a[N + 10];\npair p[N + 10];\n\nbool check(int v) {\n\treturn (v & (v - 1)) == 0;\n}\n\nvoid work() {\n\tcin >> n >> x;\n\n\tif (!check(n) || (x && n < 4)) {\n\t\tcout << -1 << '\\n';\n\t\treturn;\n\t}\n\n\tif (!x) {\n\t\tfor (int i = 0; i < n; ++i) {\n\t\t\tfor (int j = 0; j < n; ++j) {\n\t\t\t\tcout << (i ^ j) << (j + 1 == n ? '\\n' : ' ');\n\t\t\t}\n\t\t}\n\t\treturn;\n\t}\n\n\tfor (int i = 0; i <= n; ++i) vis[i] = in_o[i] = e[i] = o[i] = a[i] = 0;\n\n\tint cnt = 0;\n\tfor (int i = 0; i < n; ++i) {\n\t\tif (vis[i]) continue;\n\t\tint j = i ^ x;\n\t\tvis[i] = vis[j] = 1;\n\t\tp[++cnt] = {i, j};\n\t}\n\n\tfor (int i = 1; i <= n / 4; ++i) in_o[p[i].first] = in_o[p[i].second] = 1;\n\n\tint ec = 0, oc = 0;\n\tfor (int i = 0; i < n; ++i) {\n\t\tif (in_o[i]) o[++oc] = i;\n\t\telse e[++ec] = i;\n\t}\n\n\tec = oc = 0;\n\tfor (int i = 0; i < n; ++i) {\n\t\tif (i & 1) a[i] = o[++oc];\n\t\telse a[i] = e[++ec];\n\t}\n\n\tfor (int i = 0; i < n; ++i) {\n\t\tfor (int j = 0; j < n; ++j) {\n\t\t\tint v = a[i] ^ a[j];\n\t\t\tif ((i & 1) && (j & 1)) v ^= x;\n\t\t\tcout << v << (j + 1 == n ? '\\n' : ' ');\n\t\t}\n\t}\n}\n\nint main() {\n\tios::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n\n\tint T; cin >> T;\n\twhile (T--) work();\n\n\treturn 0;\n}\n\n```\n\n## [2249E1 - String (Easy Version)](https://codeforces.com/contest/2249/problem/E1 \"Codeforces Round 1112 (Div. 1)\")\n\n* Idea:[131241](https://codeforces.com/profile/131241 \"Specialist 131241\")\n* Solution:[131241](https://codeforces.com/profile/131241 \"Specialist 131241\")\n\n**Hint 1**\n\nTreat a prefix of length $$$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 \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}$$$.\n\n**Code**\n\n```\n#include\n#define int long long\nusing namespace std;\nint lg;\nint q,L,R,n,k,d[10000005],t[100],b[100];\nchar c[10000005],s[10000005];\nstruct node{\n\tint f[65][65];\n}sum[65][65];\nnode merge(node a,node b,int v=1){\n\tfor(int i=0;i=0;i--){\n\t\tint j=sp[i];\n\t\tif(!j) continue;\n\t\tans+=sum[i][j].f[(x+(k-1)*cnt)%k][(y+(k-1)*cnt)%k];\n\t\tif(xnow){\n\t\t\tint px=(cnt-1+lowbit*(k-1))%k,py=cnt%k;\n\t\t\tif(px==x&&py==y){\n\t\t\t\tans++;\n\t\t\t}\n\t\t}\n\t\txnow+=b[i]*j;\n\t\tcnt+=j;\n\t\tlowbit=i;\n\t}\n\treturn ans;\n}\nint gs(int r,int x){\n // Count occurrences of the pattern \"x\" in the range [0, r].\n\tr++;\n\tint sum=r/k,pl=r/k*k,y=calc(pl);\n\twhile(pl=k?y-=k:0;\n\t}\n\treturn sum;\n}\nvoid print(char c){\n\tif(c<10){\n\t\tcout<r){\n\t\treturn ;\n\t}\n\tfor(int i=0;i1){\n // If more than one residue occurs, the pattern cannot appear.\n\t\treturn ;\n\t}\n\tint m=0,pl=k,pr=k,p=0;\n\tif(cnt){\n // At least one adjacent difference is not 1.\n\t\tfor(int i=1;i=n) pr=n-i;\n\t\t\ts[m++]=c[i];\n\t\t}\n\n // Handle the two partial boundary blocks separately.\n\t\tint bl=(r/k)*k,br=(l/k+1)*k-1;\n\t\tif(br-l+1k) continue;\n\t\t\tif(szl>k>>q;\n\n // sum[i][j] stores counts of length-2 substrings in the first j * k^i characters.\n\tb[0]=1;lg=0;\n\tfor(int i=1;i<=k;i++){\n\t\tfor(int j=1;j1e17){\n\t\t\tbreak;\n\t\t}\n\t\tsum[p][1]=sum[p-1][k];\n // Concatenate the k shifted blocks.\n\t\tfor(int i=2;i<=k;i++){\n\t\t\tsum[p][i]=merge(sum[p][i-1],sum[p-1][k],i-1);\n\t\t\t// Count substrings that cross a block boundary directly.\n\t\t\tsum[p][i].f[calc(b[p]*(i-1)-1)][calc(b[p]*(i-1))]++;\n\t\t}\n\t}\n\n\twhile(q--){\n\t\tcin>>L>>R>>n;\n\t\tcin>>c;\n\t\tfor(int i=0;ir_i)\n\\quad\\text{and}\\quad\n(m-j+1v_i).\n$$\n\nBoth ranks are now known. Whether this element fits depends only on $i$, $j$, and the fixed $m$, not on which elements fill the other positions.\n\n### Greedy check for one length\n\nScan the original elements from left to right. Let $k$ be the number already selected, initially $0$.\n\nThe next selected element would have:\n\n- left rank $k+1$;\n- right rank $m-k$.\n\nIf the current element allows both ranks, select it and increment $k$. Otherwise, skip it. Stop successfully as soon as $k=m$. If the scan ends earlier, this target length is impossible.\n\nThe rule is simply: **fill each position with the earliest element that fits.** Choosing an earlier element leaves at least as much room for everything that comes next.\n\nImportantly, compute right ranks using the **target length $m$**, not the length of the partially built subsequence. We are checking whether that partial choice can become a valid length-$m$ subsequence.\n\n### Why the greedy check works\n\nSuppose a valid length-$m$ subsequence exists, with original indices\n\n$$s_1s_{j-1}\\ge g_{j-1},$$\n\nelement $s_j$ is still available after the previous greedy choice. It is eligible for position $j$, since its ranks are still $j$ and $m-j+1$. Therefore greedy finds an eligible element no later than $s_j$, giving $g_j\\le s_j$.\n\nBy induction, greedy selects all $m$ elements whenever a valid subsequence of that length exists. Conversely, when greedy selects $m$ elements, they are in original order and each passed both rank checks, so the resulting subsequence is valid.\n\nThus the check succeeds **if and only if** length $m$ is feasible.\n\n### Find the maximum\n\nTry target lengths in descending order:\n\n$$n,n-1,\\ldots,1.$$\n\nThe first successful check gives the maximum possible length. If every positive length fails, output $0$: deleting everything is always allowed.\n\nDon't binary-search the answer. Feasibility is **not monotonic**: deleting an element changes the ranks of other elements and can make them invalid. The fourth sample makes this concrete—length $2$ works, but length $1$ does not. Smaller does not automatically mean easier here.\n\n### Complexity\n\nEach target length takes $O(n)$ time, and there are at most $n$ targets:\n\n- **Time:** $O(n^2)$ per test case.\n- **Space:** $O(n)$ to store the intervals, with $O(1)$ additional space for each check.\n\nAcross all test cases, $\\sum n\\le 5000$, so\n\n$$\\sum n^2\\le \\left(\\sum n\\right)^2\\le 25{,}000{,}000.$$\n\nThat comfortably fits the limits. All ranks and endpoints fit in `int`; remember that the forbidden intervals include both endpoints.", "solution": "#include \nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n}\n\nint main() {\n setIO();\n\n int t;\n cin >> t;\n while (t--) {\n int n;\n cin >> n;\n vector> a(n);\n for (auto& [l, r, u, v] : a) {\n cin >> l >> r >> u >> v;\n }\n\n auto feasible = [&](int m) {\n int k = 0;\n for (const auto& [l, r, u, v] : a) {\n int left = k + 1;\n int right = m - k;\n if ((left < l || left > r) &&\n (right < u || right > v)) {\n if (++k == m) return true;\n }\n }\n return false;\n };\n\n int answer = 0;\n for (int m = n; m >= 1; --m) {\n if (feasible(m)) {\n answer = m;\n break;\n }\n }\n cout << answer << '\\n';\n }\n return 0;\n}" }, "id": "msg_0b252dcdc0019meVUBHN4QAzJu", "sessionID": "ses_f4dad64c3ffepbEdQ7a3zGKrNi" }, "parts": [ { "type": "step-start", "id": "prt_0b252e7f5001uk52dzcKW36Jm3", "sessionID": "ses_f4dad64c3ffepbEdQ7a3zGKrNi", "messageID": "msg_0b252dcdc0019meVUBHN4QAzJu" }, { "type": "tool", "tool": "StructuredOutput", "callID": "call_ANUDBpoOWVtHYBfZ0jgPox6T", "state": { "status": "completed", "input": { "status": "success", "reason": null, "hints": [ { "order": 1, "content": "The right rank depends on the final subsequence length, which makes choosing elements immediately awkward. What becomes simpler if you fix that length first?" }, { "order": 2, "content": "Try every target length $m$. Once $m$ is fixed, an element placed at position $j$ has known ranks $j$ and $m-j+1$, so you can check whether it fits that position in constant time." }, { "order": 3, "content": "Build a subsequence of the chosen length from left to right. For its next position $j$, take the earliest remaining element that satisfies both restrictions. Could choosing a later eligible element ever leave you more options afterward?" }, { "order": 4, "content": "The greedy choice is safe: replace the element at position $j$ in a feasible completion with the earlier eligible element. Its ranks stay the same, and every later element of that completion is still available. Thus one scan is enough to test a fixed $m$." }, { "order": 5, "content": "Test $m=n,n-1,\\ldots,1$, returning the first length whose greedy scan selects $m$ elements; otherwise return $0$. This takes $O(n^2)$ time. Don't binary-search: the fourth sample has a valid subsequence of length $2$ but none of length $1$, so feasibility is not monotonic." } ], "editorial": "The annoying part is the right rank: it depends on how many elements we will keep in total. **Fix that total first**, and the problem becomes a simple greedy scan.\n\n### Fix the target length\n\nSuppose we want a subsequence of exactly $m$ elements. If original element $i$ occupies position $j$, it must satisfy\n\n$$\n(jr_i)\n\\quad\\text{and}\\quad\n(m-j+1v_i).\n$$\n\nBoth ranks are now known. Whether this element fits depends only on $i$, $j$, and the fixed $m$, not on which elements fill the other positions.\n\n### Greedy check for one length\n\nScan the original elements from left to right. Let $k$ be the number already selected, initially $0$.\n\nThe next selected element would have:\n\n- left rank $k+1$;\n- right rank $m-k$.\n\nIf the current element allows both ranks, select it and increment $k$. Otherwise, skip it. Stop successfully as soon as $k=m$. If the scan ends earlier, this target length is impossible.\n\nThe rule is simply: **fill each position with the earliest element that fits.** Choosing an earlier element leaves at least as much room for everything that comes next.\n\nImportantly, compute right ranks using the **target length $m$**, not the length of the partially built subsequence. We are checking whether that partial choice can become a valid length-$m$ subsequence.\n\n### Why the greedy check works\n\nSuppose a valid length-$m$ subsequence exists, with original indices\n\n$$s_1s_{j-1}\\ge g_{j-1},$$\n\nelement $s_j$ is still available after the previous greedy choice. It is eligible for position $j$, since its ranks are still $j$ and $m-j+1$. Therefore greedy finds an eligible element no later than $s_j$, giving $g_j\\le s_j$.\n\nBy induction, greedy selects all $m$ elements whenever a valid subsequence of that length exists. Conversely, when greedy selects $m$ elements, they are in original order and each passed both rank checks, so the resulting subsequence is valid.\n\nThus the check succeeds **if and only if** length $m$ is feasible.\n\n### Find the maximum\n\nTry target lengths in descending order:\n\n$$n,n-1,\\ldots,1.$$\n\nThe first successful check gives the maximum possible length. If every positive length fails, output $0$: deleting everything is always allowed.\n\nDon't binary-search the answer. Feasibility is **not monotonic**: deleting an element changes the ranks of other elements and can make them invalid. The fourth sample makes this concrete—length $2$ works, but length $1$ does not. Smaller does not automatically mean easier here.\n\n### Complexity\n\nEach target length takes $O(n)$ time, and there are at most $n$ targets:\n\n- **Time:** $O(n^2)$ per test case.\n- **Space:** $O(n)$ to store the intervals, with $O(1)$ additional space for each check.\n\nAcross all test cases, $\\sum n\\le 5000$, so\n\n$$\\sum n^2\\le \\left(\\sum n\\right)^2\\le 25{,}000{,}000.$$\n\nThat comfortably fits the limits. All ranks and endpoints fit in `int`; remember that the forbidden intervals include both endpoints.", "solution": "#include \nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n}\n\nint main() {\n setIO();\n\n int t;\n cin >> t;\n while (t--) {\n int n;\n cin >> n;\n vector> a(n);\n for (auto& [l, r, u, v] : a) {\n cin >> l >> r >> u >> v;\n }\n\n auto feasible = [&](int m) {\n int k = 0;\n for (const auto& [l, r, u, v] : a) {\n int left = k + 1;\n int right = m - k;\n if ((left < l || left > r) &&\n (right < u || right > v)) {\n if (++k == m) return true;\n }\n }\n return false;\n };\n\n int answer = 0;\n for (int m = n; m >= 1; --m) {\n if (feasible(m)) {\n answer = m;\n break;\n }\n }\n cout << answer << '\\n';\n }\n return 0;\n}" }, "output": "Structured output captured successfully.", "metadata": { "valid": true }, "title": "Structured Output", "time": { "start": 1789698218808, "end": 1789698218811 } }, "metadata": { "openai": { "itemId": "fc_0384f573279516b0016aaca07ce08c87d0a8ad3b156c9413ec" } }, "id": "prt_0b252e7f8001klbSBx2YOAwpRL", "sessionID": "ses_f4dad64c3ffepbEdQ7a3zGKrNi", "messageID": "msg_0b252dcdc0019meVUBHN4QAzJu" }, { "reason": "tool-calls", "type": "step-finish", "tokens": { "total": 27708, "input": 571, "output": 1537, "reasoning": 0, "cache": { "write": 0, "read": 25600 } }, "cost": 0, "id": "prt_0b2539bd70011J3TyzDs9vcdEH", "sessionID": "ses_f4dad64c3ffepbEdQ7a3zGKrNi", "messageID": "msg_0b252dcdc0019meVUBHN4QAzJu" } ] } ] }