{ "info": { "id": "ses_f534230b3ffegKs0rWwiBFYJC4", "slug": "nimble-lagoon", "projectID": "global", "directory": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IGBdqu", "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IGBdqu", "title": "Nudge content generation", "agent": "nudge-generation", "model": { "id": "gpt-5.6-luna", "providerID": "openai", "variant": "max" }, "version": "1.18.30", "summary": { "additions": 0, "deletions": 0, "files": 0 }, "cost": 0, "tokens": { "input": 20055, "output": 1661, "reasoning": 5617, "cache": { "read": 39040, "write": 0 } }, "time": { "created": 1789604515660, "updated": 1789604806323 } }, "messages": [ { "info": { "role": "user", "time": { "created": 1789604515678 }, "agent": "nudge-generation", "model": { "providerID": "openai", "modelID": "gpt-5.6-luna", "variant": "max" }, "system": "You create Codeforces learning content for Nudge. Your job is to understand the best approach for the problem and teach it clearly in your own words, using existing solutions to inform your reasoning.\n\nStart with a brief, targeted web search for existing solutions so relevant editorials are in context before substantial independent problem-solving. Use any supplied tutorial links first; otherwise search by contest ID and problem index. Prefer the official tutorial/editorial, and read relevant explanations rather than relying on search snippets. Look at accessible alternatives when useful, especially if the official explanation is missing, unclear, or unnecessarily complicated. Accepted submissions can also help clarify an approach. Respect supplied source lookup statuses and avoid repeatedly retrying blocked sources. If no usable source is accessible, proceed from the supplied statement.\n\nTreat existing solutions as a starting point, not a constraint. If an approach is clear and suitable, build on it without unnecessary re-derivation. If an explanation has gaps, seems incorrect, or uses a harder method than necessary, reason through it and derive a better or simpler approach when worthwhile. Choose a correct approach that is easy to teach and fits the constraints. Verify it against the supplied statement and edge cases before writing the hints, editorial, and C++ solution.\n\nThe editorial must be self-contained: explain the key insight, algorithm, why it works, complexity, and implementation details a learner needs, filling in steps the source skips. Keep research and verification concise; spend the output on teaching the solution, not narrating the research process. Follow the user prompt and output schema exactly.", "format": { "type": "json_schema", "schema": { "type": "object", "properties": { "status": { "type": "string", "enum": [ "success", "unsolvable" ], "description": "Set to 'success' when the supplied statement defines a solvable programming problem. Set to 'unsolvable' only if the statement is fundamentally incomplete, contradictory, or depends on an inaccessible required resource." }, "reason": { "anyOf": [ { "type": "string", "minLength": 1, "maxLength": 2000 }, { "type": "null" } ], "description": "If status is 'unsolvable', provide a detailed failure message to show users. Name the concrete statement/resource blocker and any relevant source lookup/access status, such as missing official tutorial, 403/Cloudflare challenge, 404, or unavailable accepted submissions. Otherwise null." }, "hints": { "anyOf": [ { "type": "array", "minItems": 5, "maxItems": 5, "description": "Exactly 5 progressive hints, each building on the last.", "items": { "type": "object", "properties": { "order": { "type": "integer", "minimum": 1, "maximum": 5, "description": "Hint number, 1 through 5." }, "content": { "type": "string", "minLength": 1, "maxLength": 10000, "description": "Markdown hint text." } }, "required": [ "order", "content" ], "additionalProperties": false } }, { "type": "null" } ], "description": "If status is 'success', provide exactly 5 progressive hints. Otherwise null." }, "editorial": { "anyOf": [ { "type": "string", "minLength": 1, "maxLength": 100000 }, { "type": "null" } ], "description": "If status is 'success', a prose editorial explaining the solution in Nudge's own voice, with no links, citations, or source/research notes. Otherwise null." }, "solution": { "anyOf": [ { "type": "string", "minLength": 1, "maxLength": 200000 }, { "type": "null" } ], "description": "If status is 'success', a complete C++ solution. Otherwise null." } }, "required": [ "status", "reason", "hints", "editorial", "solution" ], "additionalProperties": false, "title": "problem_response", "description": "Submit the generated content for the problem, or report if it is unsolvable." }, "retryCount": 2 }, "id": "msg_0acbdcf5e001j3NKztZRAhSbbO", "sessionID": "ses_f534230b3ffegKs0rWwiBFYJC4" }, "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 2263B: \"Min Matrices\".\n\nKnown problem metadata:\n- Contest ID: 2263\n- Index: B\n- Name: Min Matrices\n- Rating: unknown / unrated\n- Tags: constructive algorithms\n- Problem URL: https://codeforces.com/contest/2263/problem/B\n\nUse the rating and tags as weak signals only. The supplied statement is the source of truth.\n\nProblem Statement:\n\n
B. Min Matrices
time limit per test
1.5 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Farmer John heard from Elsie that Bessie's favorite number is $$$k$$$, so he wants to surprise her with a handcrafted present.

For a 2D matrix $$$B$$$, let $$$f(B)$$$ denote the set of the minimum element of each row and each column of $$$B$$$.

Farmer John wants you to show him an $$$n \\times n$$$ matrix $$$A$$$ consisting of each number from $$$1$$$ to $$$n^2$$$ exactly once such that $$$|f(A)| = k$$$, or state that it is impossible.

Input

Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \\le n \\le 1000, 0 \\le k \\le 2n$$$) — the size of the matrix and the goal value of $$$|f(A)|$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$1000$$$.

Output

If there is no such matrix, print $$$-1$$$. Otherwise, print $$$n$$$ lines with $$$n$$$ integers each — an $$$n \\times n$$$ matrix that satisfies the conditions of the problem.

If there are multiple solutions, you may output any of them.

Example
Input
5
3 0
3 5
5 5
4 3
1 1
Output
-1\n8 5 9\n6 3 7\n2 1 4\n16 14 17 15 3\n25 22 5 23 24\n8 1 9 6 7\n4 18 21 19 20\n12 10 13 2 11\n-1\n1
Note

In the first test case, we can see that it is impossible to construct such a $$$3 \\times 3$$$ matrix where $$$f(A)$$$ is empty.

In the second test case, we can see the minimums of the rows are $$$[5, 3, 1]$$$ respectively, and the minimums of the columns are $$$[2, 1, 4]$$$ respectively. Therefore, $$$f(A) = \\{1, 2, 3, 4, 5\\}$$$, so $$$|f(A)| = 5$$$ as desired.

\n\n\n\nSource lookup status:\n\n- Tutorial (en): https://codeforces.com/blog/entry/156688 loaded (200 OK; title: \"Codeforces Round 1120 (Div 1, Div 2) Editorial - Codeforces\"; mentions 2263B)\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- If the problem is solvable, return `status: \"success\"`, `reason: null`, and fill in `hints`, `editorial`, and `solution`.\n- Return `status: \"unsolvable\"`, a short `reason`, and set `hints`, `editorial`, and `solution` to null only when the supplied statement is fundamentally incomplete, contradictory, or dependent on an inaccessible resource needed to define the task. Missing research, high rating, uncertainty, or \"can't guarantee AC\" is not a reason to return unsolvable.\n- If you return `status: \"unsolvable\"`, the `reason` is shown directly to users. Be concrete: say what exact statement/resource blocker stopped you, and include relevant source lookup status if it mattered (for example: official tutorial missing, Codeforces returned 403/Cloudflare challenge, tutorial page 404, accepted submissions unavailable).\n- Each hint must be JUST the hint text. No \"Hint 1:\" or subtitles. The UI adds those automatically.\n- Do not start the editorial with an \"# Editorial\" heading. The UI already adds that section. Feel free to add other headers as needed though.\n- Use styling in the editorial if needed! Just that first header is no good but any subheaders or bolding or whatever is good!\n\nFor the C++ solution, you MUST use this template and work around it:\n\n```cpp\n#include \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_0acbdcf5f0018yR7SvC8755BCU", "sessionID": "ses_f534230b3ffegKs0rWwiBFYJC4", "messageID": "msg_0acbdcf5e001j3NKztZRAhSbbO" } ] }, { "info": { "parentID": "msg_0acbdcf5e001j3NKztZRAhSbbO", "role": "assistant", "mode": "nudge-generation", "agent": "nudge-generation", "variant": "max", "path": { "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IGBdqu", "root": "/" }, "cost": 0, "tokens": { "total": 9580, "input": 2931, "output": 38, "reasoning": 83, "cache": { "write": 0, "read": 6528 } }, "modelID": "gpt-5.6-luna", "providerID": "openai", "time": { "created": 1789604515816, "completed": 1789604522232 }, "finish": "tool-calls", "id": "msg_0acbdcfe8001KZCdyxqm41kBVm", "sessionID": "ses_f534230b3ffegKs0rWwiBFYJC4" }, "parts": [ { "type": "step-start", "id": "prt_0acbdd964001nZR1RpkJu1XnAD", "sessionID": "ses_f534230b3ffegKs0rWwiBFYJC4", "messageID": "msg_0acbdcfe8001KZCdyxqm41kBVm" }, { "type": "reasoning", "text": "**Planning targeted web research**", "time": { "start": 1789604518246, "end": 1789604520226 }, "metadata": { "openai": { "itemId": "rs_0010c4e84f3238f6016aab32a62a3487d1a95ccf35c9bd20d5", "reasoningEncryptedContent": "gAAAAABqqzKoI3hmpbTD3VdrQXYmjlOQPidU4wN9aOep8dkE7K_8aOBj1CCeeKA1Xu5qKdvfhUo2J6QsBLLJcnwxvxw2S_v85plHIvOJjTmonqPgNqZ_46SS00FSNCbtzya6bKPUdN04A3NHlCscfXaScguX-_XkPY7owcXkABHbZXSnwD9Ihqh9IJsDps4tmhjrkiJnM3z0jys-qfTEYJiW8_GDlYRnsYEHWUeqm5uD3rjKH084vGNDpE2MC-NWIXbrqR6OIdHLkDHzLrQsKCD5GNFnIjAexFDa5KrLFzRBNd6Zp-y2J-uMZV20wFBuS1fFUdJxMEkAIh0KyTHXKUeJtvKSxaK1Zw86y4-pECDGyuPqwBTtDTEna2fYiYCSwlfOVkin8hAkfeMBsk2n2ruJw2XAWU-ZK9w_ig-VMYOlpGvLN_Y_wlO9926ttHJjYqL8fEkBSkDvafiAKy0HnQSehZx6RhLwpo-CWDnkk04RZ349lhuYyjShLmq9grfiG2btfAZwRGyfzXSzXu02uuJbSyrXlLnbkzcEtPGHs8e8cyuihzUlCNuu9GGjS1or-XUoD5SHI5mIcvyHjeB1KF-l8wzNXdKYryO1UgC-P-9YoMwuz8Y5GLng54QI3UXmv2eUk0qs5_IJ8JCaKiDqkCdVDx2gu4JAUUqUdULdWqgixb0kowDMRgge4kAU3oD2q7Ekm-qgj-d99DCbvJ-r91coVeKwO__dkCj0L3Q2-Ztt5ipR3Eu2cnwh8wZe2fiZU0oOxeDVHQxxecy3lfO3VZvBFxvo5RCZhy5u9TOWi9l9MtERUCYqjGeQUMZcotr6x4_yt9PaNIborCO5akkZn0_5EkLgzSSlaoS5IzLedQZslo_ppQnSrHDGYtnaTMPlP0I-s9a9f_2vhapWVRnThTRnCKO7sR4qo4ThiyyUTyIZ1SM8APNQY20fZkxvFr0KjMYrXs_ft5dqtieH4ZCwkpFjKSyLNZTWw7et16ikhz7vHoa9afSL7NW6e3emFs-N58iMDIX4_AHsK6USy51jdTyhW8RAWzomAWtdynQUm4IjpJ0owHrN7L4WwzG76XTgJDN-OUB7zCYlknY2R_y41UC_9NcmtCt-kVd191f9dwVfmiYOautJPW3ks1oe2EiXOQFeCy6obBiHD2RQKW4QrO3T1bSxumDSVkQMUQW3ZhZDDAY7Hssqp63Lf_HjS6bNED4z9nFuIqX8sCS7y9uhPvo1c7d4LSdpCPrOuOT-h8PfngqbmmL_gdXGVGnStFPef6Zaws9wxGkPO2gd05WtxWgQCnDYVEVTVwH1tmoAJMqQApw1_uHD5jHGlJe1HDyDIVyufNMwunMTaUly-Rf8g0BVR5XD6zf0MFH2pEGcp6QR-_f1tOIpB4Y29cJrc4TiW278pRoUeJZkQy0yL7DTOTrbgucc7XX0rBKN5jLRirBziUgv2n1UL8alzWZx5V_cfr3uekPRykW45ECfAAUbBGQ51Y99Hodt_vn8UW1wyr0Saf2gZPhgVL3tENzsix-FsD4uVTtHIgL-VQ8CS2IDEZVANUlc6Sxyk7zsaNVRedWXTj8L6DNX3WXUnIeIeDsD4wUK5WKhx_8SWGNGf0LpaHoIySOkvHqGH8RCiZFK9w6BK7HsBOhTIX3sRKURZ7rB0LppqO1UEIpWOzQnn0Kr3ZG15DZDFG6nSdEHZtyJzkhTcDqQEGLMabcqBfjEEy6RGKOqyZT4gIyaBwZiXUhrFFnwuhCOeNQo5qrelzMnAo-fdSdBYIrzIUKbNc0Woz-U_XkvDYWDEKDu1tiaGoWhDl7b_duXRlqQnmu6lCztSLreRXq27cvboXuLKzRx5njKAwyz16Wzp4R2UQboV8Pyi1dfJxVTiUqegQ==" } }, "id": "prt_0acbdd966001JDmkUHHNNPxjEr", "sessionID": "ses_f534230b3ffegKs0rWwiBFYJC4", "messageID": "msg_0acbdcfe8001KZCdyxqm41kBVm" }, { "type": "tool", "tool": "webfetch", "callID": "call_KSUxYoRQaurQlBuVOltZbPkU", "state": { "status": "completed", "input": { "url": "https://codeforces.com/blog/entry/156688", "format": "markdown", "timeout": 30 }, "output": "---\nimage: https://codeforces.com/codeforces.org/s/30864/images/codeforces-telegram-square-1024x1024.png\ntitle: Codeforces Round 1120 (Div 1, Div 2) Editorial - Codeforces\ndescription: Codeforces. 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lumibons\") → [IOI 2027 Call for Tasks](https://codeforces.com/blog/entry/156793) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/30864/images/icons/comment-12x12.png \"New comment(s)\")\n* [OmarAfterMidnight](https://codeforces.com/profile/OmarAfterMidnight \"Pupil OmarAfterMidnight\") → [VoidHub Update: Relive CPC Contests as If You Were There](https://codeforces.com/blog/entry/156807) ![Text created or updated](//codeforces.com/codeforces.org/s/30864/images/icons/x-update-12x12.png \"Text created or updated\")\n* [Grizoo](https://codeforces.com/profile/Grizoo \"Candidate Master Grizoo\") → [\\[GYM\\] Aleppo Collegiate Programming Contest 2026](https://codeforces.com/blog/entry/156707) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/30864/images/icons/comment-12x12.png \"New comment(s)\")\n* [pllk](https://codeforces.com/profile/pllk \"Grandmaster pllk\") → [Competitive Programmer's Handbook — a new book on competitive 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\"Newbie maruf004\") → [Will Meta Hacker Cup 2026 Happen?](https://codeforces.com/blog/entry/156789) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/30864/images/icons/comment-12x12.png \"New comment(s)\")\n\n| | [Detailed →](https://codeforces.com/recent-actions) |\n| | --------------------------------------------------- |\n\n* [kondasujay2](https://codeforces.com/profile/kondasujay2)\n* [Blog](https://codeforces.com/blog/kondasujay2)\n* [Teams](https://codeforces.com/teams/with/kondasujay2)\n* [Submissions](https://codeforces.com/submissions/kondasujay2)\n* [Groups](https://codeforces.com/groups/with/kondasujay2)\n* [Contests](https://codeforces.com/contests/with/kondasujay2)\n* [Problemsetting](https://codeforces.com/contests/writer/kondasujay2)\n\n### [kondasujay2's blog](https://codeforces.com/blog/kondasujay2)\n\n[ Codeforces Round 1120 (Div 1, Div 2) Editorial ](https://codeforces.com/blog/entry/156688) \n\nBy [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\"), [history](https://codeforces.com/topic/157332/en14), 4 days ago, ![In English](//codeforces.com/codeforces.org/s/30864/images/flags/24/gb.png \"In English\") \n\nThank you to everyone who participated! \n\nWe are sorry for the poor testcases on C1, along with C1 and C2's solutions being so similar to the recent Div3 E. \n\n[2263A - Min Max Game](https://codeforces.com/contest/2263/problem/A \"Codeforces Round 1120 (Div. 2)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include \nusing namespace std;\n\nvoid tc() {\n int n; cin >> n;\n int cnt = 0;\n vector a(n);\n for(int i = 0; i < n; i++) {\n cin >> a[i];\n cnt += a[i];\n }\n if(cnt >= n - cnt) {\n cout << \"Bessie\\n\";\n } else {\n cout << \"Elsie\\n\";\n }\n}\n\nint main() {\n ios::sync_with_stdio(false), cin.tie(nullptr);\n int t; cin >> t;\n while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263B - Min Matrices](https://codeforces.com/contest/2263/problem/B \"Codeforces Round 1120 (Div. 2)\")\n\nAuthor: [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include\n \nusing namespace std;\n \nvoid solve() {\n int n, k; cin >> n >> k;\n int x = k-n+1;\n if(!(1 <= x && x <= n)) { cout << -1; return; }\n vector> ans(n, vector(n));\n ans[0][0] = 1;\n int on = 2;\n for(int i = 1; i < x; i++) {\n ans[0][i] = on; on++;\n ans[i][0] = on; on++;\n }\n for(int i = x; i < n; i++) {\n ans[i][i] = on; on++;\n }\n\n for(int i = 0; i < n; i++) {\n for(int j = 0; j < n; j++) {\n if(!ans[i][j]) {\n ans[i][j] = on; on++;\n }\n }\n }\n\n for(int j = 0; j < n; j++) {\n if(j) cout << \"\\n\";\n for(auto &i : ans[j]) cout << i << \" \";\n }\n}\n\nint main() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n int t; cin >> t;\n while(t--) {\n solve();\n if(t) cout << \"\\n\";\n }\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263C1 - Floor of MEX (Easy Version)](https://codeforces.com/contest/2263/problem/C1 \"Codeforces Round 1120 (Div. 2)\") / [2262A1 - Floor of MEX (Easy Version)](https://codeforces.com/contest/2262/problem/A1 \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [CutSandstone](https://codeforces.com/profile/CutSandstone \"Candidate Master CutSandstone\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include \nusing namespace std;\n\nvoid tc() {\n int n; cin >> n;\n vector b(n + 1);\n for(int i = 1; i <= n; i++) cin >> b[i];\n vector> goodivs;\n vector> badivs;\n for(int i = 1; i <= n; i++) {\n badivs.push_back({b[i] * i, (b[i] + 1) * i - 1});\n for(int j = 0; j < b[i]; j++) {\n goodivs.push_back({j * i, (j + 1) * i - 1});\n }\n }\n vector d(n + 1);\n for(auto [l, r] : badivs) {\n d[min(n, l)]++;\n d[min(n, r + 1)]--;\n }\n int csm = 0;\n vector gnums;\n for(int i = 0; i < n; i++) {\n csm += d[i];\n if(csm == 0) {\n gnums.push_back(i);\n }\n }\n cout << gnums.size() << endl;\n for(int i : gnums) cout << i << \" \";\n cout << endl;\n}\n\nint main() {\n ios::sync_with_stdio(false), cin.tie(nullptr);\n int t; cin >> t;\n while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263C2 - Floor of MEX (Hard Version)](https://codeforces.com/contest/2263/problem/C2 \"Codeforces Round 1120 (Div. 2)\") / [2262A2 - Floor of MEX (Hard Version)](https://codeforces.com/contest/2262/problem/A2 \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include \nusing namespace std;\n\nconst int MOD = 1e9 + 7;\n\nvoid tc() {\n int n; cin >> n;\n vector b(n + 1);\n for(int i = 1; i <= n; i++) cin >> b[i];\n vector> goodivs;\n vector> badivs;\n for(int i = 1; i <= n; i++) {\n if(b[i] > (n - 1) / i + 1) {\n cout << 0 << endl;\n return;\n }\n badivs.push_back({b[i] * i, (b[i] + 1) * i - 1});\n for(int j = 0; j < b[i]; j++) {\n goodivs.push_back({j * i, (j + 1) * i - 1});\n }\n }\n vector d(n + 1);\n for(auto [l, r] : badivs) {\n d[min(n, l)]++;\n d[min(n, r + 1)]--;\n }\n int csm = 0;\n vector gnums;\n for(int i = 0; i < n; i++) {\n csm += d[i];\n if(csm == 0) {\n gnums.push_back(i);\n }\n }\n set> st;\n sort(goodivs.begin(), goodivs.end(), [&] (pair iv, pair iv2) {\n return iv.second - iv.first < iv2.second - iv2.first;\n });\n for(auto [l, r] : goodivs) {\n auto it = st.lower_bound({l, 0});\n if(it != st.end() && it->second <= r) {\n continue;\n }\n st.insert({l, r});\n }\n\n vector> impivs(st.begin(), st.end());\n int l = 0, r = 0;\n vector dp(impivs.size() + 1);\n dp[0] = 1;\n int poss = 1;\n for(int i : gnums) {\n while(l < impivs.size() && impivs[l].second < i) {\n poss += MOD - dp[l];\n poss %= MOD;\n l++;\n }\n while(r < impivs.size() && impivs[r].first <= i) r++;\n dp[r] += poss;\n dp[r] %= MOD;\n poss += poss;\n poss %= MOD;\n }\n cout << dp[impivs.size()] << endl;\n}\n\nint main() {\n ios::sync_with_stdio(false), cin.tie(nullptr);\n int t; cin >> t;\n while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263D - Culling Game](https://codeforces.com/contest/2263/problem/D \"Codeforces Round 1120 (Div. 2)\") / [2262B - Culling Game](https://codeforces.com/contest/2262/problem/B \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include \nusing namespace std;\n\nusing ll = long long;\n\nstruct BIT {\n vector tr;\n BIT(int n) : tr(n + 1) {}\n void add(int k, ll x) {\n while(k < tr.size()) {\n tr[k] += x;\n k += k&-k;\n }\n }\n ll sum(int k) {\n ll res = 0;\n while(k > 0) {\n res += tr[k];\n k -= k&-k;\n }\n return res;\n }\n};\n\nvoid tc() {\n int n; cin >> n;\n vector a(n), p(n);\n for(int i = 0; i < n; i++) {\n cin >> a[i];\n }\n for(int i = 0; i < n; i++) {\n cin >> p[i]; p[i]--;\n }\n reverse(p.begin(), p.end());\n BIT bit(n);\n set imp; \n vector ans(n);\n for(int i = 0; i < n; i++) {\n int k = p[i];\n auto it = imp.upper_bound(k);\n int pimp;\n if(it == imp.begin() || bit.sum(k) - bit.sum(*prev(it)) < a[k]) {\n imp.insert(k);\n pimp = k;\n } else {\n pimp = *prev(it);\n }\n bit.add(k + 1, a[k]);\n while(imp.upper_bound(k) != imp.end()) {\n int j = *imp.upper_bound(k);\n if(bit.sum(j) - bit.sum(pimp) >= a[j]) {\n imp.erase(j);\n } else {\n break;\n }\n }\n ans[i] = imp.size() - 1;\n }\n reverse(ans.begin(), ans.end());\n for(int i : ans) cout << i << \" \";\n cout << endl;\n}\n\nint main() {\n ios::sync_with_stdio(false), cin.tie(nullptr);\n int t; cin >> t;\n while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263E - Traveling the World](https://codeforces.com/contest/2263/problem/E \"Codeforces Round 1120 (Div. 2)\") / [2262C - Traveling the World](https://codeforces.com/contest/2262/problem/C \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include \nusing namespace std;\n\nusing ll = long long;\n\nconst int MOD = 1e9 + 7;\n\nvoid tc() {\n int n; cin >> n;\n vector a(n);\n for(int i = 0; i < n; i++) cin >> a[i];\n int k = n / 2;\n\n int val = 1;\n for(int i = 1; i <= k; i++) {\n val = (ll)i * val % MOD;\n }\n for(int i = 1; i <= k - (n % 2 == 0 ? 2 : 1); i++) {\n val = (ll)i * val % MOD;\n }\n\n set xposs;\n xposs.insert(a[n - 1] - a[n - 2]);\n xposs.insert(a[n - 1] - a[n - 3]);\n\n int ans = 0;\n for(ll x : xposs) {\n vector b;\n for(int i = 0; i < k; i++) {\n b.push_back(i * x);\n }\n for(int i = 0; i < n - k; i++) {\n b.push_back(a[n - 1] - i * x);\n }\n sort(b.begin(), b.end());\n if(a == b) {\n ans += val;\n if(ans >= MOD) ans -= MOD;\n }\n }\n cout << ans << endl;\n}\n\nint main() {\n ios::sync_with_stdio(false), cin.tie(nullptr);\n int t; cin >> t;\n while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263F - PLUSworld](https://codeforces.com/contest/2263/problem/F \"Codeforces Round 1120 (Div. 2)\") / [2262D - PLUSworld](https://codeforces.com/contest/2262/problem/D \"Codeforces Round 1120 (Div. 1)\")\n\nAuthors: [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") & [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include \nusing namespace std;\n\nvoid tc() {\n int n; cin >> n;\n\n vector a(n), b(n);\n for(int i = 0; i < n; i++) cin >> a[i], a[i]--;\n for(int i = 0; i < n; i++) cin >> b[i], b[i]--;\n\n int badcnt = 0;\n for(int i = 0; i < n; i++) {\n badcnt += a[i] != b[i];\n }\n if(badcnt == 0) {\n cout << \"0 1\\n\";\n return;\n }\n\n // mark everything which will ever be reached\n vector useful(n);\n for(int i = 0; i < n; i++) {\n if(a[i] == b[i]) continue;\n int x = i;\n while(!useful[x]) {\n useful[x] = true;\n x = b[x];\n }\n }\n\n // anything with b[i] == i once we reach we can't reach anything else, \n // at most one of them can exit\n vector ends;\n for(int i = 0; i < n; i++) {\n if(useful[i] && b[i] == i) ends.push_back(i);\n }\n\n if(ends.size() > 1) {\n cout << \"-1\\n\";\n return;\n }\n\n vector> adj(n), radj(n);\n for(int i = 0; i < n; i++) {\n if(a[i] > b[i]) {\n cout << \"-1\\n\";\n return;\n }\n for(int j = a[i]; j <= b[i]; j++) {\n adj[i].push_back(j);\n radj[j].push_back(i);\n }\n }\n\n // tarjan algorithm, from cp-algorithms\n vector st; // - stack holding the unclaimed vertices\n vector roots(n, -1); // - keeps track of the SCC roots of the vertices\n int timer = 0; // - dfs timestamp counter\n vector t_in(n, -1); // - keeps track of the dfs timestamp of the vertices\n vector t_low(n, -1); // - keeps track of the lowest t_in of unclaimed vertices\n // reachable in the subtree\n\n // implements the tarjan algorithm for strongly connected components\n auto dfs = [&](auto&& self, int s) -> void {\n t_low[s] = t_in[s] = timer++;\n st.push_back(s);\n\n for (auto u : adj[s]) {\n if (t_in[u] == -1) { // tree-edge\n self(self, u);\n t_low[s] = min(t_low[s], t_low[u]);\n } else if (roots[u] == -1) { // back-edge, cross-edge or forward-edge to an unclaimed vertex\n t_low[s] = min(t_low[s], t_in[u]);\n }\n }\n\n if (t_low[s] == t_in[s]) { // vertex is a root\n while (true) {\n int u = st.back();\n st.pop_back();\n roots[u] = s; // claims the vertex\n if (u == s)\n break;\n }\n }\n };\n\n // applies the tarjan algorithm to all the vertices\n for (int v = 0; v < n; v++) {\n if (t_in[v] == -1) {\n dfs(dfs, v);\n }\n }\n vector> comps(n);\n for(int v = 0; v < n; v++) {\n comps[roots[v]].push_back(v);\n }\n\n vector p(n, -1);\n vector sz(n);\n for(int i = 0; i < n; i++) {\n if(a[i] != b[i]) {\n sz[roots[i]]++;\n }\n if(useful[i] && roots[i] != roots[b[i]]) {\n if(p[roots[b[i]]] != -1) {\n cout << \"-1\\n\";\n return;\n }\n p[roots[b[i]]] = roots[i];\n }\n }\n vector chain;\n int x = roots[ends[0]];\n int csz = 0;\n while(x != -1) {\n chain.push_back(x);\n csz += sz[x];\n x = p[x];\n }\n if(csz != badcnt) {\n cout << \"-1\\n\";\n return;\n }\n reverse(chain.begin(), chain.end());\n\n {\n x = comps[chain[0]][0];\n vector ops;\n auto op1 = [&] () {\n ops.push_back(1);\n a[x]++;\n };\n auto op2 = [&] () {\n ops.push_back(2);\n x = a[x];\n };\n auto solve_scc = [&] () {\n int rt = roots[x];\n vector comp = comps[rt];\n\n vector p2(n, -1);\n // only one thing in comp case\n auto rdfs = [&] (auto&& self, int s) -> void {\n for(int u : radj[s]) {\n if(roots[u] != rt) continue;\n if(p2[u] == -1) {\n p2[u] = s;\n self(self, u);\n }\n }\n };\n\n // root a tree\n rdfs(rdfs, comp[0]);\n\n for(int i = 0; i < comp.size(); i++) {\n while(x != comp[i]) {\n while(a[x] < max(p2[x], comp[i])) op1();\n op2();\n }\n while(a[x] < b[x]) op1();\n op2();\n }\n };\n for(int i = 0; i < chain.size(); i++) solve_scc();\n\n cout << ops.size() << \" \" << comps[chain[0]][0] + 1 << endl;\n for(int i : ops) cout << i << \" \";\n cout << endl;\n\n for(int i = 0; i < n; i++) assert(a[i] == b[i]);\n }\n}\n\nint main() {\n ios::sync_with_stdio(false), cin.tie(nullptr);\n int t; cin >> t;\n while(t--) tc();\n}\n\n```\n\n**Magic**\n\nWith this, you can prove a significantly simpler solution. We will add this proof later.\n\n**Magic Solution**\n\n```\n// Note: AI generated solution\n#include \nusing namespace std;\n\nint main() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n\n int t;\n cin >> t;\n\n while (t--) {\n int n;\n cin >> n;\n\n vector a(n + 1), b(n + 1);\n for (int i = 1; i <= n; i++) {\n cin >> a[i];\n }\n for (int i = 1; i <= n; i++) {\n cin >> b[i];\n }\n\n int start = 1, remaining = 0;\n bool possible = true;\n\n for (int i = 1; i <= n; i++) {\n if (a[i] > b[i]) {\n possible = false;\n }\n if (a[i] < b[i]) {\n if (remaining == 0) {\n start = i;\n }\n remaining++;\n }\n }\n\n if (!possible) {\n cout << -1 << '\\n';\n continue;\n }\n\n vector operations;\n int position = start;\n\n while (remaining > 0) {\n if (a[position] < b[position]) {\n while (a[position] < b[position] &&\n (a[position] >= position ||\n a[a[position]] == b[a[position]])) {\n a[position]++;\n operations.push_back(1);\n }\n\n if (a[position] == b[position]) {\n remaining--;\n }\n }\n\n if (remaining == 0 || a[position] == position) {\n break;\n }\n\n operations.push_back(2);\n position = a[position];\n }\n\n if (remaining > 0) {\n cout << -1 << '\\n';\n continue;\n }\n\n cout << operations.size() << ' ' << start << '\\n';\n for (int operation : operations) {\n cout << operation << ' ';\n }\n cout << '\\n';\n }\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2262E - Paired Bracket Sequences](https://codeforces.com/contest/2262/problem/E \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include \nusing namespace std;\n\nint MOD;\n\nvoid madd(int& a, int b) {\n a += b;\n if(a >= MOD) a -= MOD;\n}\n\nvoid msub(int& a, int b) {\n a -= b;\n if(a < 0) a += MOD;\n}\n\nint mmul(int a, int b) {\n return 1LL * a * b % MOD;\n}\n\nint bpow(int x, int y) {\n return y == 0 ? 1 : mmul((y % 2) ? x : 1, bpow(mmul(x, x), y / 2));\n}\n\nvoid tc() {\n int n; cin >> n >> MOD;\n vector f(2 * n + 1), invf(2 * n + 1);\n f[0] = 1;\n for(int i = 1; i <= 2 * n; i++) {\n f[i] = mmul(f[i - 1], i);\n }\n invf[2 * n] = bpow(f[2 * n], MOD - 2);\n for(int i = 2 * n; i >= 1; i--) {\n invf[i - 1] = mmul(invf[i], i);\n }\n auto F = [&] (int n, int k) -> int {\n return mmul(f[n], mmul(invf[k + 1], invf[n - k]));\n };\n auto choose = [&] (int n, int k) -> int {\n return mmul(f[n], mmul(invf[k], invf[n - k]));\n };\n vector> p(n + 1, vector(n + 1));\n for(int i = 0; i <= n; i++) {\n p[0][i] = mmul(F(2 * i, i), F(2 * i, i));\n }\n for(int i = 1; i <= n; i++) {\n for(int j = 0; j <= n; j++) {\n for(int k = 0; j + k <= n; k++) {\n madd(p[i][j + k], mmul(p[i - 1][j], p[0][k]));\n }\n }\n }\n vector ans(n + 1);\n for(int i = 0; i <= n; i++) {\n ans[i] = mmul(F(2 * n, i), p[i][n - i]);\n }\n for(int i = n - 1; i >= 0; i--) {\n for(int j = i + 1; j <= n; j++) {\n msub(ans[i], mmul(ans[j], choose(j, i)));\n }\n }\n for(int i = 0; i <= n; i++) {\n cout << ans[i] << \" \";\n }\n cout << endl;\n}\n\nint main() {\n ios::sync_with_stdio(false), cin.tie(nullptr);\n int t; cin >> t; \n while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2262F - Rank Removal](https://codeforces.com/contest/2262/problem/F \"Codeforces Round 1120 (Div. 1)\")\n\nAuthors: [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") & [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include \nusing namespace std;\n\nconst int MXN = 400;\n\ntemplate int compute_inv(const vector>& mat, vector>& inv) {\n int n = mat.size();\n vector empty(n, true);\n vector> basis(n);\n\n for(int i = 0; i < n; i++) {\n bitset cur = mat[i];\n bitset curinv;\n curinv[i] = 1;\n for(int j = 0; j < n; j++) {\n if(cur[j]) {\n if(empty[j]) {\n basis[j] = cur;\n inv[j] = curinv;\n empty[j] = false;\n break;\n } else {\n cur ^= basis[j];\n curinv ^= inv[j];\n }\n }\n }\n }\n for(int i = 0; i < n; i++) {\n if(empty[i]) return 0;\n }\n\n for(int i = n - 1; i >= 1; i--) {\n for(int j = 0; j < i; j++) {\n if(basis[j][i]) {\n basis[j] ^= basis[i];\n inv[j] ^= inv[i];\n }\n }\n }\n return 1;\n}\n\nbool try_removal(vector> cells, vector>& mat, vector>& inv, vector& row_cnt) {\n int n = mat.size(), k = cells.size();\n for(auto [x, y] : cells) {\n if(mat[x][y] == 0) {\n return false;\n }\n }\n assert(k < 4);\n vector> K(k); \n for(int i = 0; i < k; i++) {\n for(int j = 0; j < k; j++) {\n K[i][j] = (i == j) ^ inv[cells[i].second][cells[j].first];\n }\n }\n vector> invK(cells.size());\n if(!compute_inv(K, invK)) {\n return false;\n }\n for(auto [x, y] : cells) {\n mat[x][y] = 0;\n row_cnt[x]--;\n }\n\n auto old = inv;\n for(int i = 0; i < n; i++) {\n bitset<3> coef;\n for(int alpha = 0; alpha < k; alpha++) {\n int r = cells[alpha].first;\n\n if(old[i][r]) {\n coef ^= invK[alpha];\n }\n }\n\n // delta = coef * (V^T * old)\n for(int beta = 0; beta < k; beta++) {\n if(coef[beta]) {\n int c = cells[beta].second;\n inv[i] ^= old[c];\n }\n }\n }\n return true;\n}\n\nvector compute_matching(const vector>& mat) {\n int n = mat.size();\n vector> adj(n);\n for(int i = 0; i < n; i++) {\n for(int j = 0; j < n; j++) {\n if(mat[i][j]) adj[i].push_back(j);\n }\n }\n\n vector match(n, -1);\n vector done(n);\n\n function dfs = [&] (int s) {\n for(int u : adj[s]) {\n if(!done[u]) {\n done[u] = true;\n\n if(match[u] == -1 || dfs(match[u])) {\n match[u] = s;\n return true;\n }\n }\n }\n return false;\n };\n\n for(int i = 0; i < n; i++) {\n fill(done.begin(), done.end(), false);\n dfs(i);\n }\n\n for(int i = 0; i < n; i++) {\n assert(match[i] != -1);\n }\n return match;\n}\n\nvoid tc() {\n int n, m; cin >> n >> m;\n\n vector> mat(n);\n vector row_cnt(n);\n for(int i = 0; i < m; i++) {\n int x, y; cin >> x >> y;\n x--, y--;\n mat[x][y] = 1;\n row_cnt[x]++;\n }\n\n vector> inv(n);\n compute_inv(mat, inv);\n\n int sm = m;\n int left = (sm + n - 1) / n;\n cout << left << '\\n';\n\n while(sm > 0) {\n vector> omat = mat;\n if(sm <= n) {\n for(int i = 0; i < n; i++) {\n for(int j = 0; j < n; j++) {\n mat[i][j] = 0;\n }\n }\n row_cnt.assign(n, 0);\n } else if(sm <= 2 * n) {\n vector matching = compute_matching(mat);\n for(int i = 0; i < n; i++) {\n for(int j = 0; j < n; j++) {\n mat[i][j] = 0;\n }\n }\n row_cnt.assign(n, 0);\n for(int i = 0; i < sm - n; i++) {\n mat[matching[i]][i] = 1;\n row_cnt[matching[i]]++;\n }\n } else if(sm <= 3 * n) {\n vector matching = compute_matching(mat);\n int above = 0;\n for(int i = 0; i < n; i++) {\n for(int j = matching[i] + 1; j < n; j++) {\n above += mat[j][i];\n }\n }\n\n vector> nmat(n);\n vector nrow_cnt(n);\n int ones_left = sm - n;\n for(int i = 0; i < n; i++) {\n nmat[matching[i]][i] = 1;\n nrow_cnt[matching[i]]++;\n ones_left--;\n }\n // make matrix either upper or lower triangular, whichever one\n // removes a sufficient number of ones\n if(above >= ones_left) {\n for(int i = 0; i < n; i++) {\n for(int j = matching[i] + 1; j < n; j++) {\n if(mat[j][i]) {\n ones_left--;\n nmat[j][i] = 1;\n nrow_cnt[j]++;\n if(ones_left == 0) break;\n }\n }\n if(ones_left == 0) break;\n }\n } else {\n for(int i = 0; i < n; i++) {\n for(int j = 0; j < matching[i]; j++) {\n if(mat[j][i]) {\n ones_left--;\n nmat[j][i] = 1;\n nrow_cnt[j]++;\n if(ones_left == 0) break;\n }\n }\n if(ones_left == 0) break;\n }\n }\n mat = nmat;\n row_cnt = nrow_cnt;\n } else {\n int csm = sm;\n if(n % 2) {\n // odd case (remove either 1 or 3)\n int dup_row = -1;\n for(int i = 0; i < n; i++) {\n if(row_cnt[i] >= 2) {\n dup_row = i;\n break;\n }\n }\n assert(dup_row != -1);\n\n vector cols;\n for(int i = 0; i < n; i++) {\n if(mat[dup_row][i]) cols.push_back(i);\n }\n\n assert(cols.size() >= 2);\n\n for(int i = 0; i < n; i++) {\n if(try_removal({{i, cols[0]}}, mat, inv, row_cnt)) {\n csm--;\n break;\n }\n if(try_removal({{i, cols[1]}}, mat, inv, row_cnt)) {\n csm--;\n break;\n }\n }\n\n if(csm == sm) {\n for(int i = 0; i < n; i++) {\n if(i != dup_row) {\n // remove L shape\n if(try_removal(\n {{i, cols[0]}, \n {dup_row, cols[0]}, \n {dup_row, cols[1]}},\n mat, inv, row_cnt)) {\n csm -= 3;\n break;\n }\n }\n }\n assert(sm != csm);\n }\n }\n\n while(csm > sm - n) {\n // remove 2 at a time\n int dup_row = -1;\n for(int i = 0; i < n; i++) {\n if(row_cnt[i] >= 3) {\n dup_row = i;\n break;\n }\n }\n\n vector cols;\n for(int i = 0; i < n; i++) {\n if(mat[dup_row][i]) cols.push_back(i);\n }\n\n assert(cols.size() >= 3);\n\n if(!try_removal({{dup_row, cols[0]}, {dup_row, cols[1]}}, mat, inv, row_cnt)) {\n if(!try_removal({{dup_row, cols[1]}, {dup_row, cols[2]}}, mat, inv, row_cnt)) {\n if(!try_removal({{dup_row, cols[0]}, {dup_row, cols[2]}}, mat, inv, row_cnt)) {\n assert(false);\n }\n }\n }\n\n csm -= 2;\n }\n assert(csm == sm - n);\n }\n\n vector> move;\n for(int i = 0; i < n; i++) {\n for(int j = 0; j < n; j++) {\n if(omat[i][j] == 1 && mat[i][j] == 0) {\n move.push_back({i, j});\n }\n }\n }\n cout << move.size() << \"\\n\";\n for(auto [x, y] : move) cout << x + 1 << \" \" << y + 1 << \"\\n\";\n sm -= n;\n\n int csm = 0;\n for(int i = 0; i < n; i++) {\n csm += row_cnt[i];\n }\n assert(csm == max(0, sm));\n }\n}\n\nint main() {\n int t; cin >> t;\n while(t--) tc();\n}\n\n```\n\n**Tutorial 2**\n\nCredit to [BurnedChicken](https://codeforces.com/profile/BurnedChicken \"International Grandmaster BurnedChicken\") for finding this in testing!\n\nWe will first prove that for any row/column of a rank $$$n$$$ matrix, we can zero out everything in that row/column except one, and maintain rank $$$n$$$. This is because rank $$$n$$$ means the determinant is $$$1$$$, so there are exactly an odd number of perfect matchings in total (read Solution 1 for more understanding). Therefore at least $$$1$$$ of the ones in the row/column must be in an odd number of perfect matchings (since if they were all even then the total number would be even). This means if we delete all other ones but that one, then we would keep rank $$$n$$$.\n\nThis actually all you need to prove to solve the problem. Note that after we clear out one row except one number, we can clear out that column except that one number (even partially) and the determinant will not change (and the same works when clearing out a column with the leftover row). \n\nFor every turn, we want to try to find $$$n$$$ things to remove without changing the rank. Therefore, we begin by finding the one in the biggest row/column we can keep while removing everything else. Then we remove everything in the column/row of that one, and the everything in the row/column into the queue. If the queue + the current removals is big enough to reach $$$n$$$ removals, then we can just remove the ones in the queue. If the queue is too small, then we repeat one the next biggest row/column. Because the queue contains the row/column with the most ones, when the current removals will never exceed $$$n$$$ if the the queue + current removals is less than $$$n$$$. Once we run out of row/columns to clear, we can start removing the ones we excluded removing, since we have to reduce the rank at this point.\n\nTo find the one we must keep, you have to utilize the inverse, since we can clear out the row/column on $$$M\\_{i, j}$$$ if $$$M^{-1}\\_{j, i} = 1$$$ (proof is by Cramer's Rule). We can keep track of the inverse after this clear out operation in $$$O(n^2 / w)$$$ if done properly (using something similar to Woodbury Matrix Identity, look at Solution 1 for details). If implemented properly, this results in a $$$O(n^3 / w)$$$ solution.\n\n**Solution 2**\n\n```\n#include \nusing namespace std;\n \nconst int SZ = 400;\n \n \ntemplate int compute_inv(const vector>& mat, vector>& inv) {\n int n = mat.size();\n vector empty(n, true);\n vector> basis(n);\n \n for(int i = 0; i < n; i++) {\n bitset cur = mat[i];\n bitset curinv;\n curinv[i] = 1;\n for(int j = 0; j < n; j++) {\n if(cur[j]) {\n if(empty[j]) {\n basis[j] = cur;\n inv[j] = curinv;\n empty[j] = false;\n break;\n } else {\n cur ^= basis[j];\n curinv ^= inv[j];\n }\n }\n }\n }\n for(int i = 0; i < n; i++) {\n if(empty[i]) return 0;\n }\n \n for(int i = n - 1; i >= 1; i--) {\n for(int j = 0; j < i; j++) {\n if(basis[j][i]) {\n basis[j] ^= basis[i];\n inv[j] ^= inv[i];\n }\n }\n }\n return 1;\n}\n \nvoid tc() {\n int n, m; cin >> n >> m;\n vector> mat(n);\n for(int i = 0; i < m; i++) {\n int x, y; cin >> x >> y;\n x--, y--;\n mat[x][y] = 1;\n }\n \n vector> inv(n);\n compute_inv(mat, inv);\n vector cnts(2 * n);\n for(int i = 0; i < n; i++) {\n for(int j = 0; j < n; j++) {\n cnts[i] += mat[i][j];\n cnts[n + j] += mat[i][j];\n }\n }\n queue> mustq, freeq;\n vector> centers;\n vector>> moves;\n while(true) {\n while(mustq.size() + freeq.size() < n) {\n int best_rc = 0;\n for(int rc = 0; rc < 2 * n; rc++) {\n if(cnts[rc] >= cnts[best_rc]) best_rc = rc;\n }\n if(cnts[best_rc] == 0) {\n break;\n }\n \n int x = -1, y = -1;\n if(best_rc < n) {\n x = best_rc;\n for(int j = 0; j < n; j++) {\n if(mat[x][j] && inv[j][x]) {\n y = j;\n break;\n }\n }\n } else {\n y = best_rc - n;\n for(int j = 0; j < n; j++) {\n if(mat[j][y] && inv[y][j]) {\n x = j;\n break;\n }\n }\n }\n assert(x != -1 && y != -1);\n \n centers.push_back({x, y});\n vector> must, free;\n auto erase = [&] (int a, int b) {\n mat[a][b] = 0;\n cnts[a]--;\n cnts[b + n]--;\n };\n for(int j = 0; j < n; j++) {\n if(mat[x][j] && j != y) {\n must.push_back({x, j});\n erase(x, j);\n }\n }\n for(int j = 0; j < n; j++) {\n if(mat[j][y] && j != x) {\n free.push_back({j, y});\n erase(j, y);\n }\n }\n erase(x, y);\n assert(cnts[x] == 0 && cnts[n + y] == 0);\n \n if(best_rc < n) swap(must, free);\n for(auto u : must) mustq.push(u);\n for(auto u : free) freeq.push(u);\n \n bitset row = inv[y];\n for(int j = 0; j < n; j++) {\n if(cnts[j + n] > 0 && inv[j][x]) {\n inv[j] ^= row;\n }\n }\n }\n int left = n;\n vector> move;\n while(left > 0 && mustq.size() > 0) {\n left--;\n move.push_back(mustq.front());\n mustq.pop();\n }\n while(left > 0 && freeq.size() > 0) {\n left--;\n move.push_back(freeq.front());\n freeq.pop();\n }\n while(left > 0 && centers.size() > 0) {\n left--;\n move.push_back(centers.back());\n centers.pop_back();\n }\n moves.push_back(move);\n if(centers.size() == 0) break;\n }\n cout << moves.size() << \"\\n\";\n for(auto move : moves) {\n cout << move.size() << \"\\n\";\n for(auto [u, v] : move) cout << u + 1 << \" \" << v + 1 << \"\\n\";\n }\n}\n \nint main() {\n ios::sync_with_stdio(false), cin.tie(nullptr);\n int t; cin >> t;\n while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n\n...108135 bytes truncated...\n\nThe tool call succeeded but the output was truncated. Full output saved to: /Users/zayd/.local/share/opencode/tool-output/tool_0acbde8ee001wiSY58dayIdJGF\nUse the Task tool to have explore agent process this file with Grep and Read (with offset/limit). Do NOT read the full file yourself - delegate to save context.", "metadata": { "truncated": true, "outputPath": "/Users/zayd/.local/share/opencode/tool-output/tool_0acbde8ee001wiSY58dayIdJGF" }, "title": "https://codeforces.com/blog/entry/156688 (text/markdown; charset=utf-8)", "time": { "start": 1789604520437, "end": 1789604522226 } }, "metadata": { "openai": { "itemId": "fc_0010c4e84f3238f6016aab32a8208087d182504370701e6d63" } }, "id": "prt_0acbde124001thZkwlanmuiHSK", "sessionID": "ses_f534230b3ffegKs0rWwiBFYJC4", "messageID": "msg_0acbdcfe8001KZCdyxqm41kBVm" }, { "reason": "tool-calls", "type": "step-finish", "tokens": { "total": 9580, "input": 2931, "output": 38, "reasoning": 83, "cache": { "write": 0, "read": 6528 } }, "cost": 0, "id": "prt_0acbde8f5001bLUeoxn0zHE51s", "sessionID": "ses_f534230b3ffegKs0rWwiBFYJC4", "messageID": "msg_0acbdcfe8001KZCdyxqm41kBVm" } ] }, { "info": { "parentID": "msg_0acbdcf5e001j3NKztZRAhSbbO", "role": "assistant", "mode": "nudge-generation", 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"ses_f534230b3ffegKs0rWwiBFYJC4", "messageID": "msg_0acbde8fa0013ucPr4qQ4pYCQ7" }, { "type": "reasoning", "text": "**Preparing explore agent for file analysis**", "time": { "start": 1789604528226, "end": 1789604528242 }, "metadata": { "openai": { "itemId": "rs_04352f08cbd43965016aab32ab414487d18c8f3438c0b49927", "reasoningEncryptedContent": 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} }, "id": "prt_0acbe00620014TXxrFj796HxYl", "sessionID": "ses_f534230b3ffegKs0rWwiBFYJC4", "messageID": "msg_0acbde8fa0013ucPr4qQ4pYCQ7" }, { "type": "tool", "tool": "task", "callID": "call_cxmtsHY9vqqsCMJQftf3mEfq", "state": { "status": "completed", "input": { "description": "Analyze matrix construction", "prompt": "Inspect the fetched tutorial output at /Users/zayd/.local/share/opencode/tool-output/tool_0acbde8ee001wiSY58dayIdJGF using Grep/Read (do not read it all). Focus on the 2263B Min Matrices section and the supplied statement. Analyze the construction, derive a concise correctness proof, identify feasibility conditions and edge cases, and report any pitfalls in the shown code. Do not edit files; return only findings for a learning-content answer.", "subagent_type": "explore" }, "output": "\n\nFindings from `/Users/zayd/.local/share/opencode/tool-output/tool_0acbde8ee001wiSY58dayIdJGF`:\n\n### Construction\n\nThe code sets:\n\n```cpp\nx = k - n + 1\n```\n\nIt creates `k` special “anchor” cells:\n\n- `(0,0)` gets `1`.\n- For `1 <= i < x`, `(0,i)` and `(i,0)` get the next values.\n- For `x <= i < n`, `(i,i)` gets the next values.\n- Every remaining cell is filled afterward.\n\nThe anchor count is:\n\n\\[\n1+2(x-1)+(n-x)=n+x-1=k.\n\\]\n\nAll anchors receive values `1..k`; every other cell receives a larger value.\n\n### Correctness proof\n\nEvery row and column contains at least one anchor:\n\n- Row/column `0` contains `(0,0)`.\n- For `i < x`, row `i` has `(i,0)` and column `i` has `(0,i)`.\n- For `i >= x`, both row and column `i` have `(i,i)`.\n\nSince all non-anchors are larger than all anchors, each row/column minimum is an anchor. Every anchor appears in `f(A)`:\n\n- `(0,0)` is both a row and column minimum.\n- `(i,0)` is the minimum of row `i`.\n- `(0,i)` is the minimum of column `i`.\n- `(i,i)` is the minimum of both row and column `i`.\n\nThus `f(A)` consists of exactly the `k` distinct anchor values. The fill process also assigns every number from `1` through `n²` exactly once.\n\n### Feasibility\n\nLet `R` and `C` be the sets of row and column minima.\n\n- `|R| = n`, because all matrix entries are distinct.\n- `|C| = n`.\n- The global minimum belongs to both `R` and `C`, so `R ∩ C` is nonempty.\n\nTherefore:\n\n\\[\nn \\le |R\\cup C| \\le 2n-1.\n\\]\n\nEvery value in this range is constructible, so the exact feasibility condition is:\n\n\\[\n\\boxed{n \\le k \\le 2n-1}.\n\\]\n\nThe code’s condition `1 <= x <= n` is equivalent to this.\n\n### Edge cases\n\n- `k = n`: `x = 1`; the anchors form the main diagonal, giving exactly `n` distinct minima.\n- `k = 2n-1`: `x = n`; anchors occupy the first row and first column.\n- `n = 1`: only `k = 1` is possible.\n- `k = 0` and `k = 2n` are correctly rejected; in particular, `2n` is impossible because the global minimum is necessarily shared by one row and one column.\n\n### Code pitfalls\n\n- The order is essential: all anchors must be assigned before the remaining cells. Otherwise, non-anchor cells could become minima.\n- The zero-based boundary `i < x` versus `i >= x` is deliberate; changing either to an inclusive counterpart causes an off-by-one error.\n- `-1` is printed without a trailing newline, but this is harmless under Codeforces’ token-based checker.\n- `int` is safe under the stated `n <= 1000` constraints; values reach only `n²`.\n- The fetched editorial has no actual explanation here: the B section says “Tutorial is loading...” at lines 175–177, with only the code shown at lines 181–223.\n\n", "metadata": { "parentSessionId": "ses_f534230b3ffegKs0rWwiBFYJC4", "sessionId": "ses_f5341f7f8ffeU2OS4rxlL1i1Vk", "model": { "modelID": "gpt-5.6-luna", "providerID": "openai" }, "truncated": false }, "title": "Analyze matrix construction", "time": { "start": 1789604530190, "end": 1789604682798 } }, "metadata": { "openai": { "itemId": "fc_04352f08cbd43965016aab32b02c8487d19104499912481ee3" } }, "id": "prt_0acbe0074001FWVVlDiLHOKxWy", "sessionID": "ses_f534230b3ffegKs0rWwiBFYJC4", "messageID": "msg_0acbde8fa0013ucPr4qQ4pYCQ7" }, { "reason": "tool-calls", "type": "step-finish", "tokens": { "total": 24597, "input": 15643, "output": 131, "reasoning": 247, "cache": { "write": 0, "read": 8576 } }, "cost": 0, "id": "prt_0acc05c2f001ihRb76wodIQ9nj", "sessionID": "ses_f534230b3ffegKs0rWwiBFYJC4", "messageID": "msg_0acbde8fa0013ucPr4qQ4pYCQ7" } ] }, { "info": { "parentID": "msg_0acbdcf5e001j3NKztZRAhSbbO", "role": "assistant", "mode": "nudge-generation", "agent": "nudge-generation", "variant": "max", "path": { "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IGBdqu", "root": "/" }, "cost": 0, "tokens": { "total": 32196, "input": 1481, "output": 1492, "reasoning": 5287, "cache": { "write": 0, "read": 23936 } }, "modelID": "gpt-5.6-luna", "providerID": "openai", "time": { "created": 1789604682804, "completed": 1789604806315 }, "finish": "tool-calls", "structured": { "status": "success", "reason": null, "hints": [ { "order": 1, "content": "Let $R$ be the set of row minima and $C$ the set of column minima. Each set has $n$ values, but they can overlap. Which matrix value is guaranteed to belong to both sets?" }, { "order": 2, "content": "The globally smallest entry is the minimum of its row and its column, so $R \\cap C$ is nonempty. This gives $n \\le |R \\cup C| \\le 2n-1$. The remaining question is how to construct every value in that interval." }, { "order": 3, "content": "Reserve some cells for the smallest values, call them anchors, and put larger values everywhere else. If every row and column contains an anchor, all line minima come from anchors. Arrange the anchors so that every anchor is itself the minimum of at least one row or column." }, { "order": 4, "content": "Set $x=k-n+1$. Use $(0,0)$, then $(0,i)$ and $(i,0)$ for $1 \\le i < x$, and finally $(i,i)$ for $x \\le i < n$. Count these cells and check that they cover every row and column." }, { "order": 5, "content": "If $x$ is outside $[1,n]$, print $-1$. Otherwise assign $1,2,\\ldots,k$ to the anchor cells, scan the matrix and assign the remaining values to every empty cell. The anchor count is $1+2(x-1)+(n-x)=k$, so the line-minimum set has exactly the required size." } ], "editorial": "Let $R$ be the set of minimum values of the rows, and let $C$ be the corresponding set for the columns. Because every matrix entry is distinct, minima from different rows are distinct, so $|R|=n$. Similarly, $|C|=n$.\n\nThe smallest entry in the whole matrix is the minimum of both its row and its column. Therefore $R \\cap C$ is nonempty, and\n\n$$n \\le |R \\cup C| \\le 2n-1.$$\n\nSince $f(A)=R \\cup C$, every $k$ outside this interval is impossible. The interesting part is showing that every value inside it can be built.\n\n**Anchor cells**\n\nCall a cell an anchor. We will put the values $1$ through $k$ in the anchors and all larger values in the other cells. Then any row or column containing an anchor has an anchor as its minimum.\n\nSet\n\n$$x=k-n+1.$$\n\nFor a feasible $k$, we have $1 \\le x \\le n$. Choose these anchor cells using zero-based indices:\n\n- $(0,0)$;\n- $(0,i)$ and $(i,0)$ for every $1 \\le i < x$;\n- $(i,i)$ for every $x \\le i < n$.\n\nThe number of anchors is\n\n$$1+2(x-1)+(n-x)=n+x-1=k.$$\n\nThey cover every row and column. Row and column $0$ contain $(0,0)$. For $1 \\le i < x$, row $i$ contains $(i,0)$ and column $i$ contains $(0,i)$. For $x \\le i < n$, both row $i$ and column $i$ contain $(i,i)$.\n\nAssign $1$ through $k$ to these cells first, then fill every remaining cell with $k+1$ through $n^2$. Thus every line has an anchor and every non-anchor is larger than every anchor. No non-anchor can be a line minimum.\n\nAlso, every anchor is guaranteed to appear among the minima: $(0,0)$ is the minimum of row and column $0$; $(0,i)$ is the minimum of column $i$; $(i,0)$ is the minimum of row $i$; and $(i,i)$ is the minimum of both its row and column. Since the anchor values are all distinct, exactly $k$ values occur in $f(A)$.\n\n**Algorithm**\n\n1. Compute $x=k-n+1$. If $x<1$ or $x>n$, print $-1$.\n2. Create an empty matrix and assign consecutive values to the anchor cells above.\n3. Scan all cells and assign the next consecutive value to every empty cell.\n4. Print the matrix.\n\nThe scan assigns every number from $1$ to $n^2$ exactly once. The construction and output both take $O(n^2)$ time and the matrix uses $O(n^2)$ memory.", "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, k;\n cin >> n >> k;\n\n int x = k - n + 1;\n if (x < 1 || x > n) {\n cout << -1 << '\\n';\n continue;\n }\n\n vector> a(n, vector(n, 0));\n int value = 1;\n\n a[0][0] = value++;\n for (int i = 1; i < x; ++i) {\n a[0][i] = value++;\n a[i][0] = value++;\n }\n for (int i = x; i < n; ++i) {\n a[i][i] = value++;\n }\n\n for (int i = 0; i < n; ++i) {\n for (int j = 0; j < n; ++j) {\n if (a[i][j] == 0) {\n a[i][j] = value++;\n }\n }\n }\n\n for (int i = 0; i < n; ++i) {\n for (int j = 0; j < n; ++j) {\n if (j) cout << ' ';\n cout << a[i][j];\n }\n cout << '\\n';\n }\n }\n}\n" }, "id": "msg_0acc05c340010chbHOkP97xV5j", "sessionID": "ses_f534230b3ffegKs0rWwiBFYJC4" }, "parts": [ { "type": 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} }, "id": "prt_0acc1d277001CaxJSlsX4GqEjK", "sessionID": "ses_f534230b3ffegKs0rWwiBFYJC4", "messageID": "msg_0acc05c340010chbHOkP97xV5j" }, { "type": "tool", "tool": "StructuredOutput", "callID": "call_yc1zD5eFmRxfHBS0x05jqKsc", "state": { "status": "completed", "input": { "status": "success", "reason": null, "hints": [ { "order": 1, "content": "Let $R$ be the set of row minima and $C$ the set of column minima. Each set has $n$ values, but they can overlap. Which matrix value is guaranteed to belong to both sets?" }, { "order": 2, "content": "The globally smallest entry is the minimum of its row and its column, so $R \\cap C$ is nonempty. This gives $n \\le |R \\cup C| \\le 2n-1$. The remaining question is how to construct every value in that interval." }, { "order": 3, "content": "Reserve some cells for the smallest values, call them anchors, and put larger values everywhere else. If every row and column contains an anchor, all line minima come from anchors. Arrange the anchors so that every anchor is itself the minimum of at least one row or column." }, { "order": 4, "content": "Set $x=k-n+1$. Use $(0,0)$, then $(0,i)$ and $(i,0)$ for $1 \\le i < x$, and finally $(i,i)$ for $x \\le i < n$. Count these cells and check that they cover every row and column." }, { "order": 5, "content": "If $x$ is outside $[1,n]$, print $-1$. Otherwise assign $1,2,\\ldots,k$ to the anchor cells, scan the matrix and assign the remaining values to every empty cell. The anchor count is $1+2(x-1)+(n-x)=k$, so the line-minimum set has exactly the required size." } ], "editorial": "Let $R$ be the set of minimum values of the rows, and let $C$ be the corresponding set for the columns. Because every matrix entry is distinct, minima from different rows are distinct, so $|R|=n$. Similarly, $|C|=n$.\n\nThe smallest entry in the whole matrix is the minimum of both its row and its column. Therefore $R \\cap C$ is nonempty, and\n\n$$n \\le |R \\cup C| \\le 2n-1.$$\n\nSince $f(A)=R \\cup C$, every $k$ outside this interval is impossible. The interesting part is showing that every value inside it can be built.\n\n**Anchor cells**\n\nCall a cell an anchor. We will put the values $1$ through $k$ in the anchors and all larger values in the other cells. Then any row or column containing an anchor has an anchor as its minimum.\n\nSet\n\n$$x=k-n+1.$$\n\nFor a feasible $k$, we have $1 \\le x \\le n$. Choose these anchor cells using zero-based indices:\n\n- $(0,0)$;\n- $(0,i)$ and $(i,0)$ for every $1 \\le i < x$;\n- $(i,i)$ for every $x \\le i < n$.\n\nThe number of anchors is\n\n$$1+2(x-1)+(n-x)=n+x-1=k.$$\n\nThey cover every row and column. Row and column $0$ contain $(0,0)$. For $1 \\le i < x$, row $i$ contains $(i,0)$ and column $i$ contains $(0,i)$. For $x \\le i < n$, both row $i$ and column $i$ contain $(i,i)$.\n\nAssign $1$ through $k$ to these cells first, then fill every remaining cell with $k+1$ through $n^2$. Thus every line has an anchor and every non-anchor is larger than every anchor. No non-anchor can be a line minimum.\n\nAlso, every anchor is guaranteed to appear among the minima: $(0,0)$ is the minimum of row and column $0$; $(0,i)$ is the minimum of column $i$; $(i,0)$ is the minimum of row $i$; and $(i,i)$ is the minimum of both its row and column. Since the anchor values are all distinct, exactly $k$ values occur in $f(A)$.\n\n**Algorithm**\n\n1. Compute $x=k-n+1$. If $x<1$ or $x>n$, print $-1$.\n2. Create an empty matrix and assign consecutive values to the anchor cells above.\n3. Scan all cells and assign the next consecutive value to every empty cell.\n4. Print the matrix.\n\nThe scan assigns every number from $1$ to $n^2$ exactly once. The construction and output both take $O(n^2)$ time and the matrix uses $O(n^2)$ memory.", "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, k;\n cin >> n >> k;\n\n int x = k - n + 1;\n if (x < 1 || x > n) {\n cout << -1 << '\\n';\n continue;\n }\n\n vector> a(n, vector(n, 0));\n int value = 1;\n\n a[0][0] = value++;\n for (int i = 1; i < x; ++i) {\n a[0][i] = value++;\n a[i][0] = value++;\n }\n for (int i = x; i < n; ++i) {\n a[i][i] = value++;\n }\n\n for (int i = 0; i < n; ++i) {\n for (int j = 0; j < n; ++j) {\n if (a[i][j] == 0) {\n a[i][j] = value++;\n }\n }\n }\n\n for (int i = 0; i < n; ++i) {\n for (int j = 0; j < n; ++j) {\n if (j) cout << ' ';\n cout << a[i][j];\n }\n cout << '\\n';\n }\n }\n}\n" }, "output": "Structured output captured successfully.", "metadata": { "valid": true }, "title": "Structured Output", "time": { "start": 1789604806298, "end": 1789604806302 } }, "metadata": { "openai": { "itemId": "fc_0656a4817fcf275d016aab33acd95c87d2bab632623cbf41e9" } }, "id": "prt_0acc1db8c001Aj50G84EIKWSFV", "sessionID": "ses_f534230b3ffegKs0rWwiBFYJC4", "messageID": "msg_0acc05c340010chbHOkP97xV5j" }, { "reason": "tool-calls", "type": "step-finish", "tokens": { "total": 32196, "input": 1481, "output": 1492, "reasoning": 5287, "cache": { "write": 0, "read": 23936 } }, "cost": 0, "id": "prt_0acc23ea8001bmySwfAISfOQ9h", "sessionID": "ses_f534230b3ffegKs0rWwiBFYJC4", "messageID": "msg_0acc05c340010chbHOkP97xV5j" } ] } ] }