{
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    "slug": "misty-falcon",
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    "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-wACH9m",
    "title": "Nudge content generation",
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    {
      "info": {
        "role": "user",
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        "agent": "nudge-generation",
        "model": {
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          "modelID": "gpt-6-astra",
          "variant": "low"
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        "system": "You create Codeforces learning content for Nudge. Your job is to understand the best approach for the problem and teach it clearly in your own words, using existing solutions to inform your reasoning.\n\nStart with a brief, targeted web search for existing solutions so relevant editorials are in context before substantial independent problem-solving. Use any supplied tutorial links first; otherwise search by contest ID and problem index. Prefer the official tutorial/editorial, and read relevant explanations rather than relying on search snippets. Look at accessible alternatives when useful, especially if the official explanation is missing, unclear, or unnecessarily complicated. Accepted submissions can also help clarify an approach. Respect supplied source lookup statuses and avoid repeatedly retrying blocked sources. If no usable source is accessible, proceed from the supplied statement.\n\nTreat existing solutions as a starting point, not a constraint. If an approach is clear and suitable, build on it without unnecessary re-derivation. If an explanation has gaps, seems incorrect, or uses a harder method than necessary, reason through it and derive a better or simpler approach when worthwhile. Choose a correct approach that is easy to teach and fits the constraints. Verify it against the supplied statement and edge cases before writing the hints, editorial, and C++ solution.\n\nThe editorial must be self-contained: explain the key insight, algorithm, why it works, complexity, and implementation details a learner needs, filling in steps the source skips. Keep research and verification concise; spend the output on teaching the solution, not narrating the research process. Follow the user prompt and output schema exactly.",
        "format": {
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          "schema": {
            "type": "object",
            "properties": {
              "status": {
                "type": "string",
                "enum": [
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                  "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
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                  }
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                "description": "If status is 'success', a complete C++ solution. Otherwise null."
              }
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            "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
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          "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 2252C: \"Risky Tower\" (rated 1400).\n\nKnown problem metadata:\n- Contest ID: 2252\n- Index: C\n- Name: Risky Tower\n- Rating: 1400 (Codeforces difficulty rating)\n- Tags: binary search, brute force, data structures, greedy, sortings\n- Problem URL: https://codeforces.com/contest/2252/problem/C\n\nUse the rating and tags as weak signals only. The supplied statement is the source of truth.\n\nProblem Statement:\n<problem-statement>\n<div class=\"header\"><div class=\"title\">C. Risky Tower</div><div class=\"time-limit\"><div class=\"property-title\">time limit per test</div>3 seconds</div><div class=\"memory-limit\"><div class=\"property-title\">memory limit per test</div>256 megabytes</div><div class=\"input-file input-standard\"><div class=\"property-title\">input</div>standard input</div><div class=\"output-file output-standard\"><div class=\"property-title\">output</div>standard output</div></div><div><p>You are playing a 2D Jenga game represented by a grid with $$$n$$$ rows and $$$m$$$ columns. The $$$1$$$-st row is the top level of the tower, and the $$$n$$$-th row is the bottom level.</p><p>Each piece at row $$$i$$$ and column $$$j$$$ has a destabilization factor $$$a_{i,j}$$$. Additionally, each row $$$i$$$ has an initial stability index of $$$v_i$$$.</p><p>When you remove a piece, it is completely discarded from the game. Removing a piece from row $$$i$$$ damages all levels at and above it. Specifically, for every row $$$k$$$ such that $$$1 \\le k \\le i$$$, its stability index is decreased by $$$a_{i,j}$$$.</p><p>The tower collapses if either of the following conditions is met: </p><ul> <li> The stability index of any level drops to $$$0$$$ or less. </li><li> Any level is left with exactly $$$0$$$ pieces (even if it is the topmost level). </li></ul><p>Find the minimum number of pieces you must remove such that the tower collapses.</p></div><div class=\"input-specification\"><div class=\"section-title\">Input</div><p>Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 10^4$$$). The description of the test cases follows.</p><p>The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n, m \\le 10^6$$$) — the number of rows and columns of the tower.</p><p>The second line contains $$$n$$$ integers $$$v_1, v_2, \\ldots, v_n$$$ ($$$1 \\le v_i \\le 10^9$$$) — the initial stability indices of each level from top to bottom.</p><p>Each of the next $$$n$$$ lines contains $$$m$$$ integers. The $$$j$$$-th integer on the $$$i$$$-th line is $$$a_{i, j}$$$ ($$$1 \\le a_{i, j} \\le 10^9$$$) — the destabilization factor of the piece at row $$$i$$$ and column $$$j$$$.</p><p>It is guaranteed that the sum of $$$n \\cdot m$$$ over all test cases does not exceed $$$10^6$$$.</p></div><div class=\"output-specification\"><div class=\"section-title\">Output</div><p>For each test case, output a single integer — the minimum number of pieces that must be removed to collapse the tower.</p></div><div class=\"sample-tests\"><div class=\"section-title\">Example</div><div class=\"sample-test\"><div class=\"input\"><div class=\"title\">Input</div><pre><div class=\"test-example-line test-example-line-even test-example-line-0\">2</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">2 3</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">10 20</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">2 2 2</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">5 5 5</div><div class=\"test-example-line test-example-line-even test-example-line-2\">3 1</div><div class=\"test-example-line test-example-line-even test-example-line-2\">100 100 100</div><div class=\"test-example-line test-example-line-even test-example-line-2\">1</div><div class=\"test-example-line test-example-line-even test-example-line-2\">2</div><div class=\"test-example-line test-example-line-even test-example-line-2\">3</div></pre></div><div class=\"output\"><div class=\"title\">Output</div><pre><div class=\"test-example-line test-example-line-odd test-example-line-1\">2</div><div class=\"test-example-line test-example-line-even test-example-line-2\">1</div></pre></div></div></div><div class=\"note\"><div class=\"section-title\">Note</div><p>In the first testcase, we have a $$$2 \\times 3$$$ tower. The initial stabilities are $$$v_1 = 10$$$ and $$$v_2 = 20$$$. The top level ($$$i=1$$$) has pieces with destabilization factors $$$2, 2, 2$$$. The bottom level ($$$i=2$$$) has pieces $$$5, 5, 5$$$.</p><p>If we remove two pieces from the bottom level, it inflicts $$$5 + 5 = 10$$$ damage to level $$$2$$$, and $$$5 + 5 = 10$$$ damage to level $$$1$$$. The remaining stability of level $$$1$$$ becomes $$$10 - 10 = 0$$$. Since its stability dropped to $$$\\le 0$$$, the tower collapses. Thus, the minimum number of pieces we must remove is $$$2$$$.</p><p>In the second testcase, $$$m = 1$$$. The rule states that the tower collapses if any level is left with exactly $$$0$$$ pieces. Since every level initially has $$$1$$$ piece, removing any $$$1$$$ piece from the tower will instantly empty a level and cause a collapse. Therefore, the minimum number of pieces to remove is $$$1$$$.</p></div>\n</problem-statement>\n\n\nSource lookup status:\n<source-lookup-status>\n- Tutorial (en): https://codeforces.com/blog/entry/155741 loaded (200 OK; title: \"Codeforces Round 1115 (Div. 2) Editorial - Codeforces\"; mentions 2252C)\n</source-lookup-status>\n\nGenerate:\n1. Five progressive hints, from a gentle nudge to the key insight.\n2. A self-contained editorial that teaches the solution clearly, including the key insight, algorithm, correctness argument, complexity, and necessary implementation details. Explain non-obvious steps without padding or a discovery-process transcript.\n3. A complete C++26 solution that gets AC on Codeforces.\n\nFormatting & Style Rules:\n- Hints and editorial: valid Markdown with LaTeX as needed (e.g., $dp[i]$, $$\\sum_{i=1}^{n} a_i$$).\n- Hints and editorial must read like Nudge's own explanation. No research notes, source notes, citations, Markdown links, URLs, or references to editorials/submissions/posts.\n- Solution: raw C++ only, no Markdown fences.\n- Keep C++ short, clean, standard, and single-file. Comments are fine when they clarify the idea.\n\nOutput strictness:\n- Return JSON matching the provided schema exactly.\n- Never include NUL (U+0000) characters in any field. Use literal $ or $$ for math delimiters, never control characters.\n- If the problem is solvable, return `status: \"success\"`, `reason: null`, and fill in `hints`, `editorial`, and `solution`.\n- Return `status: \"unsolvable\"`, a short `reason`, and set `hints`, `editorial`, and `solution` to null only when the supplied statement is fundamentally incomplete, contradictory, or dependent on an inaccessible resource needed to define the task. Missing research, high rating, uncertainty, or \"can't guarantee AC\" is not a reason to return unsolvable.\n- If you return `status: \"unsolvable\"`, the `reason` is shown directly to users. Be concrete: say what exact statement/resource blocker stopped you, and include relevant source lookup status if it mattered (for example: official tutorial missing, Codeforces returned 403/Cloudflare challenge, tutorial page 404, accepted submissions unavailable).\n- Each hint must be an object with `order` (1 through 5) and `content` (the Markdown hint text), not a plain string. The `content` must contain just the hint text, with no \"Hint 1:\" or subtitles. The UI adds those automatically.\n- Do not start the editorial with an \"# Editorial\" heading. The UI already adds that section. Feel free to add other headers as needed though.\n- Use styling in the editorial if needed! Just that first header is no good but any subheaders or bolding or whatever is good!\n\nFor the C++ solution, you MUST use this template and work around it:\n\n```cpp\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n}\n```",
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            "output": "---\nimage: https://codeforces.com/codeforces.org/s/64117/images/codeforces-telegram-square-1024x1024.png\ntitle: Codeforces Round 1115 (Div. 2) Editorial - Codeforces\ndescription: Codeforces. Programming competitions and contests, programming community\n---\n\n<!DOCTYPE HTML PUBLIC \"-//W3C//DTD HTML 4.01//EN\"> \n\n  \n[![Codeforces](//codeforces.com/codeforces.org/s/64117/images/codeforces-logo-with-telegram-555x130.png \"Codeforces\")](https://codeforces.com/) \n\n[![In English](//codeforces.com/codeforces.org/s/64117/images/flags/24/gb.png \"In English\")](https://codeforces.com/blog/entry/155741?locale=en) [![По-русски](//codeforces.com/codeforces.org/s/64117/images/flags/24/ru.png \"По-русски\")](https://codeforces.com/blog/entry/155741?locale=ru) \n\n[Enter](https://codeforces.com/enter?back=%2Fblog%2Fentry%2F155741)| [Register](https://codeforces.com/register) \n\n  \n* [Home](https://codeforces.com/)\n* [Top](https://codeforces.com/top)\n* [Catalog](https://codeforces.com/catalog)\n* [Contests](https://codeforces.com/contests)\n* [Gym](https://codeforces.com/gyms)\n* [Problemset](https://codeforces.com/problemset)\n* [Groups](https://codeforces.com/groups)\n* 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[Contests](https://codeforces.com/contests/with/Bufu)\n* [Problemsetting](https://codeforces.com/contests/writer/Bufu)\n\n### [Bufu's blog](https://codeforces.com/blog/Bufu)\n\n[ Codeforces Round 1115 (Div. 2) Editorial ](https://codeforces.com/blog/entry/155741) \n\nBy [Bufu](https://codeforces.com/profile/Bufu \"Candidate Master Bufu\"), 6 weeks ago, ![In English](//codeforces.com/codeforces.org/s/64117/images/flags/24/gb.png \"In English\") \n\nThank you for participating in Codeforces Round 1115 (Div. 2)! We hope you enjoyed the problems.\n\nThe round was prepared by [Bufu](https://codeforces.com/profile/Bufu \"Candidate Master Bufu\") and [TomitaMatei](https://codeforces.com/profile/TomitaMatei \"Master TomitaMatei\"), and coordinated by [cry](https://codeforces.com/profile/cry \"International Master cry\").\n\n---\n\n#### [2252A - Boss Fight](https://codeforces.com/contest/2252/problem/A \"Codeforces Round 1115 (Div. 2)\")\n\n**Solution**\n\nThe boss's shield permanently activates and blocks all subsequent damage only when two cards of the exact same value are played back-to-back. Importantly, the card that actually triggers the shield still deals its normal damage.\n\nThis means we want to play as many cards as possible _before_ being forced to place two identical cards adjacent to each other. We should only do this on the very last card we play.\n\nLet the most frequent card in our hand have a damage value of $$$X$$$, and let its frequency be $$$F$$$. Let the number of all _other_ cards be $$$O = n - F$$$.\n\nTo maximize the number of $$$X$$$ cards we can play, we must use the $$$O$$$ \"other\" cards as separators. By placing an \"other\" card between every $$$X$$$ card, we can form an alternating sequence like this:\n\n$$$X, \\\\text{other}\\_1, X, \\\\text{other}\\_2, \\\\ldots, X, \\\\text{other}\\_O, X$$$\n\nThis allows us to safely play $$$O + 1$$$ copies of $$$X$$$ without triggering the shield.\n\nFinally, to squeeze out the maximum possible damage, we can play exactly one more $$$X$$$ card at the very end of this sequence:\n\n$$$X, \\\\text{other}\\_1, X, \\\\text{other}\\_2, \\\\ldots, X, \\\\text{other}\\_O, X, \\\\mathbf{X}$$$\n\nThis final $$$X$$$ triggers the shield, but it still deals its normal damage. All remaining cards in our hand will deal $$$0$$$, but that doesn't matter because we have already used all of our \"other\" cards.\n\nThus, the maximum number of $$$X$$$ cards we can successfully play is $$$(O + 1) + 1 = O + 2$$$. Of course, if $$$F \\\\lt O + 2$$$, we simply play all $$$F$$$ of our $$$X$$$ cards without any issues.\n\nThe maximum number of majority cards we can play is $$$\\\\min(F, O + 2)$$$. Since we always play every single \"other\" card, the maximum total damage is:\n\n$$$\\\\text{Sum of all \"other\" cards} + X \\\\cdot \\\\min(F, O + 2)$$$\n\n**Time complexity:** $$$\\\\mathcal{O}(n)$$$.\n\n**Code (C++)**\n\n```\n#include <iostream>\n#include <vector>\n#include <map>\n#include <algorithm>\n \nusing namespace std;\n \nvoid solve() {\n    int n;\n    cin >> n;\n    long long total_sum = 0;\n    map<int, int> freq;\n    \n    int max_f = 0;\n    int maj_val = 0;\n    \n    for (int i = 0; i < n; i++) {\n        int x;\n        cin >> x;\n        total_sum += x;\n        freq[x]++;\n        if (freq[x] > max_f) {\n            max_f = freq[x];\n            maj_val = x;\n        }\n    }\n    \n    int others = n - max_f;\n    \n    int max_majority_played = min(max_f, others + 2);\n    \n    long long ans = (total_sum - 1LL * max_f * maj_val) + 1LL * max_majority_played * maj_val;\n    cout << ans << \"\\n\";\n}\n \nint main() {\n    ios_base::sync_with_stdio(false);\n    cin.tie(NULL);\n    int t; cin >> t;\n    while(t--) {\n        solve();\n    }\n    return 0;\n}\n\n```\n\n---\n\n#### [2252B - Always Changing](https://codeforces.com/contest/2252/problem/B \"Codeforces Round 1115 (Div. 2)\")\n\n**Solution**\n\nLet $$$n\\_0$$$ and $$$n\\_1$$$ be the number of `0`s and `1`s in the initial string. Let $$$k\\_0$$$ and $$$k\\_1$$$ be the number of `0`s and `1`s in the final alternating string. Let $$$c\\_0$$$ and $$$c\\_1$$$ be the number of `0`s and `1`s we deleted.\n\nBecause our deletions must strictly alternate, the difference between the number of deleted `0`s and `1`s can be at most $$$1$$$. Thus, $$$|c\\_0 - c\\_1| \\\\le 1$$$. We know that $$$c\\_0 = n\\_0 - k\\_0$$$ and $$$c\\_1 = n\\_1 - k\\_1$$$.\n\nLet $$$\\\\Delta n = n\\_0 - n\\_1$$$ and $$$\\\\Delta k = k\\_0 - k\\_1$$$. Substituting our equations, the condition $$$|c\\_0 - c\\_1| \\\\le 1$$$ translates to:\n\n$$$|\\\\Delta n - \\\\Delta k| \\\\le 1$$$\n\nSince the final string is strictly alternating, $$$\\\\Delta k$$$ can only be $$$-1$$$, $$$0$$$, or $$$1$$$. This means that if $$$|\\\\Delta n| \\\\gt 2$$$, it is mathematically impossible to satisfy the inequality, and we should output $$$-1$$$.\n\nIf it is possible ($$$|\\\\Delta n| \\\\le 2$$$), we want to maximize the length of our final string (which minimizes deletions). We can first greedily \"compress\" the initial string by removing all adjacent duplicate characters (e.g., `00011001` becomes `0101`). Let the length of this compressed alternating string be $$$L$$$, and let $$$\\\\Delta L$$$ be the difference between the number of `0`s and `1`s in it.\n\nTo ensure our sequence of deletions is valid, we must satisfy the condition $$$|\\\\Delta n - \\\\Delta k| \\\\le 1$$$, meaning our final $$$\\\\Delta k$$$ must be in the range $$$\\[\\\\Delta n - 1, \\\\Delta n + 1\\]$$$. We can safely remove any number of characters from the _beginning_ or the _end_ of our compressed alternating string, and it will remain alternating. Each character we remove shifts $$$\\\\Delta L$$$ by exactly $$$+1$$$ or $$$-1$$$.\n\nThus, to fix the parity and enter the valid range, the minimum number of characters we must drop from the ends of our compressed string is exactly the mathematical distance from $$$\\\\Delta L$$$ to the range $$$\\[\\\\Delta n - 1, \\\\Delta n + 1\\]$$$. This distance is exactly $$$\\\\max(0, |\\\\Delta n - \\\\Delta L| - 1)$$$.\n\nThe total number of deleted characters is the number of characters removed during compression $$$(n - L)$$$, plus the characters removed from the ends to fix the parity. Final answer: $$$(n - L) + \\\\max(0, |\\\\Delta n - \\\\Delta L| - 1)$$$.\n\n**Time complexity:** $$$\\\\mathcal{O}(n)$$$.\n\n**Code (C++)**\n\n```\n#include <iostream>\n#include <string>\n#include <cmath>\n#include <algorithm>\n \nusing namespace std;\n \nvoid solve() {\n    int n; cin >> n;\n    string s; cin >> s;\n \n    int n0 = 0, n1 = 0;\n    for (char c : s) {\n        if (c == '0') n0++;\n        else n1++;\n    }\n \n    int delta_n = n0 - n1;\n    if (abs(delta_n) > 2) {\n        cout << -1 << \"\\n\";\n        return;\n    }\n \n    int L = 1;\n    int L0 = (s[0] == '0' ? 1 : 0);\n    int L1 = (s[0] == '1' ? 1 : 0);\n \n    for (int i = 1; i < n; i++) {\n        if (s[i] != s[i - 1]) {\n            L++;\n            if (s[i] == '0') L0++;\n            else L1++;\n        }\n    }\n \n    int delta_L = L0 - L1;\n    int ans = (n - L) + max(0, abs(delta_n - delta_L) - 1);\n    cout << ans << \"\\n\";\n}\n \nint main() {\n    ios_base::sync_with_stdio(false);\n    cin.tie(NULL);\n    \n    int t; cin >> t;\n    while(t--) {\n        solve();\n    }\n    \n    return 0;\n}\n\n```\n\n---\n\n#### [2252C - Risky Tower](https://codeforces.com/contest/2252/problem/C \"Codeforces Round 1115 (Div. 2)\")\n\n**Solution**\n\nThe tower collapses if all $$$m$$$ pieces of any single level are removed. This guarantees that the answer is at most $$$m$$$.\n\nLet's write the condition to collapse a specific row $$$k$$$ via damage. The sum of destabilization factors of the removed pieces from rows $$$i \\\\ge k$$$ must satisfy:\n\n$$$\\\\sum a\\_{i,j} \\\\ge v\\_k$$$\n\nTo satisfy this inequality using the minimum number of pieces, we must greedily select pieces with the largest $$$a\\_{i,j}$$$ from rows at and below $$$k$$$.\n\nSince the answer will never exceed $$$m$$$, we only need to keep track of at most $$$m$$$ pieces at any time. We can process the grid from the bottom row ($$$n$$$) to the top row ($$$1$$$), maintaining a priority queue or sorted list of the largest $$$m$$$ destabilization factors $$$a\\_{i,j}$$$ seen so far.\n\nAt each row $$$k$$$, we insert its $$$m$$$ pieces into our active pool, discard all but the largest $$$m$$$ elements overall, and find the minimum number of largest elements that sum to at least $$$v\\_k$$$. The minimum number of pieces required across all rows $$$k$$$, bounded by $$$m$$$, is our answer.\n\n**Time complexity:** $$$\\\\mathcal{O}(n \\\\cdot m \\\\log m)$$$\n\n**Code (C++)**\n\n```\n#include <iostream>\n#include <vector>\n#include <algorithm>\n \nusing namespace std;\n \nvoid solve() {\n    int n, m;\n    cin >> n >> m;\n    vector<long long> v(n + 1);\n    for(int i = 1; i <= n; i++) cin >> v[i];\n    \n    vector<vector<long long>> a(n + 1, vector<long long>(m));\n    for(int i = 1; i <= n; i++) {\n        for(int j = 0; j < m; j++) {\n            cin >> a[i][j];\n        }\n    }\n    \n    int min_ans = m; \n    vector<long long> best;\n    \n    for(int k = n; k >= 1; k--) {\n        for(int j = 0; j < m; j++) {\n            best.push_back(a[k][j]);\n        }\n        \n        sort(best.rbegin(), best.rend());\n        \n        if ((int)best.size() > m) {\n            best.resize(m);\n        }\n        \n        long long current_sum = 0;\n        int pieces = 0;\n        \n        for(long long val : best) {\n            current_sum += val;\n            pieces++;\n            if (current_sum >= v[k]) {\n                min_ans = min(min_ans, pieces);\n                break;\n            }\n        }\n    }\n    \n    cout << min_ans << \"\\n\";\n}\n \nint main() {\n    ios_base::sync_with_stdio(false);\n    cin.tie(NULL);\n    int t; cin >> t;\n    while(t--) {\n        solve();\n    }\n    return 0;\n}\n\n```\n\n---\n\n#### [2252D - Array Replacement](https://codeforces.com/contest/2252/problem/D \"Codeforces Round 1115 (Div. 2)\")\n\n**Solution**\n\nLet's consider the difference array $$$d$$$, where $$$d\\_i = a\\_{i+1} - a\\_i$$$ for all $$$1 \\\\le i \\\\lt n$$$.\n\nConsider what happens to the array $$$a$$$ when we perform an operation on index $$$i$$$. The triplet $$$(a\\_{i-1}, a\\_i, a\\_{i+1})$$$ becomes $$$(a\\_{i-1}, a\\_{i-1} - a\\_i + a\\_{i+1}, a\\_{i+1})$$$. Let's see how this affects our difference array $$$d$$$:\n\n* The new $$$d\\_{i-1}$$$ becomes $$$(a\\_{i-1} - a\\_i + a\\_{i+1}) - a\\_{i-1} = a\\_{i+1} - a\\_i$$$, which is exactly the old $$$d\\_i$$$.\n* The new $$$d\\_i$$$ becomes $$$a\\_{i+1} - (a\\_{i-1} - a\\_i + a\\_{i+1}) = a\\_i - a\\_{i-1}$$$, which is exactly the old $$$d\\_{i-1}$$$.\n\nThus, performing the operation on index $$$i$$$ is exactly equivalent to **swapping** $$$d\\_{i-1}$$$ and $$$d\\_i$$$!\n\nHowever, we can only perform this operation if $$$a\\_{i-1}$$$ and $$$a\\_{i+1}$$$ have the same parity. Notice that $$$a\\_{i+1} - a\\_{i-1} = (a\\_{i+1} - a\\_i) + (a\\_i - a\\_{i-1}) = d\\_i + d\\_{i-1}$$$. For $$$a\\_{i-1}$$$ and $$$a\\_{i+1}$$$ to have the same parity, their difference $$$d\\_i + d\\_{i-1}$$$ must be an even number. The sum of two integers is even if and only if they have the **same parity**.\n\nTherefore, the condition that $$$a\\_{i-1}$$$ and $$$a\\_{i+1}$$$ have the same parity is equivalent to saying $$$d\\_{i-1}$$$ and $$$d\\_i$$$ have the same parity.\n\nThis means we can swap any two adjacent elements in the difference array $$$d$$$ as long as they share the same parity. This allows us to independently sort any contiguous subsegment of $$$d$$$ that consists entirely of even numbers, or entirely of odd numbers.\n\nTo construct the lexicographically smallest array $$$a$$$, we simply group contiguous elements of $$$d$$$ that share the same parity, sort each group in non-decreasing order, and then reconstruct the array $$$a$$$ starting from $$$a\\_1$$$.\n\n**Time complexity:** $$$\\\\mathcal{O}(n \\\\log n)$$$.\n\n**Code (C++)**\n\n```\n#include <iostream>\n#include <vector>\n#include <algorithm>\n#include <cmath>\n \nusing namespace std;\n \nvoid solve() {\n    int n;\n    cin >> n;\n    vector<long long> a(n);\n    for (int i = 0; i < n; i++) {\n        cin >> a[i];\n    }\n    \n    vector<long long> d(n - 1);\n    for (int i = 0; i < n - 1; i++) {\n        d[i] = a[i + 1] - a[i];\n    }\n    \n    for (int i = 0; i < n - 1; ) {\n        int j = i;\n        while (j < n - 1 && (d[j] & 1) == (d[i] & 1)) {\n            j++;\n        }\n        sort(d.begin() + i, d.begin() + j);\n        i = j;\n    }\n    \n    cout << a[0] << \" \";\n    long long cur = a[0];\n    for (int i = 0; i < n - 1; i++) {\n        cur += d[i];\n        cout << cur << \" \";\n    }\n    cout << \"\\n\";\n}\n \nint main() {\n    ios_base::sync_with_stdio(false);\n    cin.tie(NULL);\n    int t; cin >> t;\n    while(t--) {\n        solve();\n    }\n    return 0;\n}\n\n```\n\n---\n\n#### [2252E - Generational Triplets](https://codeforces.com/contest/2252/problem/E \"Codeforces Round 1115 (Div. 2)\")\n\n**Solution**\n\nFrom the arithmetic progression condition,\n\n$$$2b = a + c.$$$\n\nFrom the XOR condition,\n\n$$$b = a \\\\oplus c.$$$\n\nSubstituting the second equation into the first gives\n\n$$$2(a \\\\oplus c) = a + c.$$$\n\nLet $$$x$$$ be the bitwise AND of $$$a$$$ and $$$c$$$, i.e. `x = a&c`. Recall the standard bitwise addition identity\n\n$$$a + c = (a \\\\oplus c) + 2x.$$$\n\nTherefore,\n\n$$$a \\\\oplus c = 2x.$$$\n\nEquivalently, in bitwise notation, `a^c = 2(a&c)`.\n\nThus, it is enough to count pairs $$$(a,c)$$$ such that $$$a \\\\lt c \\\\le n$$$ and `a^c = 2(a&c)`.\n\nFor every such pair, $$$b$$$ is uniquely determined as $$$a \\\\oplus c$$$. Also, $$$a \\\\lt b \\\\lt c$$$ follows from the two original conditions. We only need to enforce $$$a \\\\lt c$$$: if $$$a=c$$$, then the equation becomes $$$0=2a$$$, which is impossible because $$$a \\\\ge 1$$$.\n\nWe solve this with digit DP, processing bits from $$$60$$$ down to $$$0$$$. Let the state be\n\n$$$\\\\mathrm{dp}\\[\\\\mathrm{bit}\\]\\[\\\\mathrm{cLess}\\]\\[\\\\mathrm{aLess}\\]\\[\\\\mathrm{need}\\].$$$\n\n* `bit`: the bit currently being processed.\n* `cLess`: whether the already processed prefix of $$$c$$$ is strictly smaller than the corresponding prefix of $$$n$$$.\n* `aLess`: whether the already processed prefix of $$$a$$$ is strictly smaller than the corresponding prefix of $$$c$$$.\n* `need`: the value required for the current bit of `a&c`, determined by the XOR value at the previous, more significant bit.\n\nAt each state, try all four choices for the current bits of $$$a$$$ and $$$c$$$. A transition is valid when:\n\n1. The chosen prefix of $$$c$$$ does not exceed the prefix of $$$n$$$.\n2. The chosen prefix of $$$a$$$ does not exceed the prefix of $$$c$$$.\n3. The current bit of `a&c` equals `need`.\n\nFor the next lower bit, the new value of `need` is the XOR of the current bits of $$$a$$$ and $$$c$$$. This represents the one-bit left shift in `a^c = 2(a&c)`.\n\nAfter processing all bits, the state contributes $$$1$$$ exactly when `need` is $$$0$$$ and `aLess` is true, which guarantees $$$a \\\\lt c$$$.\n\n**Time complexity:** $$$\\\\mathcal{O}(\\\\log n)$$$ per test case.\n\n**Code (C++)**\n\n```\n#include <iostream>\n#include <vector>\n#include <cstring>\n \nusing namespace std;\n \nlong long dp[65][2][2][2];\nlong long n;\n \nconst int MOD = 1000000007;\n \nlong long solve_dp(int bit, int c_less, int a_less, int need) {\n    if (bit < 0) {\n        if (need == 0 && a_less == 1) return 1;\n        return 0;\n    }\n    \n    if (dp[bit][c_less][a_less][need] != -1) {\n        return dp[bit][c_less][a_less][need];\n    }\n    \n    long long ans = 0;\n    int n_bit = (n >> bit) & 1;\n    \n    for (int a_bit = 0; a_bit <= 1; a_bit++) {\n        for (int c_bit = 0; c_bit <= 1; c_bit++) {\n            // Check constraint against n\n            if (c_less && c_bit > n_bit) continue;\n            \n            // Check constraint a <= c (since we need a < c eventually)\n            if (!a_less && a_bit > c_bit) continue;\n            \n            // Check the \"need\" condition for (a & c)\n            if (need == 1 && (a_bit & c_bit) == 0) continue;\n            if (need == 0 && (a_bit & c_bit) == 1) continue;\n            \n            int next_c_less = c_less && (c_bit == n_bit);\n            int next_a_less = a_less || (a_bit < c_bit);\n            int next_need = a_bit ^ c_bit; // The next bit's & must match this bit's ^\n            \n            ans = (ans + solve_dp(bit - 1, next_c_less, next_a_less, next_need)) % MOD;\n        }\n    }\n    \n    return dp[bit][c_less][a_less][need] = ans;\n}\n \nvoid solve() {\n    cin >> n;\n    memset(dp, -1, sizeof(dp));\n    // Start at bit 60, c is constrained by n, a is not yet less than c, need = 0\n    cout << solve_dp(60, 1, 0, 0) << \"\\n\";\n}\n \nint main() {\n    ios_base::sync_with_stdio(false);\n    cin.tie(NULL);\n    int t; cin >> t;\n    while(t--) {\n        solve();\n    }\n    return 0;\n}\n\n```\n\n---\n\n#### [2252F - Spectral Components](https://codeforces.com/contest/2252/problem/F \"Codeforces Round 1115 (Div. 2)\")\n\n**Solution**\n\nLet's solve the problem independently for a single color $$$c$$$ with $$$M$$$ vertices.\n\nFirst, consider the case where $$$k = 1$$$. The optimal component is a single vertex, which mathematically corresponds to the **centroid** of the $$$M$$$ vertices. For any edge $$$e$$$ in the tree, let $$$m\\_{\\\\mathrm{in}}\\[e\\]$$$ and $$$m\\_{\\\\mathrm{out}}\\[e\\]$$$ be the number of color $$$c$$$ vertices on either side of the edge. The number of paths from color $$$c$$$ vertices that must cross edge $$$e$$$ to reach the centroid is exactly $$$w\\[e\\] = \\\\min(m\\_{\\\\mathrm{in}}\\[e\\], m\\_{\\\\mathrm{out}}\\[e\\])$$$. Thus, the total distance to the centroid is:\n\n$$$D(\\\\mathrm{centroid}) = \\\\sum w\\[e\\]$$$\n\nwhere the sum is over all edges $$$e$$$ in the tree.\n\nWhen $$$k \\\\gt 1$$$, we must choose a connected component $$$S$$$ of $$$k$$$ vertices (which contains exactly $$$k - 1$$$ edges). Notice that if an edge $$$e$$$ is entirely inside $$$S$$$, no path from any colored vertex to $$$S$$$ needs to cross it (the distance is absorbed into the component). If an edge is outside $$$S$$$, exactly $$$w\\[e\\]$$$ paths still cross it. Thus, the cost of any valid component $$$S$$$ containing the centroid is:\n\n$$$\\\\mathrm{Cost}(S) = D(\\\\mathrm{centroid}) - \\\\sum\\_{e \\\\in S} w\\[e\\]$$$\n\nTo minimize $$$\\\\mathrm{Cost}(S)$$$, we must **maximize** the sum of $$$w\\[e\\]$$$ for the $$$k - 1$$$ edges we pick. A crucial property of tree centroids is that $$$w\\[e\\]$$$ is monotonically decreasing as you move away from the centroid. Because of this monotonicity, if we simply greedily pick the $$$k - 1$$$ edges with the globally maximum $$$w\\[e\\]$$$ values in the entire tree, they are mathematically guaranteed to form a connected component containing the centroid!\n\nHowever, calculating this naively for all colors takes $$$\\\\mathcal{O}(N^2)$$$. To achieve the required time complexity, we process all colors using **Virtual Trees** (also known as Auxiliary Trees):\n\n1. For each color $$$c$$$, build a Virtual Tree of its $$$M$$$ vertices in $$$\\\\mathcal{O}(M \\\\log M)$$$ time using LCA.\n2. Run a bottom-up DP on the Virtual Tree to find $$$m\\_{\\\\mathrm{in}}$$$ for each virtual subtree.\n3. A virtual edge $$$(u, v)$$$ represents a compressed path of $$$L = \\\\operatorname{depth}\\[v\\] - \\\\operatorname{depth}\\[u\\]$$$ real edges. Because there are no other vertices of color $$$c$$$ branching off this path, every single real edge on this path shares the exact same weight: $$$w' = \\\\min(m\\_{\\\\mathrm{in}}, m\\_c - m\\_{\\\\mathrm{in}})$$$.\n4. For color $$$c$$$, collect the pairs (weight = $$$w'$$$, count = $$$L$$$) for all virtual edges.\n5. Sort these pairs by weight descending, and greedily pick up to $$$K - 1$$$ edges to subtract from the total sum of all $$$w\\[e\\]$$$.\n\nSince the sum of sizes of all Virtual Trees over all colors is $$$\\\\mathcal{O}(n)$$$, the overall time complexity is $$$\\\\mathcal{O}(n \\\\log n)$$$.\n\n**Code (C++)**\n\n```\n#include <iostream>\n#include <vector>\n#include <algorithm>\n \nusing namespace std;\n \nconst int LOG = 19;\nint n;\nvector<int> C, K;\nvector<vector<int>> adj;\nvector<int> tin, tout, depth;\nvector<vector<int>> up;\nint timer = 0;\n \nvoid dfs(int v, int p, int d) {\n    tin[v] = ++timer;\n    depth[v] = d;\n    up[v][0] = p;\n    for (int i = 1; i < LOG; i++) {\n        up[v][i] = up[up[v][i - 1]][i - 1];\n    }\n    for (int u : adj[v]) {\n        if (u != p) dfs(u, v, d + 1);\n    }\n    tout[v] = ++timer;\n}\n \nbool is_ancestor(int u, int v) {\n    return tin[u] <= tin[v] && tout[u] >= tout[v];\n}\n \nint get_lca(int u, int v) {\n    if (is_ancestor(u, v)) return u;\n    if (is_ancestor(v, u)) return v;\n    for (int i = LOG - 1; i >= 0; i--) {\n        if (!is_ancestor(up[u][i], v)) u = up[u][i];\n    }\n    return up[u][0];\n}\n \nvoid solve() {\n    cin >> n;\n    C.assign(n + 1, 0);\n    K.assign(n + 1, 0);\n    adj.assign(n + 1, vector<int>());\n    tin.assign(n + 1, 0);\n    tout.assign(n + 1, 0);\n    depth.assign(n + 1, 0);\n    up.assign(n + 1, vector<int>(LOG, 1));\n    timer = 0;\n \n    vector<vector<int>> colors(n + 1);\n    for (int i = 1; i <= n; i++) {\n        cin >> C[i];\n        colors[C[i]].push_back(i);\n    }\n    for (int i = 1; i <= n; i++) cin >> K[i];\n    \n    for (int i = 0; i < n - 1; i++) {\n        int u, v;\n        cin >> u >> v;\n        adj[u].push_back(v);\n        adj[v].push_back(u);\n    }\n \n    dfs(1, 1, 0);\n    \n    vector<long long> ans(n + 1, -1);\n    vector<int> cnt(n + 1, 0);\n \n    for (int c = 1; c <= n; c++) {\n        if (colors[c].empty()) continue;\n        \n        int M_c = colors[c].size();\n        int K_c = K[c];\n \n        vector<int> nodes = colors[c];\n        sort(nodes.begin(), nodes.end(), [](int a, int b) { return tin[a] < tin[b]; });\n \n        int sz = nodes.size();\n        for (int i = 0; i < sz - 1; i++) {\n            nodes.push_back(get_lca(nodes[i], nodes[i+1]));\n        }\n        \n        sort(nodes.begin(), nodes.end(), [](int a, int b) { return tin[a] < tin[b]; });\n        nodes.erase(unique(nodes.begin(), nodes.end()), nodes.end());\n \n        for (int u : colors[c]) cnt[u] = 1;\n \n        vector<int> st;\n        vector<pair<int, int>> vt_edges;\n        for (int u : nodes) {\n            while (!st.empty() && !is_ancestor(st.back(), u)) st.pop_back();\n            if (!st.empty()) vt_edges.push_back({st.back(), u});\n            st.push_back(u);\n        }\n \n        long long total_dist = 0;\n        vector<pair<long long, long long>> edge_weights; // {weight, count}\n        \n        for (int i = (int)vt_edges.size() - 1; i >= 0; i--) {\n            int p = vt_edges[i].first;\n            int u = vt_edges[i].second;\n            cnt[p] += cnt[u];\n            \n            long long w = min((long long)cnt[u], (long long)M_c - cnt[u]);\n            long long L = depth[u] - depth[p];\n            total_dist += w * L;\n            edge_weights.push_back({w, L});\n        }\n        \n        sort(edge_weights.rbegin(), edge_weights.rend());\n        \n        long long saved = 0;\n        long long needed = K_c - 1;\n        for (auto& ew : edge_weights) {\n            long long take = min(needed, ew.second);\n            saved += take * ew.first;\n            needed -= take;\n            if (needed == 0) break;\n        }\n        \n        ans[c] = total_dist - saved;\n        for (int u : nodes) cnt[u] = 0;\n    }\n \n    for (int i = 1; i <= n; i++) {\n        cout << ans[i] << (i == n ? \"\" : \" \");\n    }\n    cout << \"\\n\";\n}\n \nint main() {\n    ios_base::sync_with_stdio(false);\n    cin.tie(NULL);\n    int t; cin >> t;\n    while(t--) {\n        solve();\n    }\n    return 0;\n}\n\n```\n\n![](//codeforces.com/codeforces.org/s/64117/images/icons/paperclip-16x16.png) Tutorial of [Codeforces Round 1115 (Div. 2)](https://codeforces.com/contest/2252) \n\n* [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/voteup-gray.png \"Vote: I like it\")](#)\n* +40\n* [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/votedown-gray.png \"Vote: I do not like it\")](#)\n\n* [![Author](//codeforces.com/codeforces.org/s/64117/images/blog/user_16x16.png \"Author\")](https://codeforces.com/profile/Bufu) [Bufu ](https://codeforces.com/profile/Bufu)\n* ![Publication date](//codeforces.com/codeforces.org/s/64117/images/blog/date_16x16.png \"Publication date\") 6 weeks ago\n* [![Comments](//codeforces.com/codeforces.org/s/64117/images/blog/comments_16x16.png \"Comments\")](https://codeforces.com/blog/entry/155741#comments) [76 ](https://codeforces.com/blog/entry/155741#comments)\n\n  \n![Comments](//codeforces.com/codeforces.org/s/64117/images/icons/comments-48x48.png \"Comments\") Comments (76) \n\n[Write comment?](#) \n\n| » [overi](https://codeforces.com/profile/overi \"Newbie overi\") | 6 weeks ago, [show (+1)](#) # |\n| -------------------------------------------------------------- | ----------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/overi) [overi](https://codeforces.com/profile/overi \"Newbie overi\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155741?#comment-1383494 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +36 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) A was too hard [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [BenjaminNguyen](https://codeforces.com/profile/BenjaminNguyen \"Specialist BenjaminNguyen\") | 6 weeks ago, [show](#) # |\n| ----------------------------------------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/5891261/avatar/70e8b43e379920ea.jpg) ](https://codeforces.com/profile/BenjaminNguyen) [BenjaminNguyen](https://codeforces.com/profile/BenjaminNguyen \"Specialist BenjaminNguyen\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155741?#comment-1383617 \"Link to comment\") [^](#comment-1383494 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) its alright i guess [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n* | » » [While.1](https://codeforces.com/profile/While.1 \"Newbie While.1\") | 6 weeks ago, [show](#) # |\n| ---------------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/5072827/avatar/578ea0266e326443.jpg) ](https://codeforces.com/profile/While.1) [While.1](https://codeforces.com/profile/While.1 \"Newbie While.1\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155741?#comment-1383641 \"Link to comment\") [^](#comment-1383494 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) A was a bit tricky :( [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n* | » » [TealHaVN](https://codeforces.com/profile/TealHaVN \"Pupil TealHaVN\") | 6 weeks ago, [show](#) # |\n| ------------------------------------------------------------------------ | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/5214356/avatar/c64be108f48a1e18.jpg) ](https://codeforces.com/profile/TealHaVN) [TealHaVN](https://codeforces.com/profile/TealHaVN \"Pupil TealHaVN\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155741?#comment-1383642 \"Link to comment\") [^](#comment-1383494 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) By the 30th minute, I still couldn't solve A [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n* | » » [leidianjiangjun](https://codeforces.com/profile/leidianjiangjun \"Newbie leidianjiangjun\") | 6 weeks ago, [show](#) # |\n| ---------------------------------------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/5291895/avatar/248d184ac9f15ec1.jpg) ](https://codeforces.com/profile/leidianjiangjun) [leidianjiangjun](https://codeforces.com/profile/leidianjiangjun \"Newbie leidianjiangjun\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155741?#comment-1383651 \"Link to comment\") [^](#comment-1383494 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I agree [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n* | » » [bc8h8](https://codeforces.com/profile/bc8h8 \"Specialist bc8h8\") | 6 weeks ago, [show](#) # |\n| -------------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/5145463/avatar/2eddc84d34143a00.jpg) ](https://codeforces.com/profile/bc8h8) [bc8h8](https://codeforces.com/profile/bc8h8 \"Specialist bc8h8\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155741?#comment-1383660 \"Link to comment\") [^](#comment-1383494 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I agree!!! [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n* | » » [zth413021](https://codeforces.com/profile/zth413021 \"Expert zth413021\") | 6 weeks ago, [show](#) # |\n| ---------------------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/3597335/avatar/3ec6b38e14ac1.jpg) ](https://codeforces.com/profile/zth413021) [zth413021](https://codeforces.com/profile/zth413021 \"Expert zth413021\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155741?#comment-1383671 \"Link to comment\") [^](#comment-1383494 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I agree [→](#) [Reply](#) |\n| -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n* | » » [xaero1362](https://codeforces.com/profile/xaero1362 \"Newbie xaero1362\") | 6 weeks ago, [show](#) # |\n| ---------------------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/xaero1362) [xaero1362](https://codeforces.com/profile/xaero1362 \"Newbie xaero1362\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155741?#comment-1383701 \"Link to comment\") [^](#comment-1383494 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I agree. [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n* | » » [VaiBrahma](https://codeforces.com/profile/VaiBrahma \"Expert VaiBrahma\") | 6 weeks ago, [show](#) # |\n| ---------------------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/3164338/avatar/f83e7e795800a699.jpg) ](https://codeforces.com/profile/VaiBrahma) [VaiBrahma](https://codeforces.com/profile/VaiBrahma \"Expert VaiBrahma\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155741?#comment-1383724 \"Link to comment\") [^](#comment-1383494 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/64117/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) i agree [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n* | » » [ntkoo](https://codeforces.com/profile/ntkoo \"Newbie ntkoo\") | 2 weeks ago, [show](#) # |\n| ---------------------------------------------------------------- | ------------------------ |\n\n\n...152841 bytes truncated...\n\nThe tool call succeeded but the output was truncated. Full output saved to: /Users/zayd/.local/share/opencode/tool-output/tool_0b2529523001mgdhfAXHuB5jM2\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.",
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          "type": "reasoning",
          "text": "**Choosing flat vector sort**",
          "time": {
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              "command": "python3 -c 'import random,itertools\nrandom.seed(2252)\nfor _ in range(2500):\n n=random.randint(1,4); m=random.randint(1,4)\n a=[[random.randint(1,15) for j in range(m)] for i in range(n)]\n v=[random.randint(1,65) for i in range(n)]\n best=[]; ans=m\n for i in range(n-1,-1,-1):\n  best=sorted(best+a[i],reverse=True)[:m]\n  s=0\n  for k,x in enumerate(best,1):\n   s+=x\n   if s>=v[i]: ans=min(ans,k); break\n brute=m\n for k in range(1,m):\n  found=False\n  for chosen in itertools.combinations(range(n*m),k):\n   damage=[0]*n\n   for p in chosen:\n    i,j=divmod(p,m)\n    for r in range(i+1): damage[r]+=a[i][j]\n   if any(damage[r]>=v[r] for r in range(n)):\n    found=True; break\n  if found: brute=k; break\n assert ans==brute,(n,m,a,v,ans,brute)\nprint(\"2500 random cases matched exhaustive subset enumeration.\")'"
            },
            "output": "2500 random cases matched exhaustive subset enumeration.\n",
            "metadata": {
              "output": "2500 random cases matched exhaustive subset enumeration.\n",
              "exit": 0,
              "truncated": false
            },
            "title": "python3 -c 'import random,itertools\nrandom.seed(2252)\nfor _ in range(2500):\n n=random.randint(1,4); m=random.randint(1,4)\n a=[[random.randint(1,15) for j in range(m)] for i in range(n)]\n v=[random.randint(1,65) for i in range(n)]\n best=[]; ans=m\n for i in range(n-1,-1,-1):\n  best=sorted(best+a[i],reverse=True)[:m]\n  s=0\n  for k,x in enumerate(best,1):\n   s+=x\n   if s>=v[i]: ans=min(ans,k); break\n brute=m\n for k in range(1,m):\n  found=False\n  for chosen in itertools.combinations(range(n*m),k):\n   damage=[0]*n\n   for p in chosen:\n    i,j=divmod(p,m)\n    for r in range(i+1): damage[r]+=a[i][j]\n   if any(damage[r]>=v[r] for r in range(n)):\n    found=True; break\n  if found: brute=k; break\n assert ans==brute,(n,m,a,v,ans,brute)\nprint(\"2500 random cases matched exhaustive subset enumeration.\")'",
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      "info": {
        "parentID": "msg_0b2527b85001yvAVnZJhn14GDv",
        "role": "assistant",
        "mode": "nudge-generation",
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        "variant": "low",
        "path": {
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          "status": "success",
          "reason": null,
          "hints": [
            {
              "order": 1,
              "content": "You can always collapse the tower by emptying one row. What upper bound does that give on the answer, regardless of the stability values?"
            },
            {
              "order": 2,
              "content": "Fix a row $k$ whose stability you want to reduce to zero. Only pieces in rows $k,k+1,\\ldots,n$ damage it. If you remove exactly $x$ pieces from those rows, which choices maximize the damage?"
            },
            {
              "order": 3,
              "content": "For a fixed target row, take the eligible pieces in descending order of destabilization factor. The first prefix whose sum reaches $v_k$ gives the minimum number needed. Since the answer is at most $m$, only the largest $m$ eligible values matter."
            },
            {
              "order": 4,
              "content": "Process target rows from bottom to top. Moving from row $k+1$ to row $k$ adds exactly one row of eligible pieces. Maintain the largest $m$ values seen so far; any value already discarded can never become necessary after adding more candidates."
            },
            {
              "order": 5,
              "content": "Start with `answer = m` and an empty vector. At each row, append its $m$ values, sort descending, and keep only the first $m$. Scan their prefix sums and update the answer at the first sum at least $v_k$. Each sort handles at most $2m$ values, giving $O(nm\\log(m+1))$ time. Use 64-bit prefix sums."
            }
          ],
          "editorial": "The easiest way to destroy the tower is to remove all $m$ pieces from one row. Subtle? No. Useful? Absolutely:\n\n$$\\text{answer} \\le m.$$\n\nSo we only need to find whether stability damage can collapse the tower with fewer removals.\n\n### Fix the row that breaks\n\nSuppose we want row $k$ to reach stability zero or below. A removed piece damages this row exactly when it comes from row $k$ or below it. Thus, the eligible pieces are all pieces in rows\n\n$$k,k+1,\\ldots,n.$$\n\nWe need their selected destabilization factors to sum to at least $v_k$.\n\nFor a fixed number $x$ of removals, the maximum possible damage comes from the **$x$ largest eligible values**. If we selected a smaller value while leaving a larger one unused, swapping them could only increase the damage.\n\nTherefore, after sorting the eligible values as $b_1\\ge b_2\\ge\\cdots$, the minimum number needed for this row is the first $x$ satisfying\n\n$$\\sum_{j=1}^{x}b_j\\ge v_k.$$\n\nEquality counts: stability exactly zero already collapses the tower.\n\nWe do not need to keep other rows safe during these removals. If something else collapses earlier, that only helps. Also, fewer than $m$ removals cannot empty any row, so no extra per-row selection restrictions are needed when looking for an improvement over our initial answer.\n\n### Keep only the largest $m$ values\n\nSorting every entire suffix separately would repeat far too much work. Instead, process rows **from bottom to top**.\n\nMaintain a vector `best` containing the largest $m$ values from the rows processed so far, sorted in descending order.\n\nFor each row $k$:\n\n1. Append its $m$ values to `best`.\n2. Sort `best` in descending order.\n3. Keep only the first $m$ values.\n4. Scan prefix sums. At the first sum at least $v_k$, update the answer with that prefix length.\n\nBefore appending a row, the vector has at most $m$ elements, so each sort handles at most $2m$ elements.\n\n**Why can discarded values stay discarded?** Whenever we discard a value, we retain $m$ values at least as large. All of those pieces remain eligible as we move upward. Adding more candidates cannot make the discarded value necessary among the largest $m$. Ties do not matter; equal-valued pieces are interchangeable for this calculation.\n\nIf even the sum of the retained $m$ values is below $v_k$, this row cannot improve the answer. We still have the option of emptying a row in $m$ removals.\n\n### Example\n\nIn the first sample, the bottom row contributes $[5,5,5]$. Their total is $15$, below that row's stability $20$.\n\nMoving to the top row adds $[2,2,2]$. The largest three eligible values are still $[5,5,5]$. The top row has stability $10$, and the first two values sum to $10$, so two removals are enough.\n\n### Correctness proof\n\nWe prove that the algorithm outputs the minimum number of removals.\n\n**Lemma 1: For any target row $k$ and any $x\\le m$, the largest $x$ eligible values give the maximum damage achievable with $x$ removals.**\n\nPieces above row $k$ do not damage it. Among eligible pieces, replacing a selected value with a larger unused value never decreases the damage. Repeating such replacements yields the largest $x$ eligible values, proving the claim.\n\n**Lemma 2: After processing row $k$, `best` contains the largest $m$ values from rows $k$ through $n$.**\n\nFor the bottom row, this is immediate. For every subsequent row, previously discarded values already have $m$ retained values at least as large as them. They cannot become necessary among the largest $m$ after adding the new row. Sorting the retained values together with the new row and keeping the largest $m$ therefore preserves the claim.\n\n**Theorem: The algorithm returns the optimal answer.**\n\nLet the optimal answer be $r$. Emptying a row shows that $r\\le m$.\n\nEvery candidate recorded by the algorithm guarantees collapse within that many removals: either it is the initial bound $m$, or its chosen pieces deal enough damage to some row. Hence, the returned answer cannot be smaller than $r$.\n\nIf $r<m$, an optimal sequence cannot empty a row, so some row $k$ must collapse from damage. By Lemma 1, the largest $r$ eligible values deal at least as much damage as that sequence. By Lemma 2, those values are present in `best` when row $k$ is processed. The prefix scan therefore finds a candidate of at most $r$.\n\nIf $r=m$, the initial answer already equals $r$. In both cases, the returned answer is exactly optimal.\n\n### Complexity and implementation details\n\nEach of the $n$ rows requires sorting at most $2m$ values and scanning at most $m$ values:\n\n- **Time:** $O(nm\\log(m+1))$, including the $m=1$ case.\n- **Space:** $O(nm)$ to store the grid, plus $O(n+m)$ for stability values and `best`.\n\nThe grid is stored because input arrives top-to-bottom while processing runs bottom-to-top. A flat vector avoids allocating a separate vector for every row.\n\nIndividual values fit in `int`, but prefix sums can reach $10^{15}$, so use `long long`. When $m=1$, the answer is always $1$, which the same algorithm handles naturally.",
          "solution": "#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n\n    int t;\n    cin >> t;\n    while (t--) {\n        int n, m;\n        cin >> n >> m;\n\n        vector<int> v(n), a(n * m);\n        for (int &x : v) cin >> x;\n        for (int &x : a) cin >> x;\n\n        int answer = m;\n        vector<int> best;\n        best.reserve(2 * m);\n\n        for (int i = n - 1; i >= 0; --i) {\n            for (int j = 0; j < m; ++j) {\n                best.push_back(a[i * m + j]);\n            }\n            sort(best.begin(), best.end(), greater<int>());\n            best.resize(m);\n\n            ll sum = 0;\n            for (int j = 0; j < m; ++j) {\n                sum += best[j];\n                if (sum >= v[i]) {\n                    answer = min(answer, j + 1);\n                    break;\n                }\n            }\n        }\n\n        cout << answer << '\\n';\n    }\n}\n"
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          "state": {
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              "status": "success",
              "reason": null,
              "hints": [
                {
                  "order": 1,
                  "content": "You can always collapse the tower by emptying one row. What upper bound does that give on the answer, regardless of the stability values?"
                },
                {
                  "order": 2,
                  "content": "Fix a row $k$ whose stability you want to reduce to zero. Only pieces in rows $k,k+1,\\ldots,n$ damage it. If you remove exactly $x$ pieces from those rows, which choices maximize the damage?"
                },
                {
                  "order": 3,
                  "content": "For a fixed target row, take the eligible pieces in descending order of destabilization factor. The first prefix whose sum reaches $v_k$ gives the minimum number needed. Since the answer is at most $m$, only the largest $m$ eligible values matter."
                },
                {
                  "order": 4,
                  "content": "Process target rows from bottom to top. Moving from row $k+1$ to row $k$ adds exactly one row of eligible pieces. Maintain the largest $m$ values seen so far; any value already discarded can never become necessary after adding more candidates."
                },
                {
                  "order": 5,
                  "content": "Start with `answer = m` and an empty vector. At each row, append its $m$ values, sort descending, and keep only the first $m$. Scan their prefix sums and update the answer at the first sum at least $v_k$. Each sort handles at most $2m$ values, giving $O(nm\\log(m+1))$ time. Use 64-bit prefix sums."
                }
              ],
              "editorial": "The easiest way to destroy the tower is to remove all $m$ pieces from one row. Subtle? No. Useful? Absolutely:\n\n$$\\text{answer} \\le m.$$\n\nSo we only need to find whether stability damage can collapse the tower with fewer removals.\n\n### Fix the row that breaks\n\nSuppose we want row $k$ to reach stability zero or below. A removed piece damages this row exactly when it comes from row $k$ or below it. Thus, the eligible pieces are all pieces in rows\n\n$$k,k+1,\\ldots,n.$$\n\nWe need their selected destabilization factors to sum to at least $v_k$.\n\nFor a fixed number $x$ of removals, the maximum possible damage comes from the **$x$ largest eligible values**. If we selected a smaller value while leaving a larger one unused, swapping them could only increase the damage.\n\nTherefore, after sorting the eligible values as $b_1\\ge b_2\\ge\\cdots$, the minimum number needed for this row is the first $x$ satisfying\n\n$$\\sum_{j=1}^{x}b_j\\ge v_k.$$\n\nEquality counts: stability exactly zero already collapses the tower.\n\nWe do not need to keep other rows safe during these removals. If something else collapses earlier, that only helps. Also, fewer than $m$ removals cannot empty any row, so no extra per-row selection restrictions are needed when looking for an improvement over our initial answer.\n\n### Keep only the largest $m$ values\n\nSorting every entire suffix separately would repeat far too much work. Instead, process rows **from bottom to top**.\n\nMaintain a vector `best` containing the largest $m$ values from the rows processed so far, sorted in descending order.\n\nFor each row $k$:\n\n1. Append its $m$ values to `best`.\n2. Sort `best` in descending order.\n3. Keep only the first $m$ values.\n4. Scan prefix sums. At the first sum at least $v_k$, update the answer with that prefix length.\n\nBefore appending a row, the vector has at most $m$ elements, so each sort handles at most $2m$ elements.\n\n**Why can discarded values stay discarded?** Whenever we discard a value, we retain $m$ values at least as large. All of those pieces remain eligible as we move upward. Adding more candidates cannot make the discarded value necessary among the largest $m$. Ties do not matter; equal-valued pieces are interchangeable for this calculation.\n\nIf even the sum of the retained $m$ values is below $v_k$, this row cannot improve the answer. We still have the option of emptying a row in $m$ removals.\n\n### Example\n\nIn the first sample, the bottom row contributes $[5,5,5]$. Their total is $15$, below that row's stability $20$.\n\nMoving to the top row adds $[2,2,2]$. The largest three eligible values are still $[5,5,5]$. The top row has stability $10$, and the first two values sum to $10$, so two removals are enough.\n\n### Correctness proof\n\nWe prove that the algorithm outputs the minimum number of removals.\n\n**Lemma 1: For any target row $k$ and any $x\\le m$, the largest $x$ eligible values give the maximum damage achievable with $x$ removals.**\n\nPieces above row $k$ do not damage it. Among eligible pieces, replacing a selected value with a larger unused value never decreases the damage. Repeating such replacements yields the largest $x$ eligible values, proving the claim.\n\n**Lemma 2: After processing row $k$, `best` contains the largest $m$ values from rows $k$ through $n$.**\n\nFor the bottom row, this is immediate. For every subsequent row, previously discarded values already have $m$ retained values at least as large as them. They cannot become necessary among the largest $m$ after adding the new row. Sorting the retained values together with the new row and keeping the largest $m$ therefore preserves the claim.\n\n**Theorem: The algorithm returns the optimal answer.**\n\nLet the optimal answer be $r$. Emptying a row shows that $r\\le m$.\n\nEvery candidate recorded by the algorithm guarantees collapse within that many removals: either it is the initial bound $m$, or its chosen pieces deal enough damage to some row. Hence, the returned answer cannot be smaller than $r$.\n\nIf $r<m$, an optimal sequence cannot empty a row, so some row $k$ must collapse from damage. By Lemma 1, the largest $r$ eligible values deal at least as much damage as that sequence. By Lemma 2, those values are present in `best` when row $k$ is processed. The prefix scan therefore finds a candidate of at most $r$.\n\nIf $r=m$, the initial answer already equals $r$. In both cases, the returned answer is exactly optimal.\n\n### Complexity and implementation details\n\nEach of the $n$ rows requires sorting at most $2m$ values and scanning at most $m$ values:\n\n- **Time:** $O(nm\\log(m+1))$, including the $m=1$ case.\n- **Space:** $O(nm)$ to store the grid, plus $O(n+m)$ for stability values and `best`.\n\nThe grid is stored because input arrives top-to-bottom while processing runs bottom-to-top. A flat vector avoids allocating a separate vector for every row.\n\nIndividual values fit in `int`, but prefix sums can reach $10^{15}$, so use `long long`. When $m=1$, the answer is always $1$, which the same algorithm handles naturally.",
              "solution": "#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n\n    int t;\n    cin >> t;\n    while (t--) {\n        int n, m;\n        cin >> n >> m;\n\n        vector<int> v(n), a(n * m);\n        for (int &x : v) cin >> x;\n        for (int &x : a) cin >> x;\n\n        int answer = m;\n        vector<int> best;\n        best.reserve(2 * m);\n\n        for (int i = n - 1; i >= 0; --i) {\n            for (int j = 0; j < m; ++j) {\n                best.push_back(a[i * m + j]);\n            }\n            sort(best.begin(), best.end(), greater<int>());\n            best.resize(m);\n\n            ll sum = 0;\n            for (int j = 0; j < m; ++j) {\n                sum += best[j];\n                if (sum >= v[i]) {\n                    answer = min(answer, j + 1);\n                    break;\n                }\n            }\n        }\n\n        cout << answer << '\\n';\n    }\n}\n"
            },
            "output": "Structured output captured successfully.",
            "metadata": {
              "valid": true
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            "title": "Structured Output",
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