{
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    "slug": "mighty-garden",
    "projectID": "global",
    "directory": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-PmhcCD",
    "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-PmhcCD",
    "title": "Nudge content generation",
    "agent": "nudge-generation",
    "model": {
      "id": "gpt-6-astra",
      "providerID": "openai",
      "variant": "low"
    },
    "version": "1.18.30",
    "summary": {
      "additions": 0,
      "deletions": 0,
      "files": 0
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    "cost": 0,
    "tokens": {
      "input": 24772,
      "output": 3339,
      "reasoning": 266,
      "cache": {
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      "updated": 1789260608541
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  "messages": [
    {
      "info": {
        "role": "user",
        "time": {
          "created": 1789260480257
        },
        "agent": "nudge-generation",
        "model": {
          "providerID": "openai",
          "modelID": "gpt-6-astra",
          "variant": "low"
        },
        "system": "You create Codeforces learning content for Nudge. Your job is to understand the best approach for the problem and teach it clearly in your own words, using existing solutions to inform your reasoning.\n\nStart with a brief, targeted web search for existing solutions so relevant editorials are in context before substantial independent problem-solving. Use any supplied tutorial links first; otherwise search by contest ID and problem index. Prefer the official tutorial/editorial, and read relevant explanations rather than relying on search snippets. Look at accessible alternatives when useful, especially if the official explanation is missing, unclear, or unnecessarily complicated. Accepted submissions can also help clarify an approach. Respect supplied source lookup statuses and avoid repeatedly retrying blocked sources. If no usable source is accessible, proceed from the supplied statement.\n\nTreat existing solutions as a starting point, not a constraint. If an approach is clear and suitable, build on it without unnecessary re-derivation. If an explanation has gaps, seems incorrect, or uses a harder method than necessary, reason through it and derive a better or simpler approach when worthwhile. Choose a correct approach that is easy to teach and fits the constraints. Verify it against the supplied statement and edge cases before writing the hints, editorial, and C++ solution.\n\nThe editorial must be self-contained: explain the key insight, algorithm, why it works, complexity, and implementation details a learner needs, filling in steps the source skips. Keep research and verification concise; spend the output on teaching the solution, not narrating the research process. Follow the user prompt and output schema exactly.",
        "format": {
          "type": "json_schema",
          "schema": {
            "type": "object",
            "properties": {
              "status": {
                "type": "string",
                "enum": [
                  "success",
                  "unsolvable"
                ],
                "description": "Set to 'success' when the supplied statement defines a solvable programming problem. Set to 'unsolvable' only if the statement is fundamentally incomplete, contradictory, or depends on an inaccessible required resource."
              },
              "reason": {
                "anyOf": [
                  {
                    "type": "string",
                    "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_0983c3f010011C7pcZVuLSlbMq",
        "sessionID": "ses_f67c3c110ffejR1GTeDIvVljZW"
      },
      "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 2260E: \"Cyclic Balance\".\n\nKnown problem metadata:\n- Contest ID: 2260\n- Index: E\n- Name: Cyclic Balance\n- Rating: unknown / unrated\n- Tags: binary search, constructive algorithms, dp, math\n- Problem URL: https://codeforces.com/contest/2260/problem/E\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\">E. Cyclic Balance</div><div class=\"time-limit\"><div class=\"property-title\">time limit per test</div>2 seconds</div><div class=\"memory-limit\"><div class=\"property-title\">memory limit per test</div>512 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>Consider a string $$$t=t_1t_2\\ldots t_m$$$ consisting of the characters <span class=\"tex-font-style-tt\">0</span> and <span class=\"tex-font-style-tt\">1</span>. We call the pairs of adjacent characters of the string $$$t$$$ the pairs $$$t_1t_2, t_2t_3, \\ldots, t_{m-1}t_m$$$, as well as the pair $$$t_mt_1$$$. The last pair connects the end of the string with its beginning, so exactly $$$m$$$ pairs are considered in total. If the string consists of only one character, the only pair that is considered is $$$t_1 t_1$$$.</p><p>We call a string $$$t$$$ <span class=\"tex-font-style-underline\">cyclically balanced</span> if, among its pairs of adjacent characters, the numbers of pairs <span class=\"tex-font-style-tt\">00</span>, <span class=\"tex-font-style-tt\">01</span>, <span class=\"tex-font-style-tt\">10</span>, and <span class=\"tex-font-style-tt\">11</span> are equal.</p><p>The cost of a binary string is the minimum number of characters that need to be inserted into it so that it becomes cyclically balanced. Characters may be inserted in any positions, including before the first and after the last character of the string. It is not allowed to delete or replace the original characters.</p><p>You are given a binary string $$$s$$$ and $$$q$$$ queries. In each query, indices $$$l$$$ and $$$r$$$ are given. Find the cost of the substring $$$s_l s_{l+1}\\ldots s_r$$$.</p></div><div class=\"input-specification\"><div class=\"section-title\">Input</div><p>The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\le n, q \\le 3 \\cdot 10^5$$$)&nbsp;— the length of the string and the number of queries.</p><p>The second line contains $$$s$$$ — a sequence of length $$$n$$$ consisting of the characters <span class=\"tex-font-style-tt\">0</span> and/or <span class=\"tex-font-style-tt\">1</span>.</p><p>Then follow $$$q$$$ lines; the $$$i$$$-th of them contains two integers $$$l_i$$$ and $$$r_i$$$ ($$$1 \\le l_i \\le r_i \\le n$$$)&nbsp;— the boundaries of the substring for the corresponding query.</p></div><div class=\"output-specification\"><div class=\"section-title\">Output</div><p>Print $$$q$$$ integers: the $$$i$$$-th integer should be equal to the cost of the substring from the $$$i$$$-th query.</p></div><div class=\"sample-tests\"><div class=\"section-title\">Example</div><div class=\"sample-test\"><div class=\"input\"><div class=\"title\">Input</div><pre><div class=\"test-example-line test-example-line-even test-example-line-0\">11 7</div><div class=\"test-example-line test-example-line-even test-example-line-0\">00111100000</div><div class=\"test-example-line test-example-line-even test-example-line-0\">1 8</div><div class=\"test-example-line test-example-line-even test-example-line-0\">1 1</div><div class=\"test-example-line test-example-line-even test-example-line-0\">1 2</div><div class=\"test-example-line test-example-line-even test-example-line-0\">1 4</div><div class=\"test-example-line test-example-line-even test-example-line-0\">3 6</div><div class=\"test-example-line test-example-line-even test-example-line-0\">2 7</div><div class=\"test-example-line test-example-line-even test-example-line-0\">7 11</div></pre></div><div class=\"output\"><div class=\"title\">Output</div><pre>4\n3\n2\n0\n4\n2\n7\n</pre></div></div></div>\n</problem-statement>\n\n\nSource lookup status:\n<source-lookup-status>\n- Tutorial: https://codeforces.com/blog/entry/156529 loaded (200 OK; title: \"Educational Codeforces Round 194 — Editorial - Codeforces\"; mentions 2260E)\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- 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 <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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        }
      ]
    },
    {
      "info": {
        "parentID": "msg_0983c3f010011C7pcZVuLSlbMq",
        "role": "assistant",
        "mode": "nudge-generation",
        "agent": "nudge-generation",
        "variant": "low",
        "path": {
          "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-PmhcCD",
          "root": "/"
        },
        "cost": 0,
        "tokens": {
          "total": 8532,
          "input": 8500,
          "output": 32,
          "reasoning": 0,
          "cache": {
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        },
        "modelID": "gpt-6-astra",
        "providerID": "openai",
        "time": {
          "created": 1789260480390,
          "completed": 1789260487531
        },
        "finish": "tool-calls",
        "id": "msg_0983c3f86001LDFRQdaz3n8NeO",
        "sessionID": "ses_f67c3c110ffejR1GTeDIvVljZW"
      },
      "parts": [
        {
          "type": "step-start",
          "id": "prt_0983c4fee001KJHnAlJq3DSrPV",
          "sessionID": "ses_f67c3c110ffejR1GTeDIvVljZW",
          "messageID": "msg_0983c3f86001LDFRQdaz3n8NeO"
        },
        {
          "type": "tool",
          "tool": "webfetch",
          "callID": "call_0X9dY6oXjp16nNG2HfRK99Zx",
          "state": {
            "status": "completed",
            "input": {
              "url": "https://codeforces.com/blog/entry/156529",
              "format": "markdown"
            },
            "output": "---\nimage: https://codeforces.com/codeforces.org/s/13758/images/codeforces-telegram-square-1024x1024.png\ntitle: Educational Codeforces Round 194 — 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/13758/images/codeforces-logo-with-telegram-555x130.png \"Codeforces\")](https://codeforces.com/) \n\n[![In English](//codeforces.com/codeforces.org/s/13758/images/flags/24/gb.png \"In English\")](https://codeforces.com/blog/entry/156529?locale=en) [![По-русски](//codeforces.com/codeforces.org/s/13758/images/flags/24/ru.png \"По-русски\")](https://codeforces.com/blog/entry/156529?locale=ru) \n\n[Enter](https://codeforces.com/enter?back=%2Fblog%2Fentry%2F156529)| [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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[Groups](https://codeforces.com/groups/with/BledDest)\n* [Contests](https://codeforces.com/contests/with/BledDest)\n* [Problemsetting](https://codeforces.com/contests/writer/BledDest)\n\n### [BledDest's blog](https://codeforces.com/blog/BledDest)\n\n[ Educational Codeforces Round 194 — Editorial ](https://codeforces.com/blog/entry/156529) \n\nBy [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\"), 4 days ago, translation, ![In English](//codeforces.com/codeforces.org/s/13758/images/flags/24/gb.png \"In English\") \n\n[2260A - Monocarp's Contest](https://codeforces.com/contest/2260/problem/A \"Educational Codeforces Round 194 (Rated for Div. 2)\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\nt = int(input())\nfor i in range(t):\n    n = int(input())\n    a = list(map(int, input().split()))\n    \n    cnt0 = a.count(0)\n    if cnt0 >= 2:\n        print(a[0] + a[-1])\n    else:\n        print(-1)\n\n\n```\n\n[2260B - Monocarp and Projects](https://codeforces.com/contest/2260/problem/B \"Educational Codeforces Round 194 (Rated for Div. 2)\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include<bits/stdc++.h>\n\nusing namespace std;\n\nvoid solve()\n{\n    long long x, y, k;\n    cin >> x >> y >> k;\n    long long ans = 0;\n    for(long long i = 0; i < min(y, k); i++)\n        ans += (y + i) % (x + i);\n    long long full = max(0ll, k - y);\n    ans += full * (y - x);\n    cout << ans << endl;    \n}\n\nint main()\n{\n    ios_base::sync_with_stdio(0);\n    cin.tie(0);\n    int t;\n    cin >> t;\n    for(int i = 0; i < t; i++)\n        solve();\n}\n\n\n```\n\n[2260C - Maximize XOR, Minimize Operations](https://codeforces.com/contest/2260/problem/C \"Educational Codeforces Round 194 (Rated for Div. 2)\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include<bits/stdc++.h>\n\nusing namespace std;\n\nvoid solve()\n{\n    int x, y;\n    cin >> x >> y;\n    int s = x + y;\n    for(int d = (1 << 30); d >= 1; d >>= 1)\n        if((s & d) != 0 && x >= d)\n            x -= d;\n    cout << s << \" \" << x << endl;    \n}\n\nint main()\n{\n    ios_base::sync_with_stdio(0);\n    cin.tie(0);\n    int t;\n    cin >> t;\n    for(int i = 0; i < t; i++)\n        solve();\n}\n\n\n```\n\n[2260D - Signs of Prefix Sums](https://codeforces.com/contest/2260/problem/D \"Educational Codeforces Round 194 (Rated for Div. 2)\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include<bits/stdc++.h>\n\nusing namespace std;\n\nconst int K = 3;\nconst int INF = int(1e9);\n\npair<int, int> get_segment(char c)\n{\n    if(c == '0') return {K, K};\n    if(c == '-') return {0, K - 1};\n    return {K + 1, K * 2};\n}\n\nvoid solve()\n{\n    int n;\n    string s;\n    cin >> n >> s;\n    vector<vector<int>> dp(n + 1, vector<int>(2 * K + 1, INF));\n    dp[0][K] = 0;\n    for(int i = 0; i < n; i++)\n        for(int j = 0; j < 2 * K + 1; j++)\n        {\n            if(dp[i][j] == INF) continue;\n            auto [l, r] = get_segment(s[i]);\n            for(int k = l; k <= r; k++)\n            {\n                if(k == j) continue;\n                int& d = dp[i + 1][k];\n                d = min(d, max(dp[i][j], abs(k - j)));\n            }\n        }\n    int ans = *min_element(dp[n].begin(), dp[n].end());\n    if(ans == INF) ans = -1;\n    cout << ans << endl;\n}\n\nint main()\n{\n    ios_base::sync_with_stdio(0);\n    cin.tie(0);\n    int t;\n    cin >> t;\n    for(int i = 0; i < t; i++)\n        solve();    \n}\n\n\n```\n\n[2260E - Cyclic Balance](https://codeforces.com/contest/2260/problem/E \"Educational Codeforces Round 194 (Rated for Div. 2)\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include<bits/stdc++.h>\n\nusing namespace std;\n\nint get(int c00, int c11, int cd)\n{                                      \n    int len = c00 + c11 + cd;\n    int lf = len / 4;\n    int rg = 1e8;\n    int ans = 1e8;\n    while(rg >= lf)\n    {\n        int mid = (lf + rg) / 2;\n        int need_break = max(0, c00 - mid) + max(0, c11 - mid);\n        if(need_break + cd / 2 <= mid)\n        {\n            ans = mid;\n            rg = mid - 1;\n        }\n        else\n            lf = mid + 1;\n    }\n    return ans;\n}\n\nvoid solve()\n{\n    int n, q;\n    cin >> n >> q;\n    string s;\n    cin >> s;\n    vector<int> pref1(n + 1), prefdiff(n);\n    for(int i = 0; i < n; i++)\n        pref1[i + 1] = pref1[i] + (s[i] - '0');\n    for(int i = 0; i + 1 < n; i++)\n        prefdiff[i + 1] = prefdiff[i] + (s[i] != s[i + 1]);\n\n    for(int i = 0; i < q; i++)\n    {\n        int l, r;\n        cin >> l >> r;\n        if(l == r)\n            cout << 3 << \"\\n\";\n        else\n        {\n            int c1 = pref1[r] - pref1[l - 1];\n            int cd = prefdiff[r - 1] - prefdiff[l - 1];\n            if(s[l - 1] != s[r - 1])\n                cd++;\n            int c0 = (r - l + 1) - c1;\n            int c00 = c0 - (cd / 2);\n            int c11 = c1 - (cd / 2);\n            cout << get(c00, c11, cd) * 4 - (r - l + 1) << \"\\n\";\n        }\n    }\n}\n\nint main()\n{\n    ios_base::sync_with_stdio(0);\n    cin.tie(0);\n    int t = 1;\n    //cin >> t;\n    for(int i = 0; i < t; i++)\n        solve();\n}\n\n\n```\n\n[2260F - Edge Three-Coloring](https://codeforces.com/contest/2260/problem/F \"Educational Codeforces Round 194 (Rated for Div. 2)\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include<bits/stdc++.h>\n\nusing namespace std;\n\nconst int N = 3010;\n\nvector<pair<int, int>> g[N];\nint p[N];\nint pe[N];\nint used[N];\nint n, m;\nvector<int> bad_edges;\nint cycle[N];\nint xs[N], ys[N];\n\nvoid dfs1(int x)\n{\n    for(auto [y, i] : g[x])\n        if(p[y] == -1)\n        {\n            p[y] = x;\n            pe[y] = i;\n            dfs1(y);\n        }\n        else if(p[x] != y)\n            bad_edges.push_back(i);\n}\n\nvoid go_to_root(int x)\n{\n    while(x != 0)\n    {\n        cycle[pe[x]] ^= 1;\n        x = p[x];\n    }\n}\n\nvoid dfs2(int x)\n{\n    used[x] = true;\n    for(auto [y, i] : g[x])\n        if(cycle[i] == 0 && !used[y])\n            dfs2(y); \n}\n\nvoid solve()\n{\n    cin >> n >> m;\n    bad_edges.clear();\n    for(int i = 0; i < n; i++)\n    {\n        g[i].clear();\n        p[i] = -1;\n        pe[i] = -1;\n    }\n    for(int i = 0; i < m; i++)\n    {\n        int x, y;\n        cin >> x >> y;\n        --x;\n        --y;\n        xs[i] = x;\n        ys[i] = y;\n        g[x].push_back(make_pair(y, i));\n        g[y].push_back(make_pair(x, i));\n    }\n    \n    p[0] = 0;\n    dfs1(0);\n    sort(bad_edges.begin(), bad_edges.end());\n    bad_edges.erase(unique(bad_edges.begin(), bad_edges.end()), bad_edges.end());\n\n    int k = bad_edges.size();\n    for(int mask = 1; mask < (1 << k); mask++)\n    {\n        for(int edge = 0; edge < m; edge++) cycle[edge] = 0;\n        for(int bit = 0; bit < k; bit++)\n            if((mask >> bit) & 1)\n            {                                                               \n                go_to_root(xs[bad_edges[bit]]);\n                go_to_root(ys[bad_edges[bit]]);\n                cycle[bad_edges[bit]] ^= 1;\n            }\n        for(int vertex = 0; vertex < n; vertex++)\n            used[vertex] = 0;\n        dfs2(0);\n        bool good = true;\n        for(int vertex = 0; vertex < n; vertex++)\n            if(!used[vertex])\n                good = false;\n        if(good)\n        {\n            cout << \"YES\\n\";\n            return;\n        }\n    }\n    cout << \"NO\\n\";\n}\n\nint main()\n{\n    ios_base::sync_with_stdio(0);\n    cin.tie(0);\n    int t;\n    cin >> t;\n    for(int i = 0; i < t; i++)\n        solve();    \n}\n\n\n```\n\n[2260G - Sortable Permutations](https://codeforces.com/contest/2260/problem/G \"Educational Codeforces Round 194 (Rated for Div. 2)\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include<bits/stdc++.h>\n\nusing namespace std;\n\nconst int MOD = 998244353;\n\nint add(int x, int y)\n{\n    x += y;\n    while(x >= MOD) x -= MOD;\n    while(x < 0) x += MOD;\n    return x;   \n}\n\nint mul(int x, int y)\n{\n    return (x * 1ll * y) % MOD;\n}\n\nint binpow(int x, int y)\n{\n    int z = 1;\n    while(y > 0)\n    {\n        if(y % 2 == 1) z = mul(z, x);\n        x = mul(x, x);\n        y /= 2;\n    }\n    return z;\n}\n\nint inv(int x)\n{\n    return binpow(x, MOD - 2);\n}\n\nconst int N = 200043;\n\nint mind[N];\nint fact[N];\nint ifact[N];\n\nvoid precalc()\n{\n    for(int i = 2; i < N; i++)\n        for(int j = i; j < N; j += i)\n            if(mind[j] == 0)\n                mind[j] = i;\n\n    fact[0] = 1;\n    for(int i = 1; i < N; i++)\n        fact[i] = mul(fact[i - 1], i);\n    ifact[N - 1] = inv(fact[N - 1]);\n    for(int i = N - 1; i >= 1; i--)\n        ifact[i - 1] = mul(ifact[i], i);\n}\n\nint choose(int n, int k)\n{\n    if(n < 0 || n < k || k < 0) return 0;\n    return mul(fact[n], mul(ifact[k], ifact[n - k]));\n}\n\nint n;\n\nvector<int> factorize(int x)\n{\n    vector<int> res;\n    while(x != 1)\n    {\n        res.push_back(mind[x]);\n        x /= mind[x];\n    }\n    return res;\n}\n\nvoid gen(const vector<int>& d, int i, long long cur, int m, vector<pair<long long, int>>& res)\n{\n    if(i == d.size()) res.push_back(make_pair(cur, m));\n    else\n    {\n        gen(d, i + 1, cur, m, res);\n        gen(d, i + 1, cur * 1ll * d[i], -m, res);\n    }\n} \n\nvoid solve()\n{\n    cin >> n;\n    vector<pair<pair<long long, int>, int>> edges;\n    for(int i = 1; i < n; i++)\n    {\n        auto f1 = factorize(i);\n        auto f2 = factorize(i + 1);\n        for(auto x : f2) f1.push_back(x);\n        sort(f1.begin(), f1.end());\n        f1.erase(unique(f1.begin(), f1.end()), f1.end());\n        vector<pair<long long, int>> pos;\n        gen(f1, 0, 1, 1, pos);\n        for(auto [x, s] : pos)\n        {\n            if(x == 1) continue;\n            edges.push_back(make_pair(make_pair(x, s), i));\n        }\n    }\n    \n    int ans = 1;\n    int l = 0;\n    sort(edges.begin(), edges.end());\n    while(l < edges.size())\n    {\n        int r = l;\n        while(r < edges.size() && edges[r].first.first == edges[l].first.first)\n            r++;\n        vector<int> e;\n        for(int i = l; i < r; i++)\n            e.push_back(edges[i].second);\n        long long i = edges[l].first.first;\n        int m = edges[l].first.second;\n        if(m != 0)\n        {\n            int absent = n / i;\n            int cur = mul(fact[n], ifact[n - absent]);\n            int lf = 0;\n            while(lf < e.size())\n            {\n                int rg = lf;\n                int nxt = e[lf];\n                while(rg < e.size() && e[rg] == nxt)\n                {\n                    rg++;\n                    nxt++;\n                }\n                vector<int> cnt(2);\n                for(int k = e[lf]; k <= e[rg - 1] + 1; k++)\n                {\n                    if(k % i != 0)\n                        cnt[k % 2]++;     \n                }\n                //cout << cnt[0] << \" \" << cnt[1] << endl;\n                cur = mul(cur, choose(cnt[0] + cnt[1], cnt[0]));\n                lf = rg;\n            }\n            //cout << i << \" \" << cur << endl;\n            ans = add(ans, -1 * m * add(cur, -1));\n        }\n        l = r;\n    }\n    cout << ans << endl;\n}\n\nint main()\n{\n    ios_base::sync_with_stdio(0);\n    cin.tie(0);\n    precalc();\n    int t = 1;\n    //cin >> t;\n    for(int i = 0; i < t; i++)\n        solve();\n}\n\n```\n\n![](//codeforces.com/codeforces.org/s/13758/images/icons/paperclip-16x16.png) Tutorial of [Educational Codeforces Round 194 (Rated for Div. 2)](https://codeforces.com/contest/2260) \n\n* [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/voteup-gray.png \"Vote: I like it\")](#)\n* +91\n* [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/votedown-gray.png \"Vote: I do not like it\")](#)\n\n* [![Author](//codeforces.com/codeforces.org/s/13758/images/blog/user_16x16.png \"Author\")](https://codeforces.com/profile/BledDest) [BledDest ](https://codeforces.com/profile/BledDest)\n* ![Publication date](//codeforces.com/codeforces.org/s/13758/images/blog/date_16x16.png \"Publication date\") 4 days ago\n* [![Comments](//codeforces.com/codeforces.org/s/13758/images/blog/comments_16x16.png \"Comments\")](https://codeforces.com/blog/entry/156529#comments) [41 ](https://codeforces.com/blog/entry/156529#comments)\n\n  \n![Comments](//codeforces.com/codeforces.org/s/13758/images/icons/comments-48x48.png \"Comments\") Comments (41) \n\n[Write comment?](#) \n\n| » [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\") | 4 days ago, [show (+1)](#) # |\n| ------------------------------------------------------------------------------------------ | ---------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/331814/avatar/36406af67e5558a6.jpg) ](https://codeforces.com/profile/BledDest) [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156529?#comment-1391195 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) The tutorials for problems will be available in a few minutes. [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [Askew\\_breadcrumb6](https://codeforces.com/profile/Askew%5Fbreadcrumb6 \"Newbie Askew_breadcrumb6\") | 2 days ago, [show (+1)](#) # |\n| ------------------------------------------------------------------------------------------------------- | ---------------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/2572984/avatar/6e314e22b534732f.jpg) ](https://codeforces.com/profile/Askew%5Fbreadcrumb6) [Askew\\_breadcrumb6](https://codeforces.com/profile/Askew%5Fbreadcrumb6 \"Newbie Askew_breadcrumb6\") | 2 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156529?#comment-1391575 \"Link to comment\") [^](#comment-1391195 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) In problem C, what's preventing us from having a greedy pick where, if in the sum S=X+Y there is a 0 bit, it is a result of 1 in X and 1 in Y?In other words, Why can't I have 1 ^ 1 resulting in 0 in S?Why should I be mapping the set bits of S to X and Y? Why not have a configuration where unset bits are a result of 1 bit in X and 1 bit in Y? [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |\n\n  * | » » » [amey75](https://codeforces.com/profile/amey75 \"Specialist amey75\") | 40 hours ago, [show (+1)](#) # |\n| ------------------------------------------------------------------------- | ------------------------------ |\n\n| » » » [ ![](https://codeforces.com/userpic.codeforces.org/4580660/avatar/53b3970dc02dcbd7.jpg) ](https://codeforces.com/profile/amey75) [amey75](https://codeforces.com/profile/amey75 \"Specialist amey75\") | 40 hours ago, [hide](#) [#](https://codeforces.com/blog/entry/156529?#comment-1391662 \"Link to comment\") [^](#comment-1391575 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) cause that would lead to X&Y being non zero. [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n    * | » » » » [Askew\\_breadcrumb6](https://codeforces.com/profile/Askew%5Fbreadcrumb6 \"Newbie Askew_breadcrumb6\") | 39 hours ago, [show](#) # |\n| ----------------------------------------------------------------------------------------------------------- | ------------------------- |\n\n| » » » » [ ![](https://codeforces.com/userpic.codeforces.org/2572984/avatar/6e314e22b534732f.jpg) ](https://codeforces.com/profile/Askew%5Fbreadcrumb6) [Askew\\_breadcrumb6](https://codeforces.com/profile/Askew%5Fbreadcrumb6 \"Newbie Askew_breadcrumb6\") | 39 hours ago, [hide](#) [#](https://codeforces.com/blog/entry/156529?#comment-1391668 \"Link to comment\") [^](#comment-1391662 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) My bad, thanks!! [→](#) [Reply](#) |\n| ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [Krish-006-19](https://codeforces.com/profile/Krish-006-19 \"Newbie Krish-006-19\") | 4 days ago, [show (+1)](#) # |\n| ----------------------------------------------------------------------------------- | ---------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/5268131/avatar/fa8bb1ed65c58067.jpg) ](https://codeforces.com/profile/Krish-006-19) [Krish-006-19](https://codeforces.com/profile/Krish-006-19 \"Newbie Krish-006-19\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156529?#comment-1391198 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) B felt like a bullet stuck in the ribs, nor does in come out nor was it blocked, felt suffocated! [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [dolaisubhrabijoy](https://codeforces.com/profile/dolaisubhrabijoy \"Newbie dolaisubhrabijoy\") | 4 days ago, [show](#) # |\n| ------------------------------------------------------------------------------------------------- | ----------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/5059690/avatar/a8276810d3136972.jpg) ](https://codeforces.com/profile/dolaisubhrabijoy) [dolaisubhrabijoy](https://codeforces.com/profile/dolaisubhrabijoy \"Newbie dolaisubhrabijoy\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156529?#comment-1391200 \"Link to comment\") [^](#comment-1391198 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Tomorrow is my exam—worst decision ever to register! [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n* | » » [Malay\\_07](https://codeforces.com/profile/Malay%5F07 \"Newbie Malay_07\") | 4 days ago, [show](#) # |\n| ---------------------------------------------------------------------------- | ----------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/5652393/avatar/5e5e416d875077c9.jpg) ](https://codeforces.com/profile/Malay%5F07) [Malay\\_07](https://codeforces.com/profile/Malay%5F07 \"Newbie Malay_07\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156529?#comment-1391213 \"Link to comment\") [^](#comment-1391198 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) same here bro...it sucks... [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [misytnikita](https://codeforces.com/profile/misytnikita \"Newbie misytnikita\") | 4 days ago, [show (+1)](#) # |\n| -------------------------------------------------------------------------------- | ---------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/misytnikita) [misytnikita](https://codeforces.com/profile/misytnikita \"Newbie misytnikita\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156529?#comment-1391211 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) A — easy, B, C literally 1500 [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [ntkoo](https://codeforces.com/profile/ntkoo \"Newbie ntkoo\") | 3 days ago, [show](#) # |\n| ---------------------------------------------------------------- | ----------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/ntkoo) [ntkoo](https://codeforces.com/profile/ntkoo \"Newbie ntkoo\") | 3 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156529?#comment-1391499 \"Link to comment\") [^](#comment-1391211 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) l agree [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |\n\n| » [ggq15plus2](https://codeforces.com/profile/ggq15plus2 \"Candidate Master ggq15plus2\") | 4 days ago, [show (+1)](#) # |\n| --------------------------------------------------------------------------------------- | ---------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/ggq15plus2) [ggq15plus2](https://codeforces.com/profile/ggq15plus2 \"Candidate Master ggq15plus2\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156529?#comment-1391216 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I guess I shall improve my implementation skills now, as I got WA on #2 5 times during contest while getting AC only require changing a few chars (my specific implementation was inferior to the official solution that made debugging harder, though). Submission: [389954784](https://codeforces.com/contest/2260/submission/389954784). [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [ggq15plus2](https://codeforces.com/profile/ggq15plus2 \"Candidate Master ggq15plus2\") | 4 days ago, [show](#) # |\n| ----------------------------------------------------------------------------------------- | ----------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/ggq15plus2) [ggq15plus2](https://codeforces.com/profile/ggq15plus2 \"Candidate Master ggq15plus2\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156529?#comment-1391220 \"Link to comment\") [^](#comment-1391216 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +11 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I wrongly enumerated bitmasks starting from 0, yet another trivial bug in determining connectivity prevented me from detecting it from the samples :( [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |\n* | » » [lightningmcqueen1](https://codeforces.com/profile/lightningmcqueen1 \"Newbie lightningmcqueen1\") | 4 days ago, [show (+1)](#) # |\n| ---------------------------------------------------------------------------------------------------- | ---------------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/3662492/avatar/c4abb8459ba1b874.jpg) ](https://codeforces.com/profile/lightningmcqueen1) [lightningmcqueen1](https://codeforces.com/profile/lightningmcqueen1 \"Newbie lightningmcqueen1\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156529?#comment-1391228 \"Link to comment\") [^](#comment-1391216 \"Parent comment\") \\| [←](#)Rev. 3 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) In my opinion you needed to spend more time with pencil and paper. The solution only requires 50 or so lines of code.(Assuming this was for problem C) [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n  * | » » » [ggq15plus2](https://codeforces.com/profile/ggq15plus2 \"Candidate Master ggq15plus2\") | 4 days ago, [show (+1)](#) # |\n| ------------------------------------------------------------------------------------------- | ---------------------------- |\n\n| » » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/ggq15plus2) [ggq15plus2](https://codeforces.com/profile/ggq15plus2 \"Candidate Master ggq15plus2\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156529?#comment-1391232 \"Link to comment\") [^](#comment-1391228 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Um, but this is F. [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n    * | » » » » [lightningmcqueen1](https://codeforces.com/profile/lightningmcqueen1 \"Newbie lightningmcqueen1\") | 4 days ago, [show](#) # |\n| -------------------------------------------------------------------------------------------------------- | ----------------------- |\n\n| » » » » [ ![](https://codeforces.com/userpic.codeforces.org/3662492/avatar/c4abb8459ba1b874.jpg) ](https://codeforces.com/profile/lightningmcqueen1) [lightningmcqueen1](https://codeforces.com/profile/lightningmcqueen1 \"Newbie lightningmcqueen1\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156529?#comment-1391239 \"Link to comment\") [^](#comment-1391232 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Oh nvm. You did great by the way. The tutorial for F looks like a trip to hell. [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |\n\n| » [MastPlanHai](https://codeforces.com/profile/MastPlanHai \"Candidate Master MastPlanHai\") | 4 days ago, [show](#) # |\n| ------------------------------------------------------------------------------------------ | ----------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/4275317/avatar/5af2ed576289c3e5.jpg) ](https://codeforces.com/profile/MastPlanHai) [MastPlanHai](https://codeforces.com/profile/MastPlanHai \"Candidate Master MastPlanHai\") | 4 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156529?#comment-1391223 \"Link to comment\") \\| [←](#)Rev. 3 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +4 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Solution to E without using binary search [389952685](https://codeforces.com/contest/2260/submission/389952685 \"Submission 389952685 by MastPlanHai\") [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [invoker.\\_](https://codeforces.com/profile/invoker.%5F \"Pupil invoker._\") | 4 days ago, [show (+1)](#) # |\n\n...60732 bytes truncated...\n\nThe tool call succeeded but the output was truncated. Full output saved to: /Users/zayd/.local/share/opencode/tool-output/tool_0983c5b630010FyLetjBrthdfC\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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            "output": "Title: Problem - 2032E - Codeforces\nURL: https://codeforces.com/problemset/problem/2032/E\nPublished: N/A\nAuthor: N/A\nHighlights:\nYou are given a cyclic array a with n elements, where n is odd. In each operation, you can do the following:\n...\n- Choose an index 1≤i≤n and increase ai−1 by 1, ai by 2, and ai+1 by 1. The element before the first element is the last element because this is a cyclic array.\n...\nA cyclic array is called balanced if all its elements are equal to each other.\n...\nFind any sequence of operations to make this cyclic array balanced or determine that it is impossible. Please note that you do not have to minimize the number of operations.\n...\nEach test consists of multiple test cases. The first line contains a single integer t (1≤t≤2⋅105) — the number of test cases. The description of the test cases follows.\n...\nThe first line of each test case contains a single integer n (1≤n<2⋅105, n is odd) — the length of the array a.\n...\nThe second line of each test case contains n integers a1,a2,…,an (1≤ai≤106) — the elements of the array a.\n...\nIt is guaranteed that the sum of n over all test cases does not exceed 2⋅105.\n...\n- If it is impossible to make the cyclic array balanced, output −1.\n- Otherwise, output n integers v1,v2,…,vn (0≤vi≤1018) — where vi denotes the number of operations applied to index i. It can be proved that if any solution exists, then there exists a solution under the given constraints. If there are several solutions under the given constraints, output any of them.\n...\nIn the first test case:\n...\n- After 1 operation applied to index i=2, the array a=[3,3,3].\n...\nIn the second test case:\n...\n- After 2 operations applied to index i=1, the array a=[5,4,5].\n- After 1 operation applied to index i=2, the array a=[6,6,6].\n...\nIn the third test case:\n...\n- After 2 operations applied to index i=1, the array a=[5,4,1,2,3].\n- After 3 operations applied to index i=3, the array a=[5,7,7,5,3].\n- After 2 operations applied to index i=5, the array a=[7,7,7,7,7].\n\n---\n\nTitle: N/A\nURL: https://codejeet.com/problem/maximize-cyclic-partition-score\nPublished: N/A\nAuthor: N/A\nHighlights:\n# 3743. Maximize Cyclic Partition Score\n...\nYou are given a cyclic array `nums` and an integer `k`.\n...\nPartition `nums` into at most `k` subarrays. As `nums` is cyclic, these subarrays may wrap around from the end of the array back to the beginning.\n...\nThe range of a subarray is the difference between its maximum and minimum values. The score of a partition is the sum of subarray ranges.\n...\nReturn the maximum possible score among all cyclic partitions.\n...\nUse dynamic programming on the number of extreme picks: allow up to `2 * k` picks (each partition can supply a + and a -), so the problem becomes choosing which elements are + or -.\n...\nModel a partition by its max (+`nums[i]`) and min (-`nums[i]`) contributions; selecting an element as a max adds +`nums[i]`, and selecting it as a min adds -`nums[i]`.\n...\nUse a DP state `(picks, balance)` where `picks` is the total number of + and - chosen so far (<= `2 * k`), and `balance` represents the difference between the counts of + and - currently; at each `i`, you conceptually take +, take -, or skip.\n...\nHandle cyclicity by limiting `balance` to the range `[0, 2]`. You can show that such a balance is always achievable.\n\n---\n\nTitle: E - Cyclic Password\nURL: https://atcoder.jp/contests/awc0060/tasks/awc0060_e?lang=en\nPublished: N/A\nAuthor: AtCoder Inc.\nHighlights:\n### 問題文\n...\n高橋君の会社には、1 から N までの番号が付いた N 台の端末があります。\n...\n端末 i（1 \\leq i \\leq N）には 1 桁の数字 D_i（0 以上 9 以下）が表示されており、移動先の端末番号 P_i（1 \\leq P_i \\leq N）が設定されています。\n...\n高橋君は、指定された開始端末から出発し、次の操作をちょうど K 回繰り返します。\n...\n1. 現在いる端末に表示された数字を記録する。\n...\n2. 現在いる端末の番号が i であるとき、端末 P_i へ移動する。\n...\nK 回の操作で順に記録された数字を d_1, d_2, \\ldots, d_K とすると、得られる値 V は次のように定まります。\n...\nV = \\sum_{t=1}^{K} d_t \\cdot 10^{K-t}\n...\nこれは、数字列 d_1 d_2 \\ldots d_K を 10 進表記の非負整数とみなした値です（先頭が 0 であってもよい）。\n...\nK = 0 のとき（操作を一度も行わないとき）は V = 0 です。\n...\nQ 個の問い合わせが与えられます。j 番目（1 \\leq j \\leq Q）の問い合わせでは、開始端末の番号 S_j と操作回数 K_j が指定されます。端末 S_j から出発して操作を K_j 回行ったときに得られる V を M で割った余りを求めてください。\n...\n### 制約\n...\n- 1 \\leq N \\leq 10^5\n- 1 \\leq Q \\leq 10^5\n- N + Q \\leq 1.5 \\times 10^5\n- 1 \\leq M \\leq 10^9\n- 0 \\leq D_i \\leq 9\n- 1 \\leq P_i \\leq N\n- 1 \\leq S_j \\leq N\n- 0 \\leq K_j \\leq 10^{18}\n- 入力はすべて整数である。\n...\n### Problem Statement\n...\nTakahashi's company has N terminals numbered from 1 to N.\n...\nTerminal i (1 \\leq i \\leq N) displays a single digit D_i (between 0 and 9 inclusive) and has a destination terminal number P_i (1 \\leq P_i \\leq N) configured.\n...\nTakahashi starts from a specified starting terminal and repeats the following operation exactly K times.\n...\n1. Record the digit displayed on the current terminal.\n...\n2. If the current terminal number is i, move to terminal P_i.\n...\nLet d_1, d_2, \\ldots, d_K be the digits recorded in order over the K operations. The resulting value V is defined as follows:\n...\nV = \\sum_{t=1}^{K} d_t \\cdot 10^{K-t}\n...\nThis is the value obtained by interpreting the digit string d_1 d_2 \\ldots d_K as a non-negative integer in base 10 (leading zeros are allowed).\n...\nWhen K = 0 (no operations are performed), V = 0.\n...\nYou are given Q queries. In the j-th query (1 \\leq j \\leq Q), a starting terminal number S_j and a number of operations K_j are specified. Find the remainder when V, obtained by starting from terminal S_j and performing K_j operations, is divided by M.\n...\n- 1 \\leq N \\leq 10^5\n- 1 \\leq Q \\leq 10^5\n- N + Q \\leq 1.5 \\times 10^5\n- 1 \\leq M \\leq 10^9\n- 0 \\leq D_i \\leq 9\n- 1 \\leq P_i \\leq N\n- 1 \\leq S_j \\leq N\n- 0 \\leq K_j \\leq 10^{18}\n- All input values are integers.\n...\n- The first line contains the number of terminals N, the number of queries Q, and the divisor M, separated by spaces.\n- In the following N lines, the i-th line contains the digit D_i displayed on terminal i and the destination terminal number P_i, separated by a space.\n- In the following Q lines, the j-th line contains the starting terminal number S_j and the number of operations K_j, separated by a space.\n...\nOutput Q lines.\n...\nOn the j-th line, output the remainder when V is divided by M as the answer to the j-th query.",
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            "title": "Exa Web Search: 2260E Cyclic Balance solution max c00 c11",
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              "command": "python3 -c 'from itertools import product\nbest={}\nfor L in (4,8,12):\n for t in product(\"01\", repeat=L):\n  c=[0]*4\n  for i in range(L): c[int(t[i])*2+int(t[(i+1)%L])]+=1\n  if len(set(c))!=1: continue\n  subs={\"\"}\n  for ch in t: subs |= {u+ch for u in subs if len(u)<7}\n  for u in subs:\n   if u: best[u]=min(best.get(u,100),L-len(u))\nchecked=0\nfor m in range(1,8):\n for t in product(\"01\", repeat=m):\n  s=\"\".join(t); d=sum(s[i]!=s[(i+1)%m] for i in range(m))//2\n  a=s.count(\"0\")-d; b=s.count(\"1\")-d\n  k=max(1,d,(a+d+1)//2,(b+d+1)//2,(a+b+d+2)//3)\n  assert 4*k-m==best[s],(s,k,best[s])\n  checked+=1\nprint(\"Exhaustive subsequence check passed for\",checked,\"strings.\")'",
              "timeout": 120000
            },
            "output": "Traceback (most recent call last):\n  File \"<string>\", line 18, in <module>\n    assert 4*k-m==best[s],(s,k,best[s])\n                  ~~~~^^^\nKeyError: '0000000'\n",
            "metadata": {
              "output": "Traceback (most recent call last):\n  File \"<string>\", line 18, in <module>\n    assert 4*k-m==best[s],(s,k,best[s])\n                  ~~~~^^^\nKeyError: '0000000'\n",
              "exit": 1,
              "truncated": false
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            "title": "python3 -c 'from itertools import product\nbest={}\nfor L in (4,8,12):\n for t in product(\"01\", repeat=L):\n  c=[0]*4\n  for i in range(L): c[int(t[i])*2+int(t[(i+1)%L])]+=1\n  if len(set(c))!=1: continue\n  subs={\"\"}\n  for ch in t: subs |= {u+ch for u in subs if len(u)<7}\n  for u in subs:\n   if u: best[u]=min(best.get(u,100),L-len(u))\nchecked=0\nfor m in range(1,8):\n for t in product(\"01\", repeat=m):\n  s=\"\".join(t); d=sum(s[i]!=s[(i+1)%m] for i in range(m))//2\n  a=s.count(\"0\")-d; b=s.count(\"1\")-d\n  k=max(1,d,(a+d+1)//2,(b+d+1)//2,(a+b+d+2)//3)\n  assert 4*k-m==best[s],(s,k,best[s])\n  checked+=1\nprint(\"Exhaustive subsequence check passed for\",checked,\"strings.\")'",
            "time": {
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              "command": "python3 -c 'from itertools import product\nbest={}\nfor L in (4,8,12):\n for t in product(\"01\", repeat=L):\n  c=[0]*4\n  for i in range(L): c[int(t[i])*2+int(t[(i+1)%L])]+=1\n  if len(set(c))!=1: continue\n  subs={\"\"}\n  for ch in t: subs |= {u+ch for u in subs if len(u)<6}\n  for u in subs:\n   if u: best[u]=min(best.get(u,100),L-len(u))\nchecked=0\nfor m in range(1,7):\n for t in product(\"01\", repeat=m):\n  s=\"\".join(t); d=sum(s[i]!=s[(i+1)%m] for i in range(m))//2\n  a=s.count(\"0\")-d; b=s.count(\"1\")-d\n  k=max(1,d,(a+d+1)//2,(b+d+1)//2,(a+b+d+2)//3)\n  assert 4*k-m==best[s],(s,k,best[s])\n  checked+=1\ns=\"00111100000\"\nans=[]\nfor l,r in [(1,8),(1,1),(1,2),(1,4),(3,6),(2,7),(7,11)]:\n t=s[l-1:r]; d=sum(t[i]!=t[(i+1)%len(t)] for i in range(len(t)))//2; z=t.count(\"0\"); o=len(t)-z\n ans.append(4*max(1,d,(z+1)//2,(o+1)//2,(len(t)-d+2)//3)-len(t))\nassert ans==[4,3,2,0,4,2,7]\nprint(\"Passed exhaustive checks for\",checked,\"strings and all sample queries.\")'",
              "timeout": 120000
            },
            "output": "Passed exhaustive checks for 126 strings and all sample queries.\n",
            "metadata": {
              "output": "Passed exhaustive checks for 126 strings and all sample queries.\n",
              "exit": 0,
              "truncated": false
            },
            "title": "python3 -c 'from itertools import product\nbest={}\nfor L in (4,8,12):\n for t in product(\"01\", repeat=L):\n  c=[0]*4\n  for i in range(L): c[int(t[i])*2+int(t[(i+1)%L])]+=1\n  if len(set(c))!=1: continue\n  subs={\"\"}\n  for ch in t: subs |= {u+ch for u in subs if len(u)<6}\n  for u in subs:\n   if u: best[u]=min(best.get(u,100),L-len(u))\nchecked=0\nfor m in range(1,7):\n for t in product(\"01\", repeat=m):\n  s=\"\".join(t); d=sum(s[i]!=s[(i+1)%m] for i in range(m))//2\n  a=s.count(\"0\")-d; b=s.count(\"1\")-d\n  k=max(1,d,(a+d+1)//2,(b+d+1)//2,(a+b+d+2)//3)\n  assert 4*k-m==best[s],(s,k,best[s])\n  checked+=1\ns=\"00111100000\"\nans=[]\nfor l,r in [(1,8),(1,1),(1,2),(1,4),(3,6),(2,7),(7,11)]:\n t=s[l-1:r]; d=sum(t[i]!=t[(i+1)%len(t)] for i in range(len(t)))//2; z=t.count(\"0\"); o=len(t)-z\n ans.append(4*max(1,d,(z+1)//2,(o+1)//2,(len(t)-d+2)//3)-len(t))\nassert ans==[4,3,2,0,4,2,7]\nprint(\"Passed exhaustive checks for\",checked,\"strings and all sample queries.\")'",
            "time": {
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              "end": 1789260527252
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      "info": {
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        "mode": "nudge-generation",
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          "status": "success",
          "reason": null,
          "hints": [
            {
              "order": 1,
              "content": "Walk around a binary string in a circle. Every switch from `0` to `1` must eventually be matched by a switch back. So the counts of `01` and `10` are already equal. Also, if every pair type appears $k$ times, the final length is $4k$."
            },
            {
              "order": 2,
              "content": "For one query, let $a=\\#00$, $b=\\#11$, and $d=\\#01=\\#10$, including the last-to-first pair. If the substring has $z$ zeros and $o$ ones, then $a=z-d$ and $b=o-d$. Prefix sums of ones and adjacent changes give all these values in $O(1)$."
            },
            {
              "order": 3,
              "content": "Original equal pairs are not permanent: inserting `1` into `00` changes it into `010`. This removes one `00` and adds one pair of each transition type. For a target count $k$, you must eliminate at least $\\max(0,a-k)$ original `00` pairs and $\\max(0,b-k)$ original `11` pairs. What does that force on the transition count?"
            },
            {
              "order": 4,
              "content": "For an integer $k\\ge1$, the exact feasibility condition is\n$$d+\\max(0,a-k)+\\max(0,b-k)\\le k.$$\nBreak the excess equal pairs first. Then increase both transition counts using `01` $\\to$ `0101`, and increase equal-pair counts by duplicating individual characters. If the string is still constant, first insert one opposite character to create both transition types."
            },
            {
              "order": 5,
              "content": "You can remove the binary search entirely. Use\n$$\\max(0,x)+\\max(0,y)=\\max(0,x,y,x+y).$$\nThe feasibility condition becomes $k\\ge d$, $2k\\ge a+d$, $2k\\ge b+d$, and $3k\\ge a+b+d$. Since $a+d=z$, $b+d=o$, and $a+b+d=m-d$, the minimum target is\n$$k=\\max\\left(1,d,\\left\\lceil\\frac z2\\right\\rceil,\\left\\lceil\\frac o2\\right\\rceil,\\left\\lceil\\frac{m-d}{3}\\right\\rceil\\right).$$\nReturn $4k-m$ for each query."
            }
          ],
          "editorial": "The catch is that insertion can **destroy an existing adjacent pair**. For example, `00` becomes `010` after inserting a `1`. So taking the largest current pair count and padding everything else is not necessarily optimal.\n\nWe will find the minimum possible final length, then subtract the original length. With a little algebra, every query takes $O(1)$.\n\n### 1. Count the cyclic pairs\n\nConsider a queried substring of length $m$. Write\n\n- $z$ and $o$ for its numbers of zeros and ones;\n- $a$ for its number of cyclic `00` pairs;\n- $b$ for its number of cyclic `11` pairs;\n- $d$ for its number of cyclic `01` pairs.\n\nOn a circle, every `0`-to-`1` transition is matched by a `1`-to-`0` transition. Thus `10` also occurs $d$ times.\n\nEach zero starts either a `00` or a `01` pair, and each one starts either a `11` or a `10` pair. Therefore,\n\n$$a=z-d,\\qquad b=o-d,\\qquad m=a+b+2d.$$\n\nIf the completed string has exactly $k$ pairs of each type, its length is $4k$. Our goal is to minimize $k\\ge1$.\n\nWe can reason about insertions into cyclic gaps. Inserting into the last-to-first gap is legal: put those characters after the original substring, before the string wraps back to its first character.\n\n### 2. Which target counts are possible?\n\nFix a target $k$. If $a>k$, at least $a-k$ original `00` gaps must stop contributing a `00` pair. Similarly, at least $b-k$ original `11` gaps must stop contributing a `11` pair.\n\nA gap originally connecting two zeros becomes a path from `0` to `0` after insertions. To contain no `00`, that path must visit a `1`, requiring at least one `01` and one `10`. The same reasoning applies to eliminating a `11` gap.\n\nMeanwhile, every original `01` gap still requires at least one `01`, and every original `10` gap still requires at least one `10`. These gaps are disjoint, so their requirements add up.\n\nConsequently, a necessary condition is\n\n$$d+\\max(0,a-k)+\\max(0,b-k)\\le k.$$\n\nIt is also sufficient. Here is a construction proving it.\n\n**First, remove excess equal pairs.**\n\n- Break exactly $\\max(0,a-k)$ original `00` pairs using `00` $\\to$ `010`.\n- Break exactly $\\max(0,b-k)$ original `11` pairs using `11` $\\to$ `101`.\n\nEach operation reduces the chosen equal-pair count by one and increases both transition counts by one. Now the equal-pair counts are at most $k$, and both transition counts equal\n\n$$d'=d+\\max(0,a-k)+\\max(0,b-k)\\le k.$$\n\n**Next, bring both transition counts up to $k$.**\n\nIf $d'=0$, the string is still constant. Insert one opposite character into any cyclic gap. This decreases an equal-pair count and creates one transition of each type. Since $k\\ge1$, all counts remain at most $k$.\n\nNow a `01` pair exists. Replacing it with `0101` adds one `01` and one `10`, without changing either equal-pair count. Repeat until both transition counts reach $k$.\n\n**Finally, fill the equal-pair deficits.**\n\nDuplicating a zero adds exactly one `00`; duplicating a one adds exactly one `11`. Both characters now exist, so increase these counts independently to $k$.\n\nEvery operation is an insertion. This proves the exact criterion:\n\n$$\\boxed{k\\ge1\\quad\\text{and}\\quad d+\\max(0,a-k)+\\max(0,b-k)\\le k.}$$\n\n### 3. Turn the condition into a formula\n\nThe condition is monotone, so binary search would work. But we can do better without making the implementation harder.\n\nUse the identity\n\n$$\\max(0,x)+\\max(0,y)=\\max(0,x,y,x+y).$$\n\nThen feasibility is equivalent to all four inequalities\n\n$$\n\\begin{aligned}\nd&\\le k,\\\\\na+d&\\le2k,\\\\\nb+d&\\le2k,\\\\\na+b+d&\\le3k.\n\\end{aligned}\n$$\n\nSubstituting $a+d=z$, $b+d=o$, and $a+b+d=m-d$ gives the minimum feasible integer:\n\n$$\\boxed{k=\\max\\left(1,d,\\left\\lceil\\frac z2\\right\\rceil,\\left\\lceil\\frac o2\\right\\rceil,\\left\\lceil\\frac{m-d}{3}\\right\\rceil\\right).}$$\n\nThe answer is\n\n$$\\boxed{4k-m}.$$\n\n### 4. Answer queries with prefix sums\n\nUse one-based indexing and build:\n\n- $P[i]$: the number of ones in $s_1\\ldots s_i$;\n- $D[i]$: the number of changes between adjacent characters entirely inside $s_1\\ldots s_i$.\n\nFor a query $[l,r]$:\n\n$$m=r-l+1,\\qquad o=P[r]-P[l-1],\\qquad z=m-o.$$\n\nThe number of internal changes is $D[r]-D[l]$. Add the closing pair separately:\n\n$$c=D[r]-D[l]+[s_l\\ne s_r].$$\n\nHere $[\\text{condition}]$ is $1$ when the condition is true and $0$ otherwise. Since cyclic transitions come in matched directions,\n\n$$d=\\frac c2.$$\n\nEvaluate the formula using integer ceilings: $(z+1)/2$, $(o+1)/2$, and $(m-d+2)/3$.\n\nA length-one substring works without a special case: $d=0$, $k=1$, and the answer is $3$. Remember that its single cyclic pair is the character paired with itself.\n\n### Correctness\n\nThe prefix sums compute the queried substring's exact character and cyclic transition counts, which determine $a$, $b$, and $d$. The feasibility argument proves that a balanced completion with pair count $k$ exists exactly when the stated inequalities hold. The formula chooses the smallest integer satisfying all of them. Therefore $4k$ is the minimum possible completed length, and $4k-m$ is precisely the minimum number of insertions.\n\n### Complexity\n\nPreprocessing takes $O(n)$ time and $O(n)$ memory. Each query takes $O(1)$ time, for $O(n+q)$ total time. All values fit comfortably in a 32-bit signed integer under the given constraints.",
          "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 n, q;\n    string s;\n    cin >> n >> q >> s;\n    s = \" \" + s;\n\n    vector<int> ones(n + 1), changes(n + 1);\n    for (int i = 1; i <= n; ++i) {\n        ones[i] = ones[i - 1] + (s[i] == '1');\n        changes[i] = changes[i - 1];\n        if (i > 1) changes[i] += (s[i] != s[i - 1]);\n    }\n\n    while (q--) {\n        int l, r;\n        cin >> l >> r;\n        int m = r - l + 1;\n        int o = ones[r] - ones[l - 1];\n        int z = m - o;\n        int d = (changes[r] - changes[l] + (s[l] != s[r])) / 2;\n        int k = max({1, d, (z + 1) / 2, (o + 1) / 2,\n                     (m - d + 2) / 3});\n        cout << 4 * k - m << '\\n';\n    }\n}\n"
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              "status": "success",
              "reason": null,
              "hints": [
                {
                  "order": 1,
                  "content": "Walk around a binary string in a circle. Every switch from `0` to `1` must eventually be matched by a switch back. So the counts of `01` and `10` are already equal. Also, if every pair type appears $k$ times, the final length is $4k$."
                },
                {
                  "order": 2,
                  "content": "For one query, let $a=\\#00$, $b=\\#11$, and $d=\\#01=\\#10$, including the last-to-first pair. If the substring has $z$ zeros and $o$ ones, then $a=z-d$ and $b=o-d$. Prefix sums of ones and adjacent changes give all these values in $O(1)$."
                },
                {
                  "order": 3,
                  "content": "Original equal pairs are not permanent: inserting `1` into `00` changes it into `010`. This removes one `00` and adds one pair of each transition type. For a target count $k$, you must eliminate at least $\\max(0,a-k)$ original `00` pairs and $\\max(0,b-k)$ original `11` pairs. What does that force on the transition count?"
                },
                {
                  "order": 4,
                  "content": "For an integer $k\\ge1$, the exact feasibility condition is\n$$d+\\max(0,a-k)+\\max(0,b-k)\\le k.$$\nBreak the excess equal pairs first. Then increase both transition counts using `01` $\\to$ `0101`, and increase equal-pair counts by duplicating individual characters. If the string is still constant, first insert one opposite character to create both transition types."
                },
                {
                  "order": 5,
                  "content": "You can remove the binary search entirely. Use\n$$\\max(0,x)+\\max(0,y)=\\max(0,x,y,x+y).$$\nThe feasibility condition becomes $k\\ge d$, $2k\\ge a+d$, $2k\\ge b+d$, and $3k\\ge a+b+d$. Since $a+d=z$, $b+d=o$, and $a+b+d=m-d$, the minimum target is\n$$k=\\max\\left(1,d,\\left\\lceil\\frac z2\\right\\rceil,\\left\\lceil\\frac o2\\right\\rceil,\\left\\lceil\\frac{m-d}{3}\\right\\rceil\\right).$$\nReturn $4k-m$ for each query."
                }
              ],
              "editorial": "The catch is that insertion can **destroy an existing adjacent pair**. For example, `00` becomes `010` after inserting a `1`. So taking the largest current pair count and padding everything else is not necessarily optimal.\n\nWe will find the minimum possible final length, then subtract the original length. With a little algebra, every query takes $O(1)$.\n\n### 1. Count the cyclic pairs\n\nConsider a queried substring of length $m$. Write\n\n- $z$ and $o$ for its numbers of zeros and ones;\n- $a$ for its number of cyclic `00` pairs;\n- $b$ for its number of cyclic `11` pairs;\n- $d$ for its number of cyclic `01` pairs.\n\nOn a circle, every `0`-to-`1` transition is matched by a `1`-to-`0` transition. Thus `10` also occurs $d$ times.\n\nEach zero starts either a `00` or a `01` pair, and each one starts either a `11` or a `10` pair. Therefore,\n\n$$a=z-d,\\qquad b=o-d,\\qquad m=a+b+2d.$$\n\nIf the completed string has exactly $k$ pairs of each type, its length is $4k$. Our goal is to minimize $k\\ge1$.\n\nWe can reason about insertions into cyclic gaps. Inserting into the last-to-first gap is legal: put those characters after the original substring, before the string wraps back to its first character.\n\n### 2. Which target counts are possible?\n\nFix a target $k$. If $a>k$, at least $a-k$ original `00` gaps must stop contributing a `00` pair. Similarly, at least $b-k$ original `11` gaps must stop contributing a `11` pair.\n\nA gap originally connecting two zeros becomes a path from `0` to `0` after insertions. To contain no `00`, that path must visit a `1`, requiring at least one `01` and one `10`. The same reasoning applies to eliminating a `11` gap.\n\nMeanwhile, every original `01` gap still requires at least one `01`, and every original `10` gap still requires at least one `10`. These gaps are disjoint, so their requirements add up.\n\nConsequently, a necessary condition is\n\n$$d+\\max(0,a-k)+\\max(0,b-k)\\le k.$$\n\nIt is also sufficient. Here is a construction proving it.\n\n**First, remove excess equal pairs.**\n\n- Break exactly $\\max(0,a-k)$ original `00` pairs using `00` $\\to$ `010`.\n- Break exactly $\\max(0,b-k)$ original `11` pairs using `11` $\\to$ `101`.\n\nEach operation reduces the chosen equal-pair count by one and increases both transition counts by one. Now the equal-pair counts are at most $k$, and both transition counts equal\n\n$$d'=d+\\max(0,a-k)+\\max(0,b-k)\\le k.$$\n\n**Next, bring both transition counts up to $k$.**\n\nIf $d'=0$, the string is still constant. Insert one opposite character into any cyclic gap. This decreases an equal-pair count and creates one transition of each type. Since $k\\ge1$, all counts remain at most $k$.\n\nNow a `01` pair exists. Replacing it with `0101` adds one `01` and one `10`, without changing either equal-pair count. Repeat until both transition counts reach $k$.\n\n**Finally, fill the equal-pair deficits.**\n\nDuplicating a zero adds exactly one `00`; duplicating a one adds exactly one `11`. Both characters now exist, so increase these counts independently to $k$.\n\nEvery operation is an insertion. This proves the exact criterion:\n\n$$\\boxed{k\\ge1\\quad\\text{and}\\quad d+\\max(0,a-k)+\\max(0,b-k)\\le k.}$$\n\n### 3. Turn the condition into a formula\n\nThe condition is monotone, so binary search would work. But we can do better without making the implementation harder.\n\nUse the identity\n\n$$\\max(0,x)+\\max(0,y)=\\max(0,x,y,x+y).$$\n\nThen feasibility is equivalent to all four inequalities\n\n$$\n\\begin{aligned}\nd&\\le k,\\\\\na+d&\\le2k,\\\\\nb+d&\\le2k,\\\\\na+b+d&\\le3k.\n\\end{aligned}\n$$\n\nSubstituting $a+d=z$, $b+d=o$, and $a+b+d=m-d$ gives the minimum feasible integer:\n\n$$\\boxed{k=\\max\\left(1,d,\\left\\lceil\\frac z2\\right\\rceil,\\left\\lceil\\frac o2\\right\\rceil,\\left\\lceil\\frac{m-d}{3}\\right\\rceil\\right).}$$\n\nThe answer is\n\n$$\\boxed{4k-m}.$$\n\n### 4. Answer queries with prefix sums\n\nUse one-based indexing and build:\n\n- $P[i]$: the number of ones in $s_1\\ldots s_i$;\n- $D[i]$: the number of changes between adjacent characters entirely inside $s_1\\ldots s_i$.\n\nFor a query $[l,r]$:\n\n$$m=r-l+1,\\qquad o=P[r]-P[l-1],\\qquad z=m-o.$$\n\nThe number of internal changes is $D[r]-D[l]$. Add the closing pair separately:\n\n$$c=D[r]-D[l]+[s_l\\ne s_r].$$\n\nHere $[\\text{condition}]$ is $1$ when the condition is true and $0$ otherwise. Since cyclic transitions come in matched directions,\n\n$$d=\\frac c2.$$\n\nEvaluate the formula using integer ceilings: $(z+1)/2$, $(o+1)/2$, and $(m-d+2)/3$.\n\nA length-one substring works without a special case: $d=0$, $k=1$, and the answer is $3$. Remember that its single cyclic pair is the character paired with itself.\n\n### Correctness\n\nThe prefix sums compute the queried substring's exact character and cyclic transition counts, which determine $a$, $b$, and $d$. The feasibility argument proves that a balanced completion with pair count $k$ exists exactly when the stated inequalities hold. The formula chooses the smallest integer satisfying all of them. Therefore $4k$ is the minimum possible completed length, and $4k-m$ is precisely the minimum number of insertions.\n\n### Complexity\n\nPreprocessing takes $O(n)$ time and $O(n)$ memory. Each query takes $O(1)$ time, for $O(n+q)$ total time. All values fit comfortably in a 32-bit signed integer under the given constraints.",
              "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 n, q;\n    string s;\n    cin >> n >> q >> s;\n    s = \" \" + s;\n\n    vector<int> ones(n + 1), changes(n + 1);\n    for (int i = 1; i <= n; ++i) {\n        ones[i] = ones[i - 1] + (s[i] == '1');\n        changes[i] = changes[i - 1];\n        if (i > 1) changes[i] += (s[i] != s[i - 1]);\n    }\n\n    while (q--) {\n        int l, r;\n        cin >> l >> r;\n        int m = r - l + 1;\n        int o = ones[r] - ones[l - 1];\n        int z = m - o;\n        int d = (changes[r] - changes[l] + (s[l] != s[r])) / 2;\n        int k = max({1, d, (z + 1) / 2, (o + 1) / 2,\n                     (m - d + 2) / 3});\n        cout << 4 * k - m << '\\n';\n    }\n}\n"
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