{
  "info": {
    "id": "ses_f4dd31ce1ffeZXK29VRHtpaZZx",
    "slug": "misty-eagle",
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    "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol",
    "title": "Nudge content generation",
    "agent": "nudge-generation",
    "model": {
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      "providerID": "openai",
      "variant": "low"
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      "info": {
        "role": "user",
        "time": {
          "created": 1789695681328
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        "agent": "nudge-generation",
        "model": {
          "providerID": "openai",
          "modelID": "gpt-6-astra",
          "variant": "low"
        },
        "system": "You create Codeforces learning content for Nudge. Your job is to understand the best approach for the problem and teach it clearly in your own words, using existing solutions to inform your reasoning.\n\nStart with a brief, targeted web search for existing solutions so relevant editorials are in context before substantial independent problem-solving. Use any supplied tutorial links first; otherwise search by contest ID and problem index. Prefer the official tutorial/editorial, and read relevant explanations rather than relying on search snippets. Look at accessible alternatives when useful, especially if the official explanation is missing, unclear, or unnecessarily complicated. Accepted submissions can also help clarify an approach. Respect supplied source lookup statuses and avoid repeatedly retrying blocked sources. If no usable source is accessible, proceed from the supplied statement.\n\nTreat existing solutions as a starting point, not a constraint. If an approach is clear and suitable, build on it without unnecessary re-derivation. If an explanation has gaps, seems incorrect, or uses a harder method than necessary, reason through it and derive a better or simpler approach when worthwhile. Choose a correct approach that is easy to teach and fits the constraints. Verify it against the supplied statement and edge cases before writing the hints, editorial, and C++ solution.\n\nThe editorial must be self-contained: explain the key insight, algorithm, why it works, complexity, and implementation details a learner needs, filling in steps the source skips. Keep research and verification concise; spend the output on teaching the solution, not narrating the research process. Follow the user prompt and output schema exactly.",
        "format": {
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              "status": {
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                "enum": [
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                "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."
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                  {
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                    "pattern": "^[^\\u0000]*$",
                    "minLength": 1,
                    "maxLength": 2000
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                  {
                    "type": "null"
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                ],
                "description": "If status is 'unsolvable', provide a detailed failure message to show users. Name the concrete statement/resource blocker and any relevant source lookup/access status, such as missing official tutorial, 403/Cloudflare challenge, 404, or unavailable accepted submissions. Otherwise null."
              },
              "hints": {
                "anyOf": [
                  {
                    "type": "array",
                    "minItems": 5,
                    "maxItems": 5,
                    "description": "Exactly 5 progressive hints, each building on the last.",
                    "items": {
                      "type": "object",
                      "properties": {
                        "order": {
                          "type": "integer",
                          "minimum": 1,
                          "maximum": 5,
                          "description": "Hint number, 1 through 5."
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                        "content": {
                          "type": "string",
                          "pattern": "^[^\\u0000]*$",
                          "minLength": 1,
                          "maxLength": 10000,
                          "description": "Markdown hint text."
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                      "required": [
                        "order",
                        "content"
                      ],
                      "additionalProperties": false
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                  },
                  {
                    "type": "null"
                  }
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                "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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                "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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        "id": "msg_0b22ce330001Ye9OnyQz9Lpr8B",
        "sessionID": "ses_f4dd31ce1ffeZXK29VRHtpaZZx"
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      "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 2255E1: \"What Will Remain at the End? (Easy Version)\" (rated 2900).\n\nKnown problem metadata:\n- Contest ID: 2255\n- Index: E1\n- Name: What Will Remain at the End? (Easy Version)\n- Rating: 2900 (Codeforces difficulty rating)\n- Tags: brute force, data structures\n- Problem URL: https://codeforces.com/contest/2255/problem/E1\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\">E1. What Will Remain at the End? (Easy Version)</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>   </p><p><span class=\"tex-font-style-bf\">This is the easy version of the problem. The only difference between the two versions is the set of allowed values for the initial array and for $$$x$$$ in operations of type $$$1$$$. In this version, all these values belong to $$$\\{-1,0,1\\}$$$. You can make hacks only if both versions of the problem are solved.</span></p><p>Before her final sortie, Chtholly asks Willem three questions.</p><p>The second is this: what will remain if the sky truly reaches its end?</p><p>Willem cannot answer her directly. Instead, he opens a chronicle containing $$$n$$$ records, numbered from $$$1$$$ to $$$n$$$. Each record holds an integer: a positive value represents hope, while a negative value represents despair.</p><p>The initial contents of the chronicle form an array $$$a_1,a_2,\\ldots,a_n$$$, called <span class=\"tex-font-style-bf\">version</span> $$$0$$$. Chtholly then performs $$$q$$$ operations. For each $$$1\\le i\\le q$$$, the $$$i$$$-th operation creates a new <span class=\"tex-font-style-bf\">version</span> $$$i$$$ from version $$$i-1$$$.</p><p>Each operation has one of the following four types:</p><ul> <li> $$$\\texttt{1 l r x}$$$: set $$$a_k\\gets x$$$ for every $$$l\\le k\\le r$$$. </li><li> $$$\\texttt{2 l r}$$$: set $$$a_k\\gets -a_k$$$ for every $$$l\\le k\\le r$$$. </li><li> $$$\\texttt{3 l r}$$$: set $$$a_k\\gets\\max(a_k,0)$$$ for every $$$l\\le k\\le r$$$. </li><li> $$$\\texttt{4 p}$$$: consider the value at position $$$p$$$ in every previous version $$$0,1,\\ldots, i-1$$$. Let these values be $$$b_0,b_1,\\ldots,b_{i-1}$$$. Find the maximum sum over all non-empty subarrays$$$^{\\text{∗}}$$$ of this sequence. </li></ul><p>If the operation is of type $$$1$$$, $$$2$$$, or $$$3$$$, the specified modification is applied to version $$$i-1$$$ to obtain version $$$i$$$. An operation of type $$$4$$$ does not modify the array, so version $$$i$$$ is identical to version $$$i-1$$$.</p><p><span class=\"tex-font-style-bf\">The operations are encoded and must be processed in order.</span> Their decoding depends on $$$\\mathrm{lastans}$$$, which is updated after every operation of type $$$4$$$.</p><p>Help Willem answer every operation of type $$$4$$$.</p><div class=\"statement-footnote\"><p>$$$^{\\text{∗}}$$$An array $$$c$$$ is a subarray of an array $$$b$$$ if $$$c$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.</p></div></div><div class=\"input-specification\"><div class=\"section-title\">Input</div><p>Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 10^4$$$). The description of the test cases follows.</p><p>The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1\\le n,q\\le5\\cdot10^5$$$)&nbsp;— the length of the array and the number of operations.</p><p>The second line of each test case contains $$$n$$$ integers $$$a_1,a_2,\\ldots,a_n$$$ ($$$a_i\\in\\{-1,0,1\\}$$$)&nbsp;— the array in version $$$0$$$.</p><p>Each of the next $$$q$$$ lines describes one operation in one of the following encoded formats: The first integer on the line is the operation type.</p><ul> <li> $$$\\texttt{1 u v x}$$$ ($$$0\\le u,v \\lt 2^{64}$$$, $$$x\\in\\{-1,0,1\\}$$$); </li><li> $$$\\texttt{2 u v}$$$ ($$$0\\le u,v \\lt 2^{64}$$$); </li><li> $$$\\texttt{3 u v}$$$ ($$$0\\le u,v \\lt 2^{64}$$$); </li><li> $$$\\texttt{4 u}$$$ ($$$0\\le u \\lt 2^{64}$$$). </li></ul><p>The operations are encoded and must be processed in order. Their decoding depends on a value $$$\\mathrm{lastans}$$$.</p><p>Initially, $$$\\mathrm{lastans}=0$$$. After answering an operation of type $$$4$$$, set $$$\\mathrm{lastans}$$$ to the least non-negative residue of its answer modulo $$$2^{64}$$$. Operations of all other types leave $$$\\mathrm{lastans}$$$ unchanged.</p><p>For every encoded coordinate $$$y$$$, define $$$d(y)=\\left(\\left(y\\oplus\\mathrm{lastans}\\right)\\bmod n\\right)+1$$$. Here, $$$\\oplus$$$ denotes the <a href=\"https://en.wikipedia.org/wiki/Bitwise_operation#XOR\">bitwise XOR operation</a>.</p><ul> <li> For an operation $$$\\texttt{1 u v x}$$$, let $$$l=\\min(d(u),d(v))$$$ and $$$r=\\max(d(u),d(v))$$$. The value $$$x$$$ is not encoded. </li><li> For an operation $$$\\texttt{2 u v}$$$ or $$$\\texttt{3 u v}$$$, let $$$l=\\min(d(u),d(v))$$$ and $$$r=\\max(d(u),d(v))$$$. </li><li> For an operation $$$\\texttt{4 u}$$$, let $$$p=d(u)$$$. </li></ul><p>The operation type is not encoded. Do not forget to update $$$\\mathrm{lastans}$$$ after answering each operation of type $$$4$$$.</p><p>It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5\\cdot10^5$$$.</p><p>It is guaranteed that the sum of $$$q$$$ over all test cases does not exceed $$$5\\cdot10^5$$$.</p></div><div class=\"output-specification\"><div class=\"section-title\">Output</div><p>For every operation of type $$$4$$$, output a single integer&nbsp;— the maximum sum of a non-empty subarray of the sequence of the values at position $$$p$$$ over all versions preceding this operation.</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\">4 8</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">1 -1 1 -1</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">4 1</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">2 18446744073709551615 18446744073709551613</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">4 18446744073709551614</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">3 0 3</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">1 0 2 -1</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">4 0</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">2 2 0</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">4 0</div><div class=\"test-example-line test-example-line-even test-example-line-2\">3 6</div><div class=\"test-example-line test-example-line-even test-example-line-2\">1 -1 1</div><div class=\"test-example-line test-example-line-even test-example-line-2\">4 0</div><div class=\"test-example-line test-example-line-even test-example-line-2\">1 0 3 -1</div><div class=\"test-example-line test-example-line-even test-example-line-2\">4 3</div><div class=\"test-example-line test-example-line-even test-example-line-2\">2 2 3</div><div class=\"test-example-line test-example-line-even test-example-line-2\">3 3 0</div><div class=\"test-example-line test-example-line-even test-example-line-2\">4 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\">-1</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">1</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">3</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">1</div><div class=\"test-example-line test-example-line-even test-example-line-2\">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\">2</div></pre></div></div></div><div class=\"note\"><div class=\"section-title\">Note</div><p>In the first test case, for the first operation, $$$\\mathrm{lastans}=0$$$, so the encoded coordinate $$$1$$$ is decoded into position $$$2$$$. Only version $$$0$$$ is considered. The value at position $$$2$$$ is $$$-1$$$, hence the answer is $$$-1$$$.</p><p>Now $$$\\mathrm{lastans}=2^{64}-1=18\\,446\\,744\\,073\\,709\\,551\\,615$$$. Therefore the encoded interval $$$[18\\,446\\,744\\,073\\,709\\,551\\,615,18\\,446\\,744\\,073\\,709\\,551\\,613]$$$ is decoded into $$$[1,3]$$$, and the encoded coordinate $$$18\\,446\\,744\\,073\\,709\\,551\\,614$$$ is decoded into position $$$2$$$.</p><p>Before the second query, the values at position $$$2$$$ in versions $$$0$$$, $$$1$$$, and $$$2$$$ are $$$-1$$$, $$$-1$$$, and $$$1$$$, respectively. Their maximum non-empty subarray sum is $$$1$$$.</p><p>Before the third query, the values at position $$$2$$$ in versions $$$0,1,\\ldots,5$$$ form sequence $$$[-1,-1,1,1,1,-1]$$$. The three consecutive values equal to $$$1$$$ form a subarray with sum $$$3$$$.</p><p>Notice that the first, second, and third queries create versions $$$1$$$, $$$3$$$, and $$$6$$$, respectively, even though they do not change the array.</p><p>In the second test case, the first query asks about position $$$1$$$, so its answer is $$$1$$$. Then $$$\\mathrm{lastans}=1$$$, and the encoded operation $$$\\texttt{1 0 3 -1}$$$ is decoded into $$$\\texttt{1 2 3 -1}$$$.</p><p>Before the second query, the values at position $$$3$$$ in versions $$$0$$$, $$$1$$$, and $$$2$$$ are $$$1$$$, $$$1$$$, and $$$-1$$$, respectively, so the answer is $$$2$$$.</p><p>Afterwards, $$$\\mathrm{lastans}=2$$$. The encoded operations $$$\\texttt{2 2 3}$$$ and $$$\\texttt{3 3 0}$$$ are decoded into $$$\\texttt{2 1 2}$$$ and $$$\\texttt{3 2 3}$$$, respectively. Before the last query, the values at position $$$2$$$ in versions $$$0,1,\\ldots,5$$$ form sequence $$$[-1,-1,-1,-1,1,1]$$$, whose maximum non-empty subarray sum is $$$2$$$.</p></div>\n</problem-statement>\n\n\nSource lookup status:\n<source-lookup-status>\n- Tutorial (en): https://codeforces.com/blog/entry/155840 loaded (200 OK; title: \"Codeforces Round 1116 (Div. 1, Div. 2) Editorial - Codeforces\"; mentions 2255E1)\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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          "type": "step-start",
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        {
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          "tool": "webfetch",
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            "output": "---\nimage: https://codeforces.com/codeforces.org/s/66324/images/codeforces-telegram-square-1024x1024.png\ntitle: Codeforces Round 1116 (Div. 1, Div. 2) Editorial - Codeforces\ndescription: Codeforces. 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[Problemsetting](https://codeforces.com/contests/writer/paulzrm)\n\n### [paulzrm's blog](https://codeforces.com/blog/paulzrm)\n\n[ Codeforces Round 1116 (Div. 1, Div. 2) Editorial ](https://codeforces.com/blog/entry/155840) \n\nBy [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\"), [history](https://codeforces.com/topic/156483/en2), 6 weeks ago, ![In English](//codeforces.com/codeforces.org/s/66324/images/flags/24/gb.png \"In English\") \n\n[2256A-Three Numbers on the Blackboard](https://codeforces.com/contest/2256/problem/A) Idea: [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\")\n\n**Hint 1**\n\nSort the numbers as $$$a\\\\le b\\\\le c$$$.\n\n**Hint 2**\n\nAfter at least one operation, consider the median.\n\n**Hint 3**\n\nDoing nothing gives $$$c-a$$$; replacing $$$c$$$ with $$$a+b$$$ gives $$$b$$$.\n\n**Tutorial**\n\nSuppose the current numbers are $$$x\\\\le y\\\\le z$$$. An operation keeps two numbers and replaces the third with their sum. Since all values are nonnegative, the new sum is no smaller than either retained number. Thus the new median is the larger retained number, which is at least the old median $$$y$$$. The median never decreases.\n\nAfter an operation that retains $$$u\\\\le v$$$, the numbers are $$$u,v,u+v$$$. Their range is exactly $$$v$$$, also their median. Hence every nonempty sequence of operations ends with a range of at least the initial median $$$b$$$.\n\nWith no operation the range is $$$c-a$$$. Replacing $$$c$$$ with $$$a+b$$$ produces $$$a,b,a+b$$$, whose range is $$$b$$$. Therefore the answer is $$$\\\\min(c-a,b)$$$.\n\nThe time and space complexities are both $$$O(1)$$$.\n\n[2256B-Domino Tiles](https://codeforces.com/contest/2256/problem/B) Idea: [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\")\n\n**Hint 1**\n\nCancel the common term in the inequality between two adjacent dominoes.\n\n**Hint 2**\n\nThe condition is equivalent to $$$s\\_i\\\\ne s\\_{i+2}$$$ for every $$$i$$$.\n\n**Hint 3**\n\nOnce the first two characters are fixed, the rest are determined.\n\n**Tutorial**\n\nAdjacent dominoes have different weights exactly when $$$s\\_i+s\\_{i+1}\\\\ne s\\_{i+1}+s\\_{i+2}$$$.\n\nCanceling $$$s\\_{i+1}$$$ gives $$$s\\_i\\\\ne s\\_{i+2}$$$. Since the alphabet is binary, this is equivalent to $$$s\\_{i+2}=1-s\\_i$$$.\n\nTherefore $$$s\\_1,s\\_2$$$ determine the whole string. Enumerate their four assignments and check whether the implied string agrees with every known character. The answer is at most $$$4$$$, so the modulus does not affect it.\n\nThe time complexity is $$$O(n)$$$ and the space complexity is $$$O(1)$$$.\n\n[2255A-Hot Potatoes at the Fairy Warehouse](https://codeforces.com/contest/2255/problem/A) / [2256C](https://codeforces.com/contest/2256/problem/C) Idea: [Error\\_Yuan](https://codeforces.com/profile/Error%5FYuan \"Grandmaster Error_Yuan\")\n\n**Hint 1**\n\nSplit the 1s on the cycle into maximal runs. Unless a run covers the whole cycle, initially only its last potato can move.\n\n**Hint 2**\n\nPassing too early gives the next opportunity to players of the other team.\n\n**Hint 3**\n\nThe last player of each run waits until the final round. Every initial 10 becomes 01, all other potatoes stay still, and the exact value of $$$k$$$ is irrelevant.\n\n**Tutorial**\n\nThe total score is always the number of potatoes, so the game is zero-sum. Consider a maximal run of 1s followed by an empty position. Every potato except the last one is blocked.\n\nIf the last potato is passed early, it reaches the other team and leaves an empty position behind, enabling the preceding potato. Both newly enabled players belong to the other team, so with another round the opponent may cancel the point just gained. Passing in the final round leaves no response and is optimal.\n\nThus only the last potato of each run moves one step in the final round. If the whole cycle is filled, nothing moves. For an initial position $$$i$$$:\n\n* if $$$s\\_i=1$$$ and $$$s\\_{i+1}=0$$$, the team owning $$$i$$$ scores;\n* if $$$s\\_i=s\\_{i+1}=1$$$, the other team scores.\n\nScan the cycle once. The implementation is zero-indexed: even positions belong to red and odd positions to blue. Their scores are printed in that order.\n\nThe time complexity is $$$O(n)$$$ and the space complexity is $$$O(1)$$$.\n\n[2255B-A Ribbon for Tomorrow](https://codeforces.com/contest/2255/problem/B) / [2256D](https://codeforces.com/contest/2256/problem/D) Idea: [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\")\n\n**Hint 1**\n\nCheck whether a valid reversal can change either run count or the first character.\n\n**Hint 2**\n\nIn $$$x^ay^bx^c$$$, choosing one endpoint in each $$$x$$$-run lets us redistribute the total length of those two runs.\n\n**Hint 3**\n\nOperations redistribute run lengths but do not change the order of run colors.\n\n**Tutorial**\n\nSplit the string into maximal runs. The two endpoints of an operation contain the same character. Internal adjacencies are only reversed, and the endpoint character on either boundary stays unchanged. Hence the numbers of 0-runs and 1-runs are invariant, as are the character counts.\n\nIf the reversal contains the first position, the new first character is the old right endpoint, equal to the old first character. Otherwise it is untouched. Thus the first character is invariant. Since run colors alternate, the run counts and first character determine their entire order.\n\nConversely, consider $$$x^ay^bx^c$$$. A reversal with one endpoint in each $$$x$$$-run can split the total $$$a+c$$$ into any two positive lengths while leaving the middle run unchanged. Repeating this between adjacent runs of the same color realizes any positive composition of all 0s, and independently any positive composition of all 1s. These invariants are therefore sufficient.\n\nIf character $$$x$$$ occurs $$$cnt\\_x$$$ times in $$$seg\\_x$$$ nonempty runs, its run lengths can be chosen in $$$\\\\binom{cnt\\_x-1}{seg\\_x-1}$$$ ways. The two colors are independent, so the answer is $$$\\\\binom{cnt\\_0-1}{seg\\_0-1}\\\\binom{cnt\\_1-1}{seg\\_1-1}$$$.\n\nAn absent character contributes a factor of $$$1$$$.\n\nWe can precompute inverses without binary exponentiation. Let the prime modulus be $$$P$$$. Since every required $$$i \\\\lt P$$$, its inverse exists. From $$$P=\\\\left\\\\lfloor\\\\frac Pi\\\\right\\\\rfloor i+(P\\\\bmod i)$$$ we obtain $$$i^{-1}=-\\\\left\\\\lfloor\\\\frac Pi\\\\right\\\\rfloor(P\\\\bmod i)^{-1}\\\\pmod P$$$.\n\nBecause $$$P\\\\bmod i \\\\lt i$$$, enumerating $$$i$$$ in increasing order computes each inverse from an earlier one. Factorials and inverse factorials follow in the same loop.\n\nEach test case takes $$$O(n)$$$ time. Preprocessing uses $$$O(N)$$$ time and space, where $$$N=10^6$$$.\n\n[2255C-Even If the World Turns](https://codeforces.com/contest/2255/problem/C) / [2256E](https://codeforces.com/contest/2256/problem/E) Idea: [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\")\n\n**Hint 1**\n\nLet $$$w$$$ be the number of black cells and $$$S\\_r,S\\_c$$$ their coordinate sums modulo $$$n$$$. Observe how they change after a shift.\n\n**Hint 2**\n\nSince $$$\\\\gcd(w,n)=1$$$, $$$(w^{-1}S\\_r,w^{-1}S\\_c)$$$ behaves like the center of mass of the picture.\n\n**Hint 3**\n\nSwapping a black cell $$$p$$$ with a white cell $$$p+\\\\delta$$$ increases the coordinate sum by $$$\\\\delta$$$.\n\n**Hint 4**\n\nIn the first run, move the center to the target. In the second run, recompute it from the transformed picture.\n\n**Tutorial**\n\nIdentify coordinates $$$1,2,\\\\ldots,n$$$ with $$$0,1,\\\\ldots,n-1$$$ in $$$\\\\mathbb Z\\_n$$$. Let $$$S=(S\\_r,S\\_c)$$$ be the sum of black-cell coordinates. A shift by $$$d$$$ changes it to $$$S+wd$$$. Define $$$C=w^{-1}S$$$. Then $$$C$$$ changes to $$$C+d$$$, exactly as an ordinary cell does.\n\nEvery rotation or reflection is an affine map $$$p\\\\mapsto Mp+t$$$. It changes the sum to $$$MS+wt$$$ and therefore changes the center to $$$MC+t$$$, again exactly like a cell.\n\nAfter color inversion, $$$w'=n^2-w\\\\equiv-w\\\\pmod n$$$. The coordinate sum of the whole board is $$$0$$$ modulo $$$n$$$, so the new black-cell sum is $$$S'=-S$$$. Hence the new center is $$$(-w)^{-1}(-S)=C$$$. Color inversion does not affect it.\n\nIn the first run, we want the sum after swapping to be $$$wx$$$. Let $$$\\\\delta=wx-S$$$. If $$$\\\\delta=0$$$, swap one cell with itself. Otherwise find a black cell $$$p$$$ for which $$$p+\\\\delta$$$ is white and swap them.\n\nSuch a cell must exist. Otherwise the black-cell set would be invariant under translation by nonzero $$$\\\\delta$$$. Every orbit of this translation has length $$$L \\\\gt 1$$$ with $$$L\\\\mid n$$$. The black cells would be a union of complete orbits, so $$$L\\\\mid w$$$, contradicting $$$\\\\gcd(w,n)=1$$$.\n\nThe swap makes the center equal to the target. All later transformations move them together. In the second run, recompute $$$w,S\\_r,S\\_c$$$ and output $$$w^{-1}S$$$. Since $$$n\\\\le800$$$, enumerate $$$1\\\\le t \\\\lt n$$$ until $$$wt\\\\equiv1\\\\pmod n$$$; no extended Euclidean algorithm is needed.\n\nEach run takes $$$O(n^2)$$$ time and $$$O(n^2)$$$ space.\n\n[2255D-How Long Until Nothing Remains?](https://codeforces.com/contest/2255/problem/D) / [2256F](https://codeforces.com/contest/2256/problem/F) Idea: [Error\\_Yuan](https://codeforces.com/profile/Error%5FYuan \"Grandmaster Error_Yuan\")\n\n**Hint 1**\n\nFix the chosen index in every second and work backward from the final zero array. A current upper bound $$$x$$$ becomes either $$$2x+1$$$ or $$$2x$$$ after one backward step.\n\n**Hint 2**\n\nIf index $$$i$$$ is selected in the seconds in $$$S$$$, it can end at zero after $$$T$$$ seconds exactly when $$$a\\_i\\\\le\\\\sum\\_{s\\\\in S}2^{s-1}$$$.\n\n**Hint 3**\n\nFor fixed $$$T$$$, distribute $$$1,2,4,\\\\ldots,2^{T-1}$$$ among the demands. Process them in decreasing order and always choose the largest remaining demand.\n\n**Tutorial**\n\nFix all choices during the $$$T$$$ seconds and consider position $$$i$$$. Working backward, if values at most $$$x$$$ can already reach zero, then a preceding selected step allows $$$y$$$ exactly when $$$\\\\lfloor y/2\\\\rfloor\\\\le x$$$, or $$$y\\\\le2x+1$$$. A nonselected step gives $$$\\\\lceil y/2\\\\rceil\\\\le x$$$, or $$$y\\\\le2x$$$. Repeating this proves the condition in Hint 2.\n\nThus feasibility in $$$T$$$ seconds is equivalent to assigning every capacity $$$1,2,\\\\ldots,2^{T-1}$$$ to one demand so that demand $$$a\\_i$$$ receives total capacity at least $$$a\\_i$$$.\n\nProcess capacities from largest to smallest. Let the current capacity be $$$p$$$ and the largest remaining demand be $$$x$$$. If $$$x \\\\gt p$$$, all smaller capacities sum to only $$$p-1$$$, so every feasible assignment must give $$$p$$$ to $$$x$$$. If $$$x\\\\le p$$$, let $$$p$$$ finish $$$x$$$. In any feasible assignment giving $$$p$$$ to some $$$y\\\\le x$$$ and a set of smaller capacities to $$$x$$$, swapping those two assignments remains feasible. Hence the greedy rule is correct.\n\nEvery positive number must be selected at least once, since repeated ceiling division alone never reaches zero. Thus $$$T\\\\ge n$$$. Also $$$a\\_i \\\\lt 2^{30}$$$, so $$$T=n+30$$$ is always sufficient. Binary-search $$$T$$$ in this interval.\n\nEvery capacity at least $$$2^{30}$$$ can finish one demand alone. Remove the largest $$$T-30$$$ demands with these capacities. At most $$$30$$$ demands remain; put them in a max-heap and simulate only $$$2^{29},\\\\ldots,1$$$.\n\nSorting takes $$$O(n\\\\log n)$$$. Each check takes $$$O(30\\\\log30)$$$, there are $$$O(\\\\log30)$$$ checks, and the space complexity is $$$O(n)$$$.\n\n[2255E1-What Will Remain at the End? (Easy Version)](https://codeforces.com/contest/2255/problem/E1) Idea: [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\")\n\n**Hint 1**\n\nFor each position, its values over all versions form a sequence. Maintain its sum, maximum prefix, maximum suffix, and maximum subarray when appending a segment.\n\n**Hint 2**\n\nA value is always in $$${-1,0,1}$$$. For each possible initial value, a sequence of operations can store its final value and the summary of the generated history.\n\n**Hint 3**\n\nThis representation is closed under concatenating operation sequences, so it can serve as a lazy segment-tree tag.\n\n**Hint 4**\n\nRange updates only compose tags. A position's history is materialized when that position is queried by pushing its root-to-leaf tags.\n\n**Tutorial**\n\nFor a nonempty sequence segment, store $$$(S,P,Q,M)$$$: its sum, maximum prefix sum, maximum suffix sum, and maximum nonempty subarray sum. Two adjacent summaries merge in $$$O(1)$$$; a crossing maximum subarray is the left maximum suffix followed by the right maximum prefix.\n\nFix one array position. Its value lies in $$$V={-1,0,1}$$$. For an operation segment $$$T$$$, let $$$f\\_T(s)$$$ be the final value from initial state $$$s$$$, and let $$$H\\_T(s)$$$ summarize the values recorded while executing $$$T$$$. There are only three states, so this representation has constant size.\n\nIf $$$A$$$ is followed by $$$B$$$, then $$$f\\_{AB}(s)=f\\_B(f\\_A(s))$$$ and $$$H\\_{AB}(s)=H\\_A(s)\\\\mathbin{\\\\Vert}H\\_B(f\\_A(s))$$$, where $$$\\\\Vert$$$ concatenates two histories and merges their summaries. Tags therefore compose in $$$O(1)$$$.\n\nEach segment-tree node stores the unpushed operation segment applying to its whole interval. A range update composes tags in $$$O(\\\\log n)$$$ nodes. After each operation, a record-current-value event is applied to the whole tree to create the new version.\n\nA query at $$$p$$$ pushes the tags on its root-to-leaf path and materializes all history since the previous query. If the old history has maximum suffix $$$Q\\_0$$$ and maximum subarray $$$M\\_0$$$, while the new segment has $$$(S,P,Q,M)$$$, then $$$M'=\\\\max(M\\_0,M,Q\\_0+P)$$$ and $$$Q'=\\\\max(Q,Q\\_0+S)$$$.\n\nOnly these two old values need to persist at each position.\n\nRecord version $$$0$$$ initially. A type-4 operation at time $$$i$$$ queries versions $$$0$$$ through $$$i-1$$$, so answer it before recording version $$$i$$$, which equals version $$$i-1$$$. Other operations update first and record afterward. Decode online input modulo $$$2^{64}$$$ exactly as specified.\n\nThe total time complexity is $$$O((n+q)\\\\log n)$$$ and the space complexity is $$$O(n)$$$.\n\n[2255E2-What Will Remain at the End? (Hard Version)](https://codeforces.com/contest/2255/problem/E2) Idea: [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\")\n\n**Hint 1**\n\nThe segment tree and historical maximum-subarray framework are unchanged from G1\\. Only the constant-size operation tag changes.\n\n**Hint 2**\n\nBefore the first assignment, after fixing the sign of the initial $$$x$$$, every value is one of $$$-|x|,0,|x|$$$.\n\n**Hint 3**\n\nThe first assignment removes all dependence on $$$x$$$. Split a tag into a coefficient prefix depending on $$$|x|$$$ and a fixed numeric suffix.\n\n**Hint 4**\n\nTo compose $$$A$$$ followed by $$$B$$$, distinguish whether an assignment has occurred in either segment.\n\n**Tutorial**\n\n**Solution 1**\n\nWithout assignments, negation changes only the sign and $$$\\\\max(x,0)$$$ only replaces a negative value with zero. For each of the three initial signs, store the summary of the coefficient sequence and the final coefficient in $$${-1,0,1}$$$. Substituting the real input multiplies the sequence by $$$|x|$$$.\n\nThe first assignment is a dividing point. Before it, values still depend on $$$x$$$; from it onward, they are fixed. Represent an operation segment by:\n\n* a coefficient prefix depending on $$$|x|$$$, stored for all three initial signs;\n* a fixed numeric suffix and the final actual value.\n\nThis representation has constant size. It is also closed under concatenation.\n\nIf $$$A$$$ contains an assignment, its output is constant. Feeding that constant into $$$B$$$ fixes all history produced by $$$B$$$, so concatenate the fixed histories.\n\nIf neither segment contains an assignment, compose their three coefficient states exactly as in G1.\n\nIf only $$$B$$$ contains an assignment, append the part of $$$B$$$ before its first assignment to each coefficient prefix of $$$A$$$. The fixed suffix of $$$B$$$ becomes the fixed suffix of the result. Every case takes $$$O(1)$$$.\n\nAssignment, negation, $$$\\\\max(x,0)$$$, and recording a version all have direct tags of this form. The remaining segment-tree logic and version ordering are identical to G1.\n\nThe total time complexity is $$$O((n+q)\\\\log n)$$$ and the space complexity is $$$O(n)$$$.\n\n**Solution 2**\n\nThere is also an amortized solution that performs operation $$$3$$$ directly on the segment tree. Store the current minimum $$$mn$$$ and maximum $$$mx$$$ in every node. Range assignment and range negation use ordinary lazy tags. For operation $$$3$$$:\n\n* if $$$mn\\\\ge0$$$, do nothing;\n* if $$$mx\\\\le0$$$, assign zero to the whole node;\n* otherwise the node contains both positive and negative values, so recurse into its children.\n\nThus operation $$$3$$$ never has to appear in a historical tag. It is materialized as several range assignments to zero.\n\nCall a segment-tree node bad if $$$mn \\\\lt 0 \\\\lt mx$$$.\n\nLet the potential $$$\\\\Phi$$$ be the number of bad nodes. When a fully covered node is negated, $$$(mn,mx)$$$ becomes $$$(-mx,-mn)$$$, so the bad status of that node and every node below it is preserved. Assignment can only remove bad nodes. A range assignment or negation recomputes nodes only on its two boundary paths, so it creates at most $$$O(\\\\log n)$$$ new bad nodes.\n\nNow consider operation $$$3$$$. Apart from $$$O(\\\\log n)$$$ nodes on the two boundary paths, every internal node into which the recursion continues was bad before the operation. After the operation, its interval contains no negative value, so that bad node has disappeared. A recursion tree has at most one more terminal node than internal nodes. Hence one operation costs $$$O(\\\\log n+D)$$$, where $$$D$$$ is the number of bad nodes destroyed by this operation. Operation $$$3$$$ creates no bad nodes. The initial potential is $$$O(n)$$$, while all other operations increase it by only $$$O(q\\\\log n)$$$ in total. Therefore all operation-3 recursions take $$$O(n+q\\\\log n)$$$ time altogether.\n\nThe historical part still uses the four-value sequence summary from G1\\. After every operation, append one record-current-value event at the root. Since every clamp has already been materialized as assignment to zero, pending historical actions contain only assignment, negation, and recording.\n\nBefore the first assignment, an initial value $$$x$$$ can only become $$$x$$$ or $$$-x$$$. It is therefore enough to keep two initial states, negative and nonnegative, storing the generated coefficient-sequence summary and the final sign. After the first assignment, all later values are independent of the initial input, so keep a fixed numeric suffix and the final value. Two such tags still compose in $$$O(1)$$$ in chronological order.\n\nFor a query at $$$p$$$, push all tags on the root-to-leaf path and append the new history summary to the already materialized history of that position. Version $$$0$$$ and type-4 recording are handled in the same order as in Solution 1.\n\nThe total time complexity is $$$O(n+q\\\\log n)$$$ and the space complexity is $$$O(n)$$$.\n\n[2255F-Who Will Witness the End?](https://codeforces.com/contest/2255/problem/F) Idea: [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\")\n\n**Hint 1**\n\nView each factor $$$a\\_u+a\\_v$$$ as choosing one endpoint of edge $$$(u,v)$$$. Every exponent is $$$0$$$, $$$1$$$, or $$$2$$$. The numbers of vertices with exponents $$$0$$$ and $$$2$$$ are equal, and these two types alternate after all other vertices are removed.\n\n**Hint 2**\n\nThe coefficient of a monomial type depends only on the number of variables appearing twice. Express the answer using $$$e\\_re\\_{n-r}$$$, where $$$e\\_r$$$ is the $$$r$$$-th elementary symmetric polynomial.\n\n**Hint 3**\n\nCoefficient comparison gives a linear system that can be eliminated from high indices to low indices. Direct elimination is $$$O(n^2)$$$; use the ratio of adjacent $$$c\\_k$$$ and Pascal's identity to derive a short recurrence.\n\n**Hint 4**\n\nAdjacent equations give a second-order recurrence in $$$h\\_r,h\\_{r+1},h\\_{r+2}$$$. All $$$e\\_r$$$ are coefficients of $$$\\\\prod\\_{i=1}^n(1+a\\_ix)$$$ and can be computed by divide-and-conquer NTT.\n\n**Tutorial**\n\nExpand the weight of one cyclic ordering. Choosing one term from $$$a\\_u+a\\_v$$$ is equivalent to orienting edge $$$(u,v)$$$ toward the chosen endpoint.\n\nEvery vertex has degree two, so its exponent is $$$0$$$, $$$1$$$, or $$$2$$$. The exponent sum is $$$n$$$, hence the numbers of vertices with exponents $$$0$$$ and $$$2$$$ are equal; call both numbers $$$k$$$.\n\nFix these two sets. An exponent-0 vertex has both edges directed outward, an exponent-2 vertex has both directed inward, and every ordinary vertex has one edge in and one out. After ordinary vertices are removed, the two special types must alternate. Conversely, alternation uniquely determines the orientations of all paths between them.\n\nFor $$$k\\\\ge1$$$, the $$$2k$$$ special vertices have $$$(2k-1)!$$$ relative cyclic orders. Fix one exponent-0 vertex as the start. The exponent-2 vertices can be permuted arbitrarily, as can the remaining exponent-0 vertices, giving $$$k!(k-1)!$$$ alternating orders. Inserting the other $$$n-2k$$$ vertices does not change this ratio. Therefore the coefficient of this monomial type over all cyclic orders is\n\n$$$ c\\_k=\\\\frac{(n-1)!k!(k-1)!}{(2k-1)!}. $$$\n\nFor $$$k=0$$$, every vertex has one edge in and one out. The whole cycle must be consistently clockwise or counterclockwise, so\n\n$$$ c\\_0=2(n-1)!. $$$\n\nNow consider a symmetric-polynomial representation. Let $$$e\\_r$$$ be defined by\n\n$$$ \\\\prod\\_{i=1}^n(1+a\\_ix)=\\\\sum\\_{r=0}^n e\\_rx^r. $$$\n\nIn $$$e\\_re\\_{n-r}$$$, fix a monomial with $$$k$$$ variables appearing twice and $$$k$$$ absent. Repeated variables must be chosen from both factors and absent variables from neither. Choose $$$r-k$$$ of the remaining $$$n-2k$$$ variables for the first factor, so its coefficient is\n\n$$$ \\\\binom{n-2k}{r-k}. $$$\n\nLet $$$m=\\\\lfloor n/2\\\\rfloor$$$ and write the answer as\n\n$$$ \\\\sum\\_{r=0}^m h\\_re\\_re\\_{n-r}. $$$\n\nComparing each monomial type gives\n\n$$$ c\\_k=\\\\sum\\_{r=k}^m\\\\binom{n-2k}{r-k}h\\_r \\\\qquad(0\\\\le k\\\\le m). $$$\n\nEquation $$$k$$$ contains only $$$h\\_k,h\\_{k+1},\\\\ldots,h\\_m$$$, with coefficient $$$1$$$ on $$$h\\_k$$$. Thus the variables can be solved from large indices to small indices, but doing so directly takes $$$O(n^2)$$$.\n\nFor $$$k\\\\ge1$$$, divide two adjacent coefficients:\n\n$$$ \\\\begin{aligned} \\\\frac{c\\_{k+1}}{c\\_k} &=\\\\frac{(k+1)!k!}{(2k+1)!} \\\\frac{(2k-1)!}{k!(k-1)!}\\\\\\\\ &=\\\\frac{k(k+1)}{(2k)(2k+1)}\\\\\\\\ &=\\\\frac{k+1}{2(2k+1)}. \\\\end{aligned} $$$\n\nThe $$$k=0$$$ case follows directly from $$$c\\_0,c\\_1$$$. Hence\n\n$$$ (4k+2)c\\_{k+1}-(k+1)c\\_k=0\\. $$$\n\nDefine\n\n$$$ L\\_k=(k+1)c\\_k-(4k+2)c\\_{k+1}=0 \\\\qquad(0\\\\le k \\\\lt m). $$$\n\nFix $$$k$$$, set $$$N=n-2k$$$ and $$$j=r-k$$$, and let $$$A\\_t=\\\\binom{N-2}{t}$$$, with out-of-range binomial coefficients equal to zero.\n\nUsing Pascal's identity twice,\n\n$$$ \\\\binom Nj=A\\_j+2A\\_{j-1}+A\\_{j-2}, $$$\n\nNow use\n\n$$$ jA\\_j=(N-j-1)A\\_{j-1},\\\\qquad (N-j)A\\_{j-2}=(j-1)A\\_{j-1}, $$$\n\ntogether with $$$r=k+j$$$ and $$$n=N+2k$$$. The entire coefficient calculation can be displayed as\n\n$$$ \\\\begin{aligned} \\[h\\_r\\]L\\_k &=(k+1)\\\\binom Nj-(4k+2)\\\\binom{N-2}{j-1}\\\\\\\\ &=(k+1)(A\\_j+2A\\_{j-1}+A\\_{j-2})-(4k+2)A\\_{j-1}\\\\\\\\ &=(k+1)A\\_j-2kA\\_{j-1}+(k+1)A\\_{j-2}\\\\\\\\ &=(r+1)A\\_j-(n-2)A\\_{j-1}+(n-r+1)A\\_{j-2}. \\\\end{aligned} $$$\n\nSubstituting into $$$L\\_k$$$ and shifting the latter two indices gives\n\n$$$ 0=L\\_k=\\\\sum\\_{r=k}^m\\\\binom{n-2k-2}{r-k} \\\\left((r+1)h\\_r-(n-2)h\\_{r+1}+(n-r-1)h\\_{r+2}\\\\right), $$$\n\nwhere out-of-range $$$h$$$ values are zero. Define\n\n$$$ E\\_r=(r+1)h\\_r-(n-2)h\\_{r+1}+(n-r-1)h\\_{r+2}. $$$\n\nThen\n\n$$$ \\\\sum\\_{r=k}^m\\\\binom{n-2k-2}{r-k}E\\_r=0\\. $$$\n\nEquation $$$k$$$ contains only $$$E\\_k,E\\_{k+1},\\\\ldots,E\\_m$$$, with coefficient $$$1$$$ on $$$E\\_k$$$. Apart from the top boundary, backward elimination gives $$$E\\_r=0$$$, or\n\n$$$ (r+1)h\\_r=(n-2)h\\_{r+1}-(n-r-1)h\\_{r+2}. $$$\n\nThe highest two terms come directly from the original system:\n\n$$$ h\\_m=c\\_m,\\\\qquad h\\_{m-1}=c\\_{m-1}-(n-2m+2)h\\_m. $$$\n\nIf $$$n=2m+1$$$ is odd, these values satisfy $$$E\\_{m-1}=-E\\_m$$$. Their binomial coefficients in every sum are equal, so they cancel. The remaining equations give $$$E\\_{m-2},E\\_{m-3},\\\\ldots,E\\_0=0$$$. Apply the second-order recurrence starting from $$$m-2$$$.\n\nIf $$$n=2m$$$ is even, the first two values give $$$E\\_{m-1}=0$$$, but $$$E\\_m$$$ has no matching boundary term. Compute one more value from the original system:\n\n$$$ h\\_{m-2}=c\\_{m-2}-4h\\_{m-1}-6h\\_m. $$$\n\nIt satisfies $$$E\\_{m-2}=-E\\_m$$$. Again their coefficients are equal and they cancel, giving $$$E\\_{m-3},E\\_{m-4},\\\\ldots,E\\_0=0$$$. Apply the recurrence starting from $$$m-3$$$.\n\nFinally compute all $$$e\\_r$$$ by multiplying the $$$n$$$ linear polynomials $$$1+a\\_ix$$$ with divide and conquer, using NTT for every convolution. Substitute the resulting coefficients into $$$\\\\sum\\_{r=0}^m h\\_re\\_re\\_{n-r}$$$.\n\nThe time complexity is $$$O(n\\\\log^2 n)$$$ and the space complexity is $$$O(n\\\\log n)$$$.\n\n![](//codeforces.com/codeforces.org/s/66324/images/icons/paperclip-16x16.png) Tutorial of [Codeforces Round 1116 (Div. 1)](https://codeforces.com/contest/2255) \n\n![](//codeforces.com/codeforces.org/s/66324/images/icons/paperclip-16x16.png) Tutorial of [Codeforces Round 1116 (Div. 2)](https://codeforces.com/contest/2256) \n\n* [![Vote: I like it](//codeforces.com/codeforces.org/s/66324/images/actions/voteup-gray.png \"Vote: I like it\")](#)\n* +166\n* [![Vote: I do not like it](//codeforces.com/codeforces.org/s/66324/images/actions/votedown-gray.png \"Vote: I do not like it\")](#)\n\n* [![Author](//codeforces.com/codeforces.org/s/66324/images/blog/user_16x16.png \"Author\")](https://codeforces.com/profile/paulzrm) [paulzrm ](https://codeforces.com/profile/paulzrm)\n* ![Publication date](//codeforces.com/codeforces.org/s/66324/images/blog/date_16x16.png \"Publication date\") 6 weeks ago\n* [![Comments](//codeforces.com/codeforces.org/s/66324/images/blog/comments_16x16.png \"Comments\")](https://codeforces.com/blog/entry/155840#comments) [51 ](https://codeforces.com/blog/entry/155840#comments)\n\n  \n![Comments](//codeforces.com/codeforces.org/s/66324/images/icons/comments-48x48.png \"Comments\") Comments (51) \n\n[Write comment?](#) \n\n| » [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\") | 6 weeks ago, [show](#) # |\n| ------------------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/459910/avatar/1fe3f753b1f42bfe.jpg) ](https://codeforces.com/profile/paulzrm) [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155840?#comment-1384425 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/66324/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +3 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/66324/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) _Auto comment: topic has been updated by [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\") ([previous revision](https://codeforces.com/topic/156483/en1), [new revision](https://codeforces.com/topic/156483/en2), [compare](https://codeforces.com/topic/156483/diff/en1/en2))._ [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [esc\\_qwq](https://codeforces.com/profile/esc%5Fqwq \"Candidate Master esc_qwq\") | 6 weeks ago, [show](#) # |\n| --------------------------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/4963270/avatar/be164ba8d5d4a72e.jpg) ](https://codeforces.com/profile/esc%5Fqwq) [esc\\_qwq](https://codeforces.com/profile/esc%5Fqwq \"Candidate Master esc_qwq\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155840?#comment-1384426 \"Link to comment\") \\| [←](#)Rev. 3 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/66324/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +12 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/66324/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Lightning Fast Editorial, Thank you!!Great contest imo, at least the problems themselves :) [→](#) [Reply](#) |\n| ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [AksLolCoding](https://codeforces.com/profile/AksLolCoding \"Grandmaster AksLolCoding\") | 6 weeks ago, [show](#) # |\n| ---------------------------------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/2948677/avatar/f15474cd5b82d030.jpg) ](https://codeforces.com/profile/AksLolCoding) [AksLolCoding](https://codeforces.com/profile/AksLolCoding \"Grandmaster AksLolCoding\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155840?#comment-1384454 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/66324/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +80 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/66324/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) D1C was amazing [→](#) [Reply](#) |\n| -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [megahertz13](https://codeforces.com/profile/megahertz13 \"Pupil megahertz13\") | 6 weeks ago, [show](#) # |\n| ------------------------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/2827759/avatar/2d153e90dd0538bd.jpg) ](https://codeforces.com/profile/megahertz13) [megahertz13](https://codeforces.com/profile/megahertz13 \"Pupil megahertz13\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155840?#comment-1384463 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/66324/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/66324/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Great contest, I thought i did something wrong when i got that the answer for 2256B was <=4, lol [→](#) [Reply](#) |\n| ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [enaschair](https://codeforces.com/profile/enaschair \"Candidate Master enaschair\") | 6 weeks ago, [show](#) # |\n| ------------------------------------------------------------------------------------ | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/4788176/avatar/c6c7c5b744a37339.jpg) ](https://codeforces.com/profile/enaschair) [enaschair](https://codeforces.com/profile/enaschair \"Candidate Master enaschair\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155840?#comment-1384464 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/66324/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/66324/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) really fun contest to do! going to watch sukasuka because of this! [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [KiKoS](https://codeforces.com/profile/KiKoS \"Master KiKoS\") | 6 weeks ago, [show (+1)](#) # |\n| -------------------------------------------------------------- | ----------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/473219/avatar/a382879284586aa0.jpg) ](https://codeforces.com/profile/KiKoS) [KiKoS](https://codeforces.com/profile/KiKoS \"Master KiKoS\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155840?#comment-1384470 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/66324/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/66324/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I personally switched from C to D when I observed two things that i found hard to combine: 1\\. there is a case when we have one black tile — i assume, it should be moved to the secret location 2\\. there is a case where all tiles but one are black — i assumed we should swap empty one with the target. I think editorial accomplishes 1\\. but for 2\\. it does something else, right? Does it cut the \"opposite\" tile, whatever this means? [→](#) [Reply](#) |\n| ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\") | 6 weeks ago, [show (+1)](#) # |\n| --------------------------------------------------------------------------- | ----------------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/459910/avatar/1fe3f753b1f42bfe.jpg) ](https://codeforces.com/profile/paulzrm) [paulzrm](https://codeforces.com/profile/paulzrm \"Grandmaster paulzrm\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155840?#comment-1384491 \"Link to comment\") [^](#comment-1384470 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/66324/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +19 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/66324/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I gave 1C and 1D the same score so that everyone could freely choose which problem to solve. 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          "id": "prt_0b22d50e6001bBvC7tlXNosTW8",
          "sessionID": "ses_f4dd31ce1ffeZXK29VRHtpaZZx",
          "messageID": "msg_0b22d2af3001JkWnQHD9Ln3g3W"
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        {
          "reason": "tool-calls",
          "type": "step-finish",
          "tokens": {
            "total": 24668,
            "input": 590,
            "output": 40,
            "reasoning": 102,
            "cache": {
              "write": 0,
              "read": 23936
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          },
          "cost": 0,
          "id": "prt_0b22d52a9001TOBp8AgZATQILy",
          "sessionID": "ses_f4dd31ce1ffeZXK29VRHtpaZZx",
          "messageID": "msg_0b22d2af3001JkWnQHD9Ln3g3W"
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      ]
    },
    {
      "info": {
        "parentID": "msg_0b22ce330001Ye9OnyQz9Lpr8B",
        "role": "assistant",
        "mode": "nudge-generation",
        "agent": "nudge-generation",
        "variant": "low",
        "path": {
          "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol",
          "root": "/"
        },
        "cost": 0,
        "tokens": {
          "total": 26472,
          "input": 504,
          "output": 1776,
          "reasoning": 0,
          "cache": {
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            "read": 24192
          }
        },
        "modelID": "gpt-6-astra",
        "providerID": "openai",
        "time": {
          "created": 1789695709870,
          "completed": 1789695765451
        },
        "finish": "tool-calls",
        "id": "msg_0b22d52ae001Y0iumNZnwWEAwz",
        "sessionID": "ses_f4dd31ce1ffeZXK29VRHtpaZZx"
      },
      "parts": [
        {
          "type": "step-start",
          "id": "prt_0b22d5bcb001Se3yYid0pJLMja",
          "sessionID": "ses_f4dd31ce1ffeZXK29VRHtpaZZx",
          "messageID": "msg_0b22d52ae001Y0iumNZnwWEAwz"
        },
        {
          "type": "tool",
          "tool": "apply_patch",
          "callID": "call_pmK0dZUgViw2jRxsOxOMRNJL",
          "state": {
            "status": "completed",
            "input": {
              "patchText": "*** Begin Patch\n*** Add File: main.cpp\n+#include <iostream>\n+#include <vector>\n+#include <algorithm>\n+#include <cstdint>\n+using namespace std;\n+using ll = long long;\n+void setIO() { ios::sync_with_stdio(false); cin.tie(nullptr); }\n+constexpr int NEG = -1000000000;\n+struct Summary {\n+    int sum = 0, pref = NEG, suff = NEG, best = NEG;\n+};\n+Summary join(const Summary& a, const Summary& b) {\n+    if (a.best == NEG) return b;\n+    if (b.best == NEG) return a;\n+    return {a.sum + b.sum, max(a.pref, a.sum + b.pref),\n+            max(b.suff, b.sum + a.suff),\n+            max({a.best, b.best, a.suff + b.pref})};\n+}\n+struct Tag {\n+    Summary h[3];\n+    int to[3] = {0, 1, 2};\n+    bool active = false;\n+};\n+void append(Tag& a, const Tag& b) {\n+    if (!b.active) return;\n+    if (!a.active) { a = b; return; }\n+    for (int s = 0; s < 3; ++s) {\n+        int mid = a.to[s];\n+        a.h[s] = join(a.h[s], b.h[mid]);\n+        a.to[s] = b.to[mid];\n+    }\n+}\n+struct SegTree {\n+    int size = 1;\n+    vector<Tag> tr;\n+    SegTree(int n) {\n+        while (size < n) size *= 2;\n+        tr.resize(2 * size);\n+    }\n+    void push(int v) {\n+        if (!tr[v].active) return;\n+        append(tr[2*v], tr[v]);\n+        append(tr[2*v+1], tr[v]);\n+        tr[v] = Tag{};\n+    }\n+    void update(int v, int lo, int hi, int l, int r, int type, int x) {\n+        if (l <= lo && hi <= r) {\n+            for (int s = 0; s < 3; ++s) {\n+                int& z = tr[v].to[s];\n+                if (type == 1) z = x + 1;\n+                else if (type == 2) z = 2 - z;\n+                else z = max(z, 1);\n+            }\n+            tr[v].active = true;\n+            return;\n+        }\n+        push(v);\n+        int mid = (lo + hi) / 2;\n+        if (l <= mid) update(2*v, lo, mid, l, r, type, x);\n+        if (r > mid) update(2*v+1, mid+1, hi, l, r, type, x);\n+    }\n+    void record() {\n+        for (int s = 0; s < 3; ++s) {\n+            int x = tr[1].to[s] - 1;\n+            tr[1].h[s] = join(tr[1].h[s], {x, x, x, x});\n+        }\n+        tr[1].active = true;\n+    }\n+    int query(int p, int initial) {\n+        int v = 1, lo = 0, hi = size-1;\n+        while (lo != hi) {\n+            push(v);\n+            int mid = (lo + hi) / 2;\n+            if (p <= mid) { v *= 2; hi = mid; }\n+            else { v = 2*v+1; lo = mid+1; }\n+        }\n+        return tr[v].h[initial+1].best;\n+    }\n+};\n+int main() {\n+    setIO();\n+    int t; cin >> t;\n+    while (t--) {\n+        int n, q; cin >> n >> q;\n+        vector<int> a(n);\n+        for (int& x : a) cin >> x;\n+        SegTree st(n);\n+        st.record();\n+        uint64_t lastans = 0;\n+        auto decode = [&](uint64_t y) { return int((y ^ lastans) % uint64_t(n)); };\n+        while (q--) {\n+            int type; uint64_t u; cin >> type >> u;\n+            if (type == 4) {\n+                int p = decode(u);\n+                int ans = st.query(p, a[p]);\n+                cout << ans << '\\n';\n+                lastans = static_cast<uint64_t>(ans);\n+            } else {\n+                uint64_t v; cin >> v;\n+                int l = decode(u), r = decode(v), x = 0;\n+                if (l > r) swap(l, r);\n+                if (type == 1) cin >> x;\n+                st.update(1, 0, st.size-1, l, r, type, x);\n+            }\n+            st.record();\n+        }\n+    }\n+}\n*** Add File: test.py\n+import random, subprocess\n+random.seed(42)\n+cases, expected = [], []\n+for _ in range(2000):\n+    n, q = random.randint(1, 20), random.randint(1, 100)\n+    a = [random.randint(-1, 1) for _ in range(n)]\n+    hist = [[x] for x in a]\n+    lines = [f'{n} {q}', ' '.join(map(str, a))]\n+    last = 0\n+    for i in range(q):\n+        typ = random.randint(1, 4)\n+        u, v = random.getrandbits(64), random.getrandbits(64)\n+        p = (u ^ last) % n\n+        if typ == 4:\n+            cur = best = -10**9\n+            for z in hist[p]:\n+                cur = max(z, cur + z)\n+                best = max(best, cur)\n+            expected.append(best)\n+            last = best % (1 << 64)\n+            lines.append(f'4 {u}')\n+        else:\n+            l, r = sorted([p, (v ^ last) % n])\n+            x = random.randint(-1, 1)\n+            lines.append(f'{typ} {u} {v}' + (f' {x}' if typ == 1 else ''))\n+            for j in range(l, r+1):\n+                a[j] = x if typ == 1 else -a[j] if typ == 2 else max(a[j], 0)\n+        for j in range(n): hist[j].append(a[j])\n+    cases.append('\\n'.join(lines))\n+data = str(len(cases)) + '\\n' + '\\n'.join(cases) + '\\n'\n+result = subprocess.run(['./main'], input=data, text=True, capture_output=True, check=True)\n+actual = list(map(int, result.stdout.split()))\n+assert actual == expected\n+print(f'Passed {len(cases)} random cases and {len(expected)} queries.')\n*** End Patch"
            },
            "output": "Success. Updated the following files:\nA private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/main.cpp\nA private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/test.py",
            "metadata": {
              "diff": "Index: /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/main.cpp\n===================================================================\n--- /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/main.cpp\n+++ /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/main.cpp\n@@ -0,0 +1,108 @@\n+#include <iostream>\n+#include <vector>\n+#include <algorithm>\n+#include <cstdint>\n+using namespace std;\n+using ll = long long;\n+void setIO() { ios::sync_with_stdio(false); cin.tie(nullptr); }\n+constexpr int NEG = -1000000000;\n+struct Summary {\n+    int sum = 0, pref = NEG, suff = NEG, best = NEG;\n+};\n+Summary join(const Summary& a, const Summary& b) {\n+    if (a.best == NEG) return b;\n+    if (b.best == NEG) return a;\n+    return {a.sum + b.sum, max(a.pref, a.sum + b.pref),\n+            max(b.suff, b.sum + a.suff),\n+            max({a.best, b.best, a.suff + b.pref})};\n+}\n+struct Tag {\n+    Summary h[3];\n+    int to[3] = {0, 1, 2};\n+    bool active = false;\n+};\n+void append(Tag& a, const Tag& b) {\n+    if (!b.active) return;\n+    if (!a.active) { a = b; return; }\n+    for (int s = 0; s < 3; ++s) {\n+        int mid = a.to[s];\n+        a.h[s] = join(a.h[s], b.h[mid]);\n+        a.to[s] = b.to[mid];\n+    }\n+}\n+struct SegTree {\n+    int size = 1;\n+    vector<Tag> tr;\n+    SegTree(int n) {\n+        while (size < n) size *= 2;\n+        tr.resize(2 * size);\n+    }\n+    void push(int v) {\n+        if (!tr[v].active) return;\n+        append(tr[2*v], tr[v]);\n+        append(tr[2*v+1], tr[v]);\n+        tr[v] = Tag{};\n+    }\n+    void update(int v, int lo, int hi, int l, int r, int type, int x) {\n+        if (l <= lo && hi <= r) {\n+            for (int s = 0; s < 3; ++s) {\n+                int& z = tr[v].to[s];\n+                if (type == 1) z = x + 1;\n+                else if (type == 2) z = 2 - z;\n+                else z = max(z, 1);\n+            }\n+            tr[v].active = true;\n+            return;\n+        }\n+        push(v);\n+        int mid = (lo + hi) / 2;\n+        if (l <= mid) update(2*v, lo, mid, l, r, type, x);\n+        if (r > mid) update(2*v+1, mid+1, hi, l, r, type, x);\n+    }\n+    void record() {\n+        for (int s = 0; s < 3; ++s) {\n+            int x = tr[1].to[s] - 1;\n+            tr[1].h[s] = join(tr[1].h[s], {x, x, x, x});\n+        }\n+        tr[1].active = true;\n+    }\n+    int query(int p, int initial) {\n+        int v = 1, lo = 0, hi = size-1;\n+        while (lo != hi) {\n+            push(v);\n+            int mid = (lo + hi) / 2;\n+            if (p <= mid) { v *= 2; hi = mid; }\n+            else { v = 2*v+1; lo = mid+1; }\n+        }\n+        return tr[v].h[initial+1].best;\n+    }\n+};\n+int main() {\n+    setIO();\n+    int t; cin >> t;\n+    while (t--) {\n+        int n, q; cin >> n >> q;\n+        vector<int> a(n);\n+        for (int& x : a) cin >> x;\n+        SegTree st(n);\n+        st.record();\n+        uint64_t lastans = 0;\n+        auto decode = [&](uint64_t y) { return int((y ^ lastans) % uint64_t(n)); };\n+        while (q--) {\n+            int type; uint64_t u; cin >> type >> u;\n+            if (type == 4) {\n+                int p = decode(u);\n+                int ans = st.query(p, a[p]);\n+                cout << ans << '\\n';\n+                lastans = static_cast<uint64_t>(ans);\n+            } else {\n+                uint64_t v; cin >> v;\n+                int l = decode(u), r = decode(v), x = 0;\n+                if (l > r) swap(l, r);\n+                if (type == 1) cin >> x;\n+                st.update(1, 0, st.size-1, l, r, type, x);\n+            }\n+            st.record();\n+        }\n+    }\n+}\n\nIndex: /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/test.py\n===================================================================\n--- /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/test.py\n+++ /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/test.py\n@@ -0,0 +1,34 @@\n+import random, subprocess\n+random.seed(42)\n+cases, expected = [], []\n+for _ in range(2000):\n+    n, q = random.randint(1, 20), random.randint(1, 100)\n+    a = [random.randint(-1, 1) for _ in range(n)]\n+    hist = [[x] for x in a]\n+    lines = [f'{n} {q}', ' '.join(map(str, a))]\n+    last = 0\n+    for i in range(q):\n+        typ = random.randint(1, 4)\n+        u, v = random.getrandbits(64), random.getrandbits(64)\n+        p = (u ^ last) % n\n+        if typ == 4:\n+            cur = best = -10**9\n+            for z in hist[p]:\n+                cur = max(z, cur + z)\n+                best = max(best, cur)\n+            expected.append(best)\n+            last = best % (1 << 64)\n+            lines.append(f'4 {u}')\n+        else:\n+            l, r = sorted([p, (v ^ last) % n])\n+            x = random.randint(-1, 1)\n+            lines.append(f'{typ} {u} {v}' + (f' {x}' if typ == 1 else ''))\n+            for j in range(l, r+1):\n+                a[j] = x if typ == 1 else -a[j] if typ == 2 else max(a[j], 0)\n+        for j in range(n): hist[j].append(a[j])\n+    cases.append('\\n'.join(lines))\n+data = str(len(cases)) + '\\n' + '\\n'.join(cases) + '\\n'\n+result = subprocess.run(['./main'], input=data, text=True, capture_output=True, check=True)\n+actual = list(map(int, result.stdout.split()))\n+assert actual == expected\n+print(f'Passed {len(cases)} random cases and {len(expected)} queries.')\n\n",
              "files": [
                {
                  "filePath": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/main.cpp",
                  "relativePath": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/main.cpp",
                  "type": "add",
                  "patch": "Index: /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/main.cpp\n===================================================================\n--- /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/main.cpp\n+++ /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/main.cpp\n@@ -0,0 +1,108 @@\n+#include <iostream>\n+#include <vector>\n+#include <algorithm>\n+#include <cstdint>\n+using namespace std;\n+using ll = long long;\n+void setIO() { ios::sync_with_stdio(false); cin.tie(nullptr); }\n+constexpr int NEG = -1000000000;\n+struct Summary {\n+    int sum = 0, pref = NEG, suff = NEG, best = NEG;\n+};\n+Summary join(const Summary& a, const Summary& b) {\n+    if (a.best == NEG) return b;\n+    if (b.best == NEG) return a;\n+    return {a.sum + b.sum, max(a.pref, a.sum + b.pref),\n+            max(b.suff, b.sum + a.suff),\n+            max({a.best, b.best, a.suff + b.pref})};\n+}\n+struct Tag {\n+    Summary h[3];\n+    int to[3] = {0, 1, 2};\n+    bool active = false;\n+};\n+void append(Tag& a, const Tag& b) {\n+    if (!b.active) return;\n+    if (!a.active) { a = b; return; }\n+    for (int s = 0; s < 3; ++s) {\n+        int mid = a.to[s];\n+        a.h[s] = join(a.h[s], b.h[mid]);\n+        a.to[s] = b.to[mid];\n+    }\n+}\n+struct SegTree {\n+    int size = 1;\n+    vector<Tag> tr;\n+    SegTree(int n) {\n+        while (size < n) size *= 2;\n+        tr.resize(2 * size);\n+    }\n+    void push(int v) {\n+        if (!tr[v].active) return;\n+        append(tr[2*v], tr[v]);\n+        append(tr[2*v+1], tr[v]);\n+        tr[v] = Tag{};\n+    }\n+    void update(int v, int lo, int hi, int l, int r, int type, int x) {\n+        if (l <= lo && hi <= r) {\n+            for (int s = 0; s < 3; ++s) {\n+                int& z = tr[v].to[s];\n+                if (type == 1) z = x + 1;\n+                else if (type == 2) z = 2 - z;\n+                else z = max(z, 1);\n+            }\n+            tr[v].active = true;\n+            return;\n+        }\n+        push(v);\n+        int mid = (lo + hi) / 2;\n+        if (l <= mid) update(2*v, lo, mid, l, r, type, x);\n+        if (r > mid) update(2*v+1, mid+1, hi, l, r, type, x);\n+    }\n+    void record() {\n+        for (int s = 0; s < 3; ++s) {\n+            int x = tr[1].to[s] - 1;\n+            tr[1].h[s] = join(tr[1].h[s], {x, x, x, x});\n+        }\n+        tr[1].active = true;\n+    }\n+    int query(int p, int initial) {\n+        int v = 1, lo = 0, hi = size-1;\n+        while (lo != hi) {\n+            push(v);\n+            int mid = (lo + hi) / 2;\n+            if (p <= mid) { v *= 2; hi = mid; }\n+            else { v = 2*v+1; lo = mid+1; }\n+        }\n+        return tr[v].h[initial+1].best;\n+    }\n+};\n+int main() {\n+    setIO();\n+    int t; cin >> t;\n+    while (t--) {\n+        int n, q; cin >> n >> q;\n+        vector<int> a(n);\n+        for (int& x : a) cin >> x;\n+        SegTree st(n);\n+        st.record();\n+        uint64_t lastans = 0;\n+        auto decode = [&](uint64_t y) { return int((y ^ lastans) % uint64_t(n)); };\n+        while (q--) {\n+            int type; uint64_t u; cin >> type >> u;\n+            if (type == 4) {\n+                int p = decode(u);\n+                int ans = st.query(p, a[p]);\n+                cout << ans << '\\n';\n+                lastans = static_cast<uint64_t>(ans);\n+            } else {\n+                uint64_t v; cin >> v;\n+                int l = decode(u), r = decode(v), x = 0;\n+                if (l > r) swap(l, r);\n+                if (type == 1) cin >> x;\n+                st.update(1, 0, st.size-1, l, r, type, x);\n+            }\n+            st.record();\n+        }\n+    }\n+}\n",
                  "additions": 108,
                  "deletions": 0
                },
                {
                  "filePath": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/test.py",
                  "relativePath": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/test.py",
                  "type": "add",
                  "patch": "Index: /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/test.py\n===================================================================\n--- /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/test.py\n+++ /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/test.py\n@@ -0,0 +1,34 @@\n+import random, subprocess\n+random.seed(42)\n+cases, expected = [], []\n+for _ in range(2000):\n+    n, q = random.randint(1, 20), random.randint(1, 100)\n+    a = [random.randint(-1, 1) for _ in range(n)]\n+    hist = [[x] for x in a]\n+    lines = [f'{n} {q}', ' '.join(map(str, a))]\n+    last = 0\n+    for i in range(q):\n+        typ = random.randint(1, 4)\n+        u, v = random.getrandbits(64), random.getrandbits(64)\n+        p = (u ^ last) % n\n+        if typ == 4:\n+            cur = best = -10**9\n+            for z in hist[p]:\n+                cur = max(z, cur + z)\n+                best = max(best, cur)\n+            expected.append(best)\n+            last = best % (1 << 64)\n+            lines.append(f'4 {u}')\n+        else:\n+            l, r = sorted([p, (v ^ last) % n])\n+            x = random.randint(-1, 1)\n+            lines.append(f'{typ} {u} {v}' + (f' {x}' if typ == 1 else ''))\n+            for j in range(l, r+1):\n+                a[j] = x if typ == 1 else -a[j] if typ == 2 else max(a[j], 0)\n+        for j in range(n): hist[j].append(a[j])\n+    cases.append('\\n'.join(lines))\n+data = str(len(cases)) + '\\n' + '\\n'.join(cases) + '\\n'\n+result = subprocess.run(['./main'], input=data, text=True, capture_output=True, check=True)\n+actual = list(map(int, result.stdout.split()))\n+assert actual == expected\n+print(f'Passed {len(cases)} random cases and {len(expected)} queries.')\n",
                  "additions": 34,
                  "deletions": 0
                }
              ],
              "diagnostics": {},
              "truncated": false
            },
            "title": "Success. Updated the following files:\nA private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/main.cpp\nA private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-Ypk0Ol/test.py",
            "time": {
              "start": 1789695765328,
              "end": 1789695765336
            }
          },
          "metadata": {
            "openai": {
              "itemId": "fc_05cf4308bbdca0ca016aac96e02a1887d0a96b9786e5d9f3e1"
            }
          },
          "id": "prt_0b22d5bcc001ofyYv4T2pBEoD8",
          "sessionID": "ses_f4dd31ce1ffeZXK29VRHtpaZZx",
          "messageID": "msg_0b22d52ae001Y0iumNZnwWEAwz"
        },
        {
          "reason": "tool-calls",
          "type": "step-finish",
          "tokens": {
            "total": 26472,
            "input": 504,
            "output": 1776,
            "reasoning": 0,
            "cache": {
              "write": 0,
              "read": 24192
            }
          },
          "cost": 0,
          "id": "prt_0b22e2bc9001v2ZzTvm4KNHsFR",
          "sessionID": "ses_f4dd31ce1ffeZXK29VRHtpaZZx",
          "messageID": "msg_0b22d52ae001Y0iumNZnwWEAwz"
        }
      ]
    },
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              "order": 1,
              "content": "Fix one position and forget the spatial array for a moment. Its values over time form an ordinary sequence. If you append a whole chunk of that sequence, which four numbers let you update its maximum subarray sum in constant time?"
            },
            {
              "order": 2,
              "content": "Use the standard summary: total sum, maximum non-empty prefix sum, maximum non-empty suffix sum, and maximum non-empty subarray sum. Now separate an operation into **changing the current value** and **recording the current value**. Recording happens at every position after every operation—even after a query."
            },
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              "order": 3,
              "content": "A position's current value has only three possibilities. For a batch of deferred events, store its final value and its generated-history summary for each possible starting value $s\\in\\{-1,0,1\\}$. That is only three cases, regardless of how long the batch is."
            },
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              "content": "Suppose batch $A$ happens before batch $B$. Starting from $s$, batch $B$ starts from $f_A(s)$, so the combined final value is $f_B(f_A(s))$, and its history is $H_A(s)$ followed by $H_B(f_A(s))$. This gives a constant-size, composable lazy tag. The order matters!"
            },
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              "content": "Build a segment tree over positions, storing only these lazy tags. Range modifications change the tag's three final states; recording appends one value to each of the root tag's three histories. Push older child tags followed by newer parent tags when descending. At a queried leaf, select the history corresponding to its original array value. Record version $0$ initially, and answer type $4$ **before** recording the version it creates."
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          "editorial": "The useful restriction is not that the numbers are small. It is that **there are only three possible current values**. We can describe an arbitrarily long batch of operations by checking what it does to each of those three values.\n\nThat description becomes a lazy segment-tree tag—including the history, not just the final value.\n\n## 1. Compress a chunk of history\n\nFor one position, its values across versions form a sequence. Summarize a non-empty sequence by four integers:\n\n- $S$: total sum;\n- $P$: maximum non-empty prefix sum;\n- $Q$: maximum non-empty suffix sum;\n- $M$: maximum non-empty subarray sum.\n\nIf sequence $A$ is followed by sequence $B$, their concatenation has\n\n$$\n\\begin{aligned}\nS &= S_A+S_B,\\\\\nP &= \\max(P_A,S_A+P_B),\\\\\nQ &= \\max(Q_B,S_B+Q_A),\\\\\nM &= \\max(M_A,M_B,Q_A+P_B).\n\\end{aligned}\n$$\n\nThe last formula covers subarrays entirely in either chunk and subarrays crossing the boundary.\n\nA single value $x$ has summary $(x,x,x,x)$.\n\nWe also need an empty history: modifications can change a value without recording anything yet. Use an explicit empty-summary marker, and let concatenating an empty summary return the other summary. **Do not initialize all maxima to zero.** A history consisting only of $-1$ must answer $-1$, not $0$.\n\n## 2. Separate changes from recording\n\nThink of two kinds of events:\n\n1. **Modify:** replace the current value using assignment, negation, or clamping to zero.\n2. **Record:** append the current value to the position's history.\n\nA modification generates no history by itself. After each input operation, one record event happens at **every** position.\n\nThis distinction handles untouched positions and queries correctly: their values do not change, but time still passes. A query is not a pause button.\n\n## 3. Describe an event batch for all three starting states\n\nFor a batch $T$, store the following for every $s\\in\\{-1,0,1\\}$:\n\n- $f_T(s)$: the final current value after the batch, starting from $s$;\n- $H_T(s)$: the summary of all values recorded during the batch, starting from $s$.\n\nOnly three final states and three four-integer summaries are needed.\n\nThe identity batch has $f_T(s)=s$ and empty histories.\n\n### Concatenating batches\n\nSuppose batch $A$ happens first, then batch $B$.\n\nStarting from $s$, the first batch ends at $f_A(s)$. Therefore the second batch must use its entry for that state:\n\n$$\nf_{AB}(s)=f_B(f_A(s)),\n$$\n\n$$\nH_{AB}(s)=\\operatorname{join}\\bigl(H_A(s),H_B(f_A(s))\\bigr).\n$$\n\nWe evaluate this for three starting states, so composition takes $O(1)$ time.\n\nThis is the key insight: **we can concatenate arbitrarily long operation histories without knowing a position's actual starting value yet.**\n\n### Appending a modification\n\nIf the modification is the function $g$, simply replace\n\n$$f_T(s)\\gets g(f_T(s))$$\n\nfor all three states. Existing histories stay unchanged: an update changes the present, not the past.\n\nThe three functions are\n\n$$g(v)=x,\\qquad g(v)=-v,\\qquad g(v)=\\max(v,0).$$\n\n### Appending a record event\n\nFor each starting state, append the single value $f_T(s)$ to its history:\n\n$$H_T(s)\\gets\\operatorname{join}\\bigl(H_T(s),[f_T(s)]\\bigr).$$\n\nThe final-state mapping does not change.\n\n## 4. A segment tree containing only lazy tags\n\nBuild a segment tree over the array positions. Each node stores a pending batch common to its interval.\n\nThere is no sum or maximum to merge between spatial children. Our maximum-subarray calculation runs along **time at one position**, not along neighboring positions. Consequently, there is no pull operation.\n\n### Range modification\n\nUse the usual segment-tree range traversal:\n\n- If the node is fully covered, append the modification to its tag.\n- Otherwise, push its tag to both children, clear it, and recurse into the intersecting children.\n\nWhen pushing parent batch $B$ into a child whose pending batch is $A$, replace the child tag with $AB$: **child first, parent second**.\n\nWhy that order? We always push a node before descending beneath it. Thus anything already waiting in a child is older than the batch still waiting at its parent.\n\n### Record a version\n\nAppend a record event to the root tag. This represents recording the current value at every position, in $O(1)$ time.\n\nThe positions need not currently have equal values. The tag describes what happens for all three possible starting values, and composition selects the correct case later.\n\n### Query one position\n\nPush every tag on the root-to-leaf path.\n\nIn this implementation, a leaf's tag is never cleared. It accumulates the entire event history for that position, represented for each hypothetical original value. If the actual original value is $a_p$, the answer is\n\n$$H_{\\text{leaf}}(a_p).M.$$\n\nKeeping all three cases even at leaves makes the implementation uniform; no separate per-position history accumulator is necessary.\n\n## 5. Version order and decoding\n\nThe processing order is:\n\n1. Start with identity tags and record version $0$ at the root.\n2. For each operation:\n   - Decode its coordinates using the current `lastans`.\n   - For types $1$–$3$, apply the range modification.\n   - For type $4$, query and output the answer, then update `lastans`.\n   - Record the resulting version at the root, regardless of operation type.\n\nThus a query in operation $i$ sees exactly versions $0,\\ldots,i-1$. Its duplicate version $i$ is recorded afterward.\n\nUse `uint64_t` for encoded coordinates and `lastans`. Converting the signed answer to `uint64_t` gives precisely its least non-negative residue modulo $2^{64}$; in particular, $-1$ becomes $2^{64}-1$.\n\nThe code uses zero-based positions, so decoding is simply\n\n$$p=(u\\mathbin{\\oplus}\\mathrm{lastans})\\bmod n.$$\n\nStates $-1,0,1$ are stored as indices $0,1,2$. In these indices, negation is $z\\gets2-z$, and clamping is $z\\gets\\max(z,1)$.\n\n## Correctness proof\n\n**Lemma 1: History summaries merge correctly.**\n\nA prefix of $AB$ either stays inside $A$ or includes all of $A$ and a prefix of $B$. The suffix argument is symmetric. A non-empty subarray lies inside $A$, lies inside $B$, or consists of a suffix of $A$ followed by a prefix of $B$. These are exactly the cases in the merge formulas. Empty chunks are handled as identities.\n\n**Lemma 2: Tag composition correctly represents consecutive event batches.**\n\nFor starting state $s$, batch $A$ generates $H_A(s)$ and ends at $f_A(s)$. Batch $B$ therefore generates $H_B(f_A(s))$ and ends at $f_B(f_A(s))$. Concatenating those histories using Lemma 1 gives exactly the stated composition. Appending a modification or a record event also follows its definition directly.\n\n**Lemma 3: The segment tree preserves each position's chronological event sequence.**\n\nA fully covered range modification is appended to a tag shared by precisely that node's positions. A root record event applies to every position. Before descending, pushing concatenates the older child batch with the newer parent batch, preserving order by Lemma 2. Clearing the parent removes only a batch already transferred to both children. Therefore the deferred representation remains equivalent to executing every event directly at every position.\n\n**Theorem: Every query outputs the required answer.**\n\nVersion $0$ is recorded initially, and version $i$ is recorded only after processing operation $i$. Hence a query during operation $i$ sees exactly versions $0$ through $i-1$. Pushing its path collects all those events at its leaf by Lemma 3. Selecting the leaf history for its actual initial value gives the true value sequence, and its $M$ field is the maximum non-empty subarray sum by Lemma 1.\n\n## Complexity and implementation bounds\n\nEach tag has constant size. A range modification or point query takes $O(\\log(n+1))$ time, and recording a version takes $O(1)$.\n\nThe total complexity is\n\n$$O\\bigl(n+q\\log(n+1)\\bigr)\\text{ time},\\qquad O(n)\\text{ space}.$$\n\nThe implementation pads the tree to a power of two. Extra leaves are never queried and do not affect real positions.\n\nAt most $q+1$ values are recorded, each in $\\{-1,0,1\\}$, so all finite summary values fit comfortably in a signed 32-bit integer. The empty marker is checked before arithmetic, keeping non-empty semantics explicit.",
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                  "content": "Fix one position and forget the spatial array for a moment. Its values over time form an ordinary sequence. If you append a whole chunk of that sequence, which four numbers let you update its maximum subarray sum in constant time?"
                },
                {
                  "order": 2,
                  "content": "Use the standard summary: total sum, maximum non-empty prefix sum, maximum non-empty suffix sum, and maximum non-empty subarray sum. Now separate an operation into **changing the current value** and **recording the current value**. Recording happens at every position after every operation—even after a query."
                },
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                  "content": "A position's current value has only three possibilities. For a batch of deferred events, store its final value and its generated-history summary for each possible starting value $s\\in\\{-1,0,1\\}$. That is only three cases, regardless of how long the batch is."
                },
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                  "content": "Suppose batch $A$ happens before batch $B$. Starting from $s$, batch $B$ starts from $f_A(s)$, so the combined final value is $f_B(f_A(s))$, and its history is $H_A(s)$ followed by $H_B(f_A(s))$. This gives a constant-size, composable lazy tag. The order matters!"
                },
                {
                  "order": 5,
                  "content": "Build a segment tree over positions, storing only these lazy tags. Range modifications change the tag's three final states; recording appends one value to each of the root tag's three histories. Push older child tags followed by newer parent tags when descending. At a queried leaf, select the history corresponding to its original array value. Record version $0$ initially, and answer type $4$ **before** recording the version it creates."
                }
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              "editorial": "The useful restriction is not that the numbers are small. It is that **there are only three possible current values**. We can describe an arbitrarily long batch of operations by checking what it does to each of those three values.\n\nThat description becomes a lazy segment-tree tag—including the history, not just the final value.\n\n## 1. Compress a chunk of history\n\nFor one position, its values across versions form a sequence. Summarize a non-empty sequence by four integers:\n\n- $S$: total sum;\n- $P$: maximum non-empty prefix sum;\n- $Q$: maximum non-empty suffix sum;\n- $M$: maximum non-empty subarray sum.\n\nIf sequence $A$ is followed by sequence $B$, their concatenation has\n\n$$\n\\begin{aligned}\nS &= S_A+S_B,\\\\\nP &= \\max(P_A,S_A+P_B),\\\\\nQ &= \\max(Q_B,S_B+Q_A),\\\\\nM &= \\max(M_A,M_B,Q_A+P_B).\n\\end{aligned}\n$$\n\nThe last formula covers subarrays entirely in either chunk and subarrays crossing the boundary.\n\nA single value $x$ has summary $(x,x,x,x)$.\n\nWe also need an empty history: modifications can change a value without recording anything yet. Use an explicit empty-summary marker, and let concatenating an empty summary return the other summary. **Do not initialize all maxima to zero.** A history consisting only of $-1$ must answer $-1$, not $0$.\n\n## 2. Separate changes from recording\n\nThink of two kinds of events:\n\n1. **Modify:** replace the current value using assignment, negation, or clamping to zero.\n2. **Record:** append the current value to the position's history.\n\nA modification generates no history by itself. After each input operation, one record event happens at **every** position.\n\nThis distinction handles untouched positions and queries correctly: their values do not change, but time still passes. A query is not a pause button.\n\n## 3. Describe an event batch for all three starting states\n\nFor a batch $T$, store the following for every $s\\in\\{-1,0,1\\}$:\n\n- $f_T(s)$: the final current value after the batch, starting from $s$;\n- $H_T(s)$: the summary of all values recorded during the batch, starting from $s$.\n\nOnly three final states and three four-integer summaries are needed.\n\nThe identity batch has $f_T(s)=s$ and empty histories.\n\n### Concatenating batches\n\nSuppose batch $A$ happens first, then batch $B$.\n\nStarting from $s$, the first batch ends at $f_A(s)$. Therefore the second batch must use its entry for that state:\n\n$$\nf_{AB}(s)=f_B(f_A(s)),\n$$\n\n$$\nH_{AB}(s)=\\operatorname{join}\\bigl(H_A(s),H_B(f_A(s))\\bigr).\n$$\n\nWe evaluate this for three starting states, so composition takes $O(1)$ time.\n\nThis is the key insight: **we can concatenate arbitrarily long operation histories without knowing a position's actual starting value yet.**\n\n### Appending a modification\n\nIf the modification is the function $g$, simply replace\n\n$$f_T(s)\\gets g(f_T(s))$$\n\nfor all three states. Existing histories stay unchanged: an update changes the present, not the past.\n\nThe three functions are\n\n$$g(v)=x,\\qquad g(v)=-v,\\qquad g(v)=\\max(v,0).$$\n\n### Appending a record event\n\nFor each starting state, append the single value $f_T(s)$ to its history:\n\n$$H_T(s)\\gets\\operatorname{join}\\bigl(H_T(s),[f_T(s)]\\bigr).$$\n\nThe final-state mapping does not change.\n\n## 4. A segment tree containing only lazy tags\n\nBuild a segment tree over the array positions. Each node stores a pending batch common to its interval.\n\nThere is no sum or maximum to merge between spatial children. Our maximum-subarray calculation runs along **time at one position**, not along neighboring positions. Consequently, there is no pull operation.\n\n### Range modification\n\nUse the usual segment-tree range traversal:\n\n- If the node is fully covered, append the modification to its tag.\n- Otherwise, push its tag to both children, clear it, and recurse into the intersecting children.\n\nWhen pushing parent batch $B$ into a child whose pending batch is $A$, replace the child tag with $AB$: **child first, parent second**.\n\nWhy that order? We always push a node before descending beneath it. Thus anything already waiting in a child is older than the batch still waiting at its parent.\n\n### Record a version\n\nAppend a record event to the root tag. This represents recording the current value at every position, in $O(1)$ time.\n\nThe positions need not currently have equal values. The tag describes what happens for all three possible starting values, and composition selects the correct case later.\n\n### Query one position\n\nPush every tag on the root-to-leaf path.\n\nIn this implementation, a leaf's tag is never cleared. It accumulates the entire event history for that position, represented for each hypothetical original value. If the actual original value is $a_p$, the answer is\n\n$$H_{\\text{leaf}}(a_p).M.$$\n\nKeeping all three cases even at leaves makes the implementation uniform; no separate per-position history accumulator is necessary.\n\n## 5. Version order and decoding\n\nThe processing order is:\n\n1. Start with identity tags and record version $0$ at the root.\n2. For each operation:\n   - Decode its coordinates using the current `lastans`.\n   - For types $1$–$3$, apply the range modification.\n   - For type $4$, query and output the answer, then update `lastans`.\n   - Record the resulting version at the root, regardless of operation type.\n\nThus a query in operation $i$ sees exactly versions $0,\\ldots,i-1$. Its duplicate version $i$ is recorded afterward.\n\nUse `uint64_t` for encoded coordinates and `lastans`. Converting the signed answer to `uint64_t` gives precisely its least non-negative residue modulo $2^{64}$; in particular, $-1$ becomes $2^{64}-1$.\n\nThe code uses zero-based positions, so decoding is simply\n\n$$p=(u\\mathbin{\\oplus}\\mathrm{lastans})\\bmod n.$$\n\nStates $-1,0,1$ are stored as indices $0,1,2$. In these indices, negation is $z\\gets2-z$, and clamping is $z\\gets\\max(z,1)$.\n\n## Correctness proof\n\n**Lemma 1: History summaries merge correctly.**\n\nA prefix of $AB$ either stays inside $A$ or includes all of $A$ and a prefix of $B$. The suffix argument is symmetric. A non-empty subarray lies inside $A$, lies inside $B$, or consists of a suffix of $A$ followed by a prefix of $B$. These are exactly the cases in the merge formulas. Empty chunks are handled as identities.\n\n**Lemma 2: Tag composition correctly represents consecutive event batches.**\n\nFor starting state $s$, batch $A$ generates $H_A(s)$ and ends at $f_A(s)$. Batch $B$ therefore generates $H_B(f_A(s))$ and ends at $f_B(f_A(s))$. Concatenating those histories using Lemma 1 gives exactly the stated composition. Appending a modification or a record event also follows its definition directly.\n\n**Lemma 3: The segment tree preserves each position's chronological event sequence.**\n\nA fully covered range modification is appended to a tag shared by precisely that node's positions. A root record event applies to every position. Before descending, pushing concatenates the older child batch with the newer parent batch, preserving order by Lemma 2. Clearing the parent removes only a batch already transferred to both children. Therefore the deferred representation remains equivalent to executing every event directly at every position.\n\n**Theorem: Every query outputs the required answer.**\n\nVersion $0$ is recorded initially, and version $i$ is recorded only after processing operation $i$. Hence a query during operation $i$ sees exactly versions $0$ through $i-1$. Pushing its path collects all those events at its leaf by Lemma 3. Selecting the leaf history for its actual initial value gives the true value sequence, and its $M$ field is the maximum non-empty subarray sum by Lemma 1.\n\n## Complexity and implementation bounds\n\nEach tag has constant size. A range modification or point query takes $O(\\log(n+1))$ time, and recording a version takes $O(1)$.\n\nThe total complexity is\n\n$$O\\bigl(n+q\\log(n+1)\\bigr)\\text{ time},\\qquad O(n)\\text{ space}.$$\n\nThe implementation pads the tree to a power of two. Extra leaves are never queried and do not affect real positions.\n\nAt most $q+1$ values are recorded, each in $\\{-1,0,1\\}$, so all finite summary values fit comfortably in a signed 32-bit integer. The empty marker is checked before arithmetic, keeping non-empty semantics explicit.",
              "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\nconstexpr int NEG = -1000000000;\n\nstruct Summary {\n    int sum = 0, pref = NEG, suff = NEG, best = NEG;\n};\n\nSummary join(const Summary& a, const Summary& b) {\n    if (a.best == NEG) return b;\n    if (b.best == NEG) return a;\n    return {\n        a.sum + b.sum,\n        max(a.pref, a.sum + b.pref),\n        max(b.suff, b.sum + a.suff),\n        max({a.best, b.best, a.suff + b.pref})\n    };\n}\n\nstruct Tag {\n    Summary h[3];\n    int to[3] = {0, 1, 2};\n    bool active = false;\n};\n\n// Replace a with the batch \"a, then b\".\nvoid append(Tag& a, const Tag& b) {\n    if (!b.active) return;\n    if (!a.active) {\n        a = b;\n        return;\n    }\n    for (int s = 0; s < 3; ++s) {\n        int mid = a.to[s];\n        a.h[s] = join(a.h[s], b.h[mid]);\n        a.to[s] = b.to[mid];\n    }\n}\n\nstruct SegTree {\n    int size = 1;\n    vector<Tag> tr;\n\n    SegTree(int n) {\n        while (size < n) size *= 2;\n        tr.resize(2 * size);\n    }\n\n    void push(int v) {\n        if (!tr[v].active) return;\n        append(tr[2 * v], tr[v]);\n        append(tr[2 * v + 1], tr[v]);\n        tr[v] = Tag{};\n    }\n\n    void update(int v, int lo, int hi, int l, int r, int type, int x) {\n        if (l <= lo && hi <= r) {\n            for (int s = 0; s < 3; ++s) {\n                int& z = tr[v].to[s];\n                if (type == 1) z = x + 1;\n                else if (type == 2) z = 2 - z;\n                else z = max(z, 1);\n            }\n            tr[v].active = true;\n            return;\n        }\n        push(v);\n        int mid = (lo + hi) / 2;\n        if (l <= mid) update(2 * v, lo, mid, l, r, type, x);\n        if (r > mid) update(2 * v + 1, mid + 1, hi, l, r, type, x);\n    }\n\n    void record() {\n        for (int s = 0; s < 3; ++s) {\n            int x = tr[1].to[s] - 1;\n            tr[1].h[s] = join(tr[1].h[s], {x, x, x, x});\n        }\n        tr[1].active = true;\n    }\n\n    int query(int p, int initial) {\n        int v = 1, lo = 0, hi = size - 1;\n        while (lo != hi) {\n            push(v);\n            int mid = (lo + hi) / 2;\n            if (p <= mid) {\n                v *= 2;\n                hi = mid;\n            } else {\n                v = 2 * v + 1;\n                lo = mid + 1;\n            }\n        }\n        return tr[v].h[initial + 1].best;\n    }\n};\n\nint main() {\n    setIO();\n    int t;\n    cin >> t;\n    while (t--) {\n        int n, q;\n        cin >> n >> q;\n        vector<int> a(n);\n        for (int& x : a) cin >> x;\n\n        SegTree st(n);\n        st.record(); // Version 0.\n        uint64_t lastans = 0;\n        auto decode = [&](uint64_t y) {\n            return int((y ^ lastans) % uint64_t(n));\n        };\n\n        while (q--) {\n            int type;\n            uint64_t u;\n            cin >> type >> u;\n            if (type == 4) {\n                int p = decode(u);\n                int ans = st.query(p, a[p]);\n                cout << ans << '\\n';\n                lastans = static_cast<uint64_t>(ans);\n            } else {\n                uint64_t v;\n                cin >> v;\n                int l = decode(u), r = decode(v), x = 0;\n                if (l > r) swap(l, r);\n                if (type == 1) cin >> x;\n                st.update(1, 0, st.size - 1, l, r, type, x);\n            }\n            st.record(); // Every operation creates a version.\n        }\n    }\n}\n"
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