{
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    "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-cazIjb",
    "title": "Nudge content generation",
    "agent": "nudge-generation",
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  "messages": [
    {
      "info": {
        "role": "user",
        "time": {
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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": {
          "type": "json_schema",
          "schema": {
            "type": "object",
            "properties": {
              "status": {
                "type": "string",
                "enum": [
                  "success",
                  "unsolvable"
                ],
                "description": "Set to 'success' when the supplied statement defines a solvable programming problem. Set to 'unsolvable' only if the statement is fundamentally incomplete, contradictory, or depends on an inaccessible required resource."
              },
              "reason": {
                "anyOf": [
                  {
                    "type": "string",
                    "pattern": "^[^\\u0000]*$",
                    "minLength": 1,
                    "maxLength": 2000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'unsolvable', provide a detailed failure message to show users. Name the concrete statement/resource blocker and any relevant source lookup/access status, such as missing official tutorial, 403/Cloudflare challenge, 404, or unavailable accepted submissions. Otherwise null."
              },
              "hints": {
                "anyOf": [
                  {
                    "type": "array",
                    "minItems": 5,
                    "maxItems": 5,
                    "description": "Exactly 5 progressive hints, each building on the last.",
                    "items": {
                      "type": "object",
                      "properties": {
                        "order": {
                          "type": "integer",
                          "minimum": 1,
                          "maximum": 5,
                          "description": "Hint number, 1 through 5."
                        },
                        "content": {
                          "type": "string",
                          "pattern": "^[^\\u0000]*$",
                          "minLength": 1,
                          "maxLength": 10000,
                          "description": "Markdown hint text."
                        }
                      },
                      "required": [
                        "order",
                        "content"
                      ],
                      "additionalProperties": false
                    }
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', provide exactly 5 progressive hints. Otherwise null."
              },
              "editorial": {
                "anyOf": [
                  {
                    "type": "string",
                    "pattern": "^[^\\u0000]*$",
                    "minLength": 1,
                    "maxLength": 100000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', a prose editorial explaining the solution in Nudge's own voice, with no links, citations, or source/research notes. Otherwise null."
              },
              "solution": {
                "anyOf": [
                  {
                    "type": "string",
                    "pattern": "^[^\\u0000]*$",
                    "minLength": 1,
                    "maxLength": 200000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', a complete C++ solution. Otherwise null."
              }
            },
            "required": [
              "status",
              "reason",
              "hints",
              "editorial",
              "solution"
            ],
            "additionalProperties": false,
            "title": "problem_response",
            "description": "Submit the generated content for the problem, or report if it is unsolvable."
          },
          "retryCount": 2
        },
        "id": "msg_0adfa865a001zE7E3pd1yTOGuo",
        "sessionID": "ses_f520579b7ffevMTLeJPBDbEQ1b"
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          "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 2258E: \"DivMEX\" (rated 2400).\n\nKnown problem metadata:\n- Contest ID: 2258\n- Index: E\n- Name: DivMEX\n- Rating: 2400 (Codeforces difficulty rating)\n- Tags: data structures, math, number theory\n- Problem URL: https://codeforces.com/contest/2258/problem/E\n\nUse the rating and tags as weak signals only. The supplied statement is the source of truth.\n\nProblem Statement:\n<problem-statement>\n<div class=\"header\"><div class=\"title\">E. DivMEX</div><div class=\"time-limit\"><div class=\"property-title\">time limit per test</div>3 seconds</div><div class=\"memory-limit\"><div class=\"property-title\">memory limit per test</div>256 megabytes</div><div class=\"input-file input-standard\"><div class=\"property-title\">input</div>standard input</div><div class=\"output-file output-standard\"><div class=\"property-title\">output</div>standard output</div></div><div><p>You are given an array $$$a_1, a_2, \\ldots, a_n$$$ of positive integers.</p><p>Let $$$f(l,r)$$$ for $$$1 \\leq l \\leq r \\leq n$$$ be the smallest positive integer that doesn't divide $$$\\operatorname{lcm}(a_l, a_{l+1}, \\ldots, a_r)$$$. Here, $$$\\operatorname{lcm}$$$ of an array of integers denotes the <a href=\"https://en.wikipedia.org/wiki/Least_common_multiple\">least common multiple (LCM)</a> of these integers.</p><p>Determine every positive integer $$$x$$$ such that there exists a pair of integers $$$(l,r)$$$ such that $$$1 \\leq l \\leq r \\leq n$$$ and $$$f(l,r) = x$$$.</p></div><div class=\"input-specification\"><div class=\"section-title\">Input</div><p>Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 10^4$$$). The description of the test cases follows.</p><p>The first line of each test case contains $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$), denoting the size of the array. </p><p>The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$).</p><p>It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \\cdot 10^5$$$.</p></div><div class=\"output-specification\"><div class=\"section-title\">Output</div><p>For each test case, output two lines.</p><p>The first line should contain a single integer $$$k$$$ — the number of suitable integers $$$x$$$.</p><p>The second line should contain $$$k$$$ integers $$$x_1  \\lt  x_2  \\lt  \\ldots  \\lt  x_k$$$, the suitable integers in increasing order.</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\">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-odd test-example-line-1\">1</div><div class=\"test-example-line test-example-line-even test-example-line-2\">3</div><div class=\"test-example-line test-example-line-even test-example-line-2\">2 3 1</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">5</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">4 3 1 3 4</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\">2 </div><div class=\"test-example-line test-example-line-even test-example-line-2\">3</div><div class=\"test-example-line test-example-line-even test-example-line-2\">2 3 4 </div><div class=\"test-example-line test-example-line-odd test-example-line-3\">3</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">2 3 5 </div></pre></div></div></div><div class=\"note\"><div class=\"section-title\">Note</div><p>In the first test case, the only $$$(l,r)$$$ pair is $$$(1,1)$$$, and $$$f(1,1) = 2$$$.</p><p>In the second test case, $$$(l,r)=(1,1)$$$ yields the set of divisors $$$\\{ 1,2 \\}$$$, $$$(l,r)=(1,2)$$$ yields the set of divisors $$$\\{ 1, 2, 3, 6\\}$$$, and $$$(l,r)=(2,2)$$$ yields the set of divisors $$$\\{ 1, 3 \\}$$$. This means $$$f(1,1)=3$$$, $$$f(1,2)=4$$$, and $$$f(2,2)=2$$$. Since no other $$$(l,r)$$$ pair produces a different $$$f$$$ value, the answer is $$$[2,3,4]$$$.</p><p>For the last test case, $$$f(2,2)=2$$$, $$$f(1,1)=3$$$, and $$$f(1,5)=5$$$. It can be seen that no other values can be obtained.</p></div>\n</problem-statement>\n\n\nSource lookup status:\n<source-lookup-status>\n- Tutorial (en): https://codeforces.com/blog/entry/156316 loaded (200 OK; title: \"Codeforces Round 1118 (Div. 2) Editorial - Codeforces\"; mentions 2258E)\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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        "variant": "low",
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          "type": "step-start",
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            "output": "---\nimage: https://codeforces.com/codeforces.org/s/85295/images/codeforces-telegram-square-1024x1024.png\ntitle: Codeforces Round 1118 (Div. 2) Editorial - Codeforces\ndescription: Codeforces. Programming competitions and contests, programming community\n---\n\n<!DOCTYPE HTML PUBLIC \"-//W3C//DTD HTML 4.01//EN\"> \n\n  \n[![Codeforces](//codeforces.com/codeforces.org/s/85295/images/codeforces-logo-with-telegram-555x130.png \"Codeforces\")](https://codeforces.com/) \n\n[![In English](//codeforces.com/codeforces.org/s/85295/images/flags/24/gb.png \"In English\")](https://codeforces.com/blog/entry/156316?locale=en) [![По-русски](//codeforces.com/codeforces.org/s/85295/images/flags/24/ru.png \"По-русски\")](https://codeforces.com/blog/entry/156316?locale=ru) \n\n[Enter](https://codeforces.com/enter?back=%2Fblog%2Fentry%2F156316)| [Register](https://codeforces.com/register) \n\n  \n* [Home](https://codeforces.com/)\n* [Top](https://codeforces.com/top)\n* [Catalog](https://codeforces.com/catalog)\n* [Contests](https://codeforces.com/contests)\n* [Gym](https://codeforces.com/gyms)\n* [Problemset](https://codeforces.com/problemset)\n* [Groups](https://codeforces.com/groups)\n* 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Errichto\") | 139      |\n| 6  | [adamant](https://codeforces.com/profile/adamant \"Master adamant\")                       | 136      |\n| 7  | [AmShZ](https://codeforces.com/profile/AmShZ \"Grandmaster AmShZ\")                        | 134      |\n| 8  | [maroonrk](https://codeforces.com/profile/maroonrk \"Legendary Grandmaster maroonrk\")     | 133      |\n| 9  | [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\") | 132      |\n| 10 | [qwexd](https://codeforces.com/profile/qwexd \"Master qwexd\")                             | 129      |\n\n|  | [View all →](https://codeforces.com/top-contributed) |\n|  | ---------------------------------------------------- |\n\n→ Find user \n\nHandle: \n\n→ Recent actions \n\n* [ko\\_osaga](https://codeforces.com/profile/ko%5Fosaga \"Grandmaster ko_osaga\") → [Response to \"Is IOI Becoming Too Heuristic?\"](https://codeforces.com/blog/entry/156329) ![New 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[Submissions](https://codeforces.com/submissions/ItsNotMeItsYou)\n* [Groups](https://codeforces.com/groups/with/ItsNotMeItsYou)\n* [Contests](https://codeforces.com/contests/with/ItsNotMeItsYou)\n* [Problemsetting](https://codeforces.com/contests/writer/ItsNotMeItsYou)\n\n### [ItsNotMeItsYou's blog](https://codeforces.com/blog/ItsNotMeItsYou)\n\n[ Codeforces Round 1118 (Div. 2) Editorial ](https://codeforces.com/blog/entry/156316) \n\nBy [ItsNotMeItsYou](https://codeforces.com/profile/ItsNotMeItsYou \"Master ItsNotMeItsYou\"), 3 weeks ago, ![In English](//codeforces.com/codeforces.org/s/85295/images/flags/24/gb.png \"In English\") \n\nSorry for underestimating the difficulties of the problems, especially B2 and D. We tried to serve as many cool problems as we could. And apparently, this led to some difficult ones.\n\nThank you for participating in the round. We all hope you enjoyed all of our problems. \n\n[2258A - Odd Eraser](https://codeforces.com/contest/2258/problem/A \"Codeforces Round 1118 (Div. 2)\") Idea: [ItsNotMeItsYou](https://codeforces.com/profile/ItsNotMeItsYou \"Master ItsNotMeItsYou\"), [mychecksdead](https://codeforces.com/profile/mychecksdead \"Master mychecksdead\")\n\n**Hint1**\n\nThink of the case $$$n=3$$$. Can you remove the first or the last number in any way?\n\n**Solution**\n\nWe cannot obtain a number larger than $$$\\\\gcd(a\\_1,a\\_n)$$$ because the first and last numbers in the array can't be removed. We can remove all other elements by choosing $$$\\[1, i, n\\]$$$, so the answer is $$$\\\\gcd(a\\_1, a\\_n)$$$.\n\n**Code**\n\n[388911115](https://codeforces.com/contest/2258/submission/388911115 \"Submission 388911115 by ItsNotMeItsYou\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") Good [ ](https://codeforces.com/data/like?action=like \"I like this\") Ok [ ](https://codeforces.com/data/like?action=like \"I like this\") Bad [ ](https://codeforces.com/data/like?action=like \"I like this\") Terrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2258B1 - Carrot Chopdown (Easy Version)](https://codeforces.com/contest/2258/problem/B1 \"Codeforces Round 1118 (Div. 2)\") Idea: [mychecksdead](https://codeforces.com/profile/mychecksdead \"Master mychecksdead\")\n\n**Hint1**\n\nWhat is the answer for a fixed $$$x$$$?\n\n**Solution**\n\nIterating over the values $$$x=1...m$$$, we can find the answer for each possible $$$x$$$.\n\nFor any $$$x$$$, the answer is the number of carrots of length greater than or equal to $$$x$$$, plus the number of carrots of length $$$2x$$$.\n\n**Code**\n\n[388911137](https://codeforces.com/contest/2258/submission/388911137 \"Submission 388911137 by ItsNotMeItsYou\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") Good [ ](https://codeforces.com/data/like?action=like \"I like this\") Ok [ ](https://codeforces.com/data/like?action=like \"I like this\") Bad [ ](https://codeforces.com/data/like?action=like \"I like this\") Terrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2258B2 - Carrot Chopdown (Hard Version)](https://codeforces.com/contest/2258/problem/B2 \"Codeforces Round 1118 (Div. 2)\") Idea: [mychecksdead](https://codeforces.com/profile/mychecksdead \"Master mychecksdead\") (This problem was originally designed with only $$$k=2$$$; thanks to [ItsNotMeItsYou](https://codeforces.com/profile/ItsNotMeItsYou \"Master ItsNotMeItsYou\") for expanding on the idea.)\n\n**Hint1**\n\nConsider solving the problem for a fixed $$$x$$$.\n\n**Hint2**\n\nConsider solving the problem for $$$k=2$$$.\n\n**Hint3**\n\nWhat happens if $$$k$$$ is bigger than $$$18$$$? \n\n**Hint4**\n\nWhat are the optimal $$$x$$$ values for cutting?\n\n**Solution**\n\nFor target value $$$x$$$, it is optimal to use cuts in the following order: $$$x \\\\cdot 2^{k-1}, x \\\\cdot 2^{k-2}, \\\\ldots, x \\\\cdot 1$$$.\n\nFor each $$$i$$$, $$$a\\_i=2^k \\\\cdot x$$$ adds $$$2^k$$$ to the answer. Otherwise, the contribution is $$$\\\\min(\\\\frac{a\\_i}{x}, 2^k-1)$$$. \n\nThe optimal $$$x$$$ is less than or equal to $$$\\\\frac{m}{2^k}$$$ (if less than $$$1$$$, it equals $$$1$$$, and the answer is basically the sum of all carrot lengths). Moreover, no $$$i$$$ can contribute more than $$$2^k$$$ to any $$$x$$$. Using prefix sum, determine the number of values contributing $$$y$$$ to the answer for each possible $$$x$$$. This solution runs in $$$\\\\mathcal{O}(n + m \\\\log m)$$$.\n\n**Code**\n\n[388911156](https://codeforces.com/contest/2258/submission/388911156 \"Submission 388911156 by ItsNotMeItsYou\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") Good [ ](https://codeforces.com/data/like?action=like \"I like this\") Ok [ ](https://codeforces.com/data/like?action=like \"I like this\") Bad [ ](https://codeforces.com/data/like?action=like \"I like this\") Terrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2258C - Far Cities](https://codeforces.com/contest/2258/problem/C \"Codeforces Round 1118 (Div. 2)\") Idea: [Seferoglu](https://codeforces.com/profile/Seferoglu \"Expert Seferoglu\") (This problem was originally designed simply to find the furthest node from the root; thanks to [cadmiumky](https://codeforces.com/profile/cadmiumky \"Grandmaster cadmiumky\") for expanding on the idea. Thanks to [robert9524](https://codeforces.com/profile/robert9524 \"Candidate Master robert9524\") for providing stronger tests on cutting random solutions without an adaptive grader.)\n\n**Hint1**\n\nHow to find a diameter?\n\n**Hint2**\n\nConsider the simplest algorithm.\n\n**Solution**\n\nFor each node, first determine whether it is closer to the first vertex than the longest distance seen so far. If not, just skip the node. This uses $$$n-1$$$ queries. Otherwise, while the query returns true, increase the queried distance by one. This can happen up to $$$n-1$$$ times. We can find one end of the diameter by running $$$2n - 2$$$ queries in total. Then repeat the process to find the other end of the diameter. This yields a total of $$$4n-4$$$ queries. But, as previously written, distance can increase at most $$$n-1$$$ times. So this yields a total of $$$3n-3$$$ queries.\n\n**Code**\n\n[388911205](https://codeforces.com/contest/2258/submission/388911205 \"Submission 388911205 by ItsNotMeItsYou\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") Good [ ](https://codeforces.com/data/like?action=like \"I like this\") Ok [ ](https://codeforces.com/data/like?action=like \"I like this\") Bad [ ](https://codeforces.com/data/like?action=like \"I like this\") Terrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2258D - Magic Tiles](https://codeforces.com/contest/2258/problem/D \"Codeforces Round 1118 (Div. 2)\") Idea: [ItsNotMeItsYou](https://codeforces.com/profile/ItsNotMeItsYou \"Master ItsNotMeItsYou\")\n\n**Hint1**\n\nWhat happens when a range includes another?\n\n**Hint2**\n\nA range that is included by another range is useless. What happens when we remove all such useless ranges?\n\n**Hint3**\n\nWe only need to solve the problem for chains of ranges. How?\n\n**Solution**\n\nIn this editorial we will use BigIntegers, which are simple vectors, and addition and comparison is $$$\\\\mathcal{O}(n)$$$. It can be proven that if one range includes another, we can remove the other. Also, it can be proven that only start and endpoints of the intervals are important, and let them be $$$pts$$$. Well-known $$$\\\\mathcal{O}(n^3)$$$ dp exists; $$$dp\\_i$$$ means the maximum score if we fill all of the rows through $$$pts\\_i$$$; $$$dp\\_i=\\\\max dp\\_j + 100^{100^{i-j}}$$$ if there is an interval that covers interval $$$\\[j, i)$$$. Because there are only $$$2$$$ columns, it can be proven that this solution works in $$$\\\\mathcal{O}(n^2)$$$. From another perspective, you can imagine you have chains. \n\n**Code**\n\n[388911243](https://codeforces.com/contest/2258/submission/388911243 \"Submission 388911243 by ItsNotMeItsYou\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") Good [ ](https://codeforces.com/data/like?action=like \"I like this\") Ok [ ](https://codeforces.com/data/like?action=like \"I like this\") Bad [ ](https://codeforces.com/data/like?action=like \"I like this\") Terrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n**Challenge**\n\nCan you solve for bigger constraints, such as $$$n,m \\\\le 2 \\\\cdot 10^5$$$? Thanks to [dinohaur](https://codeforces.com/profile/dinohaur \"Master dinohaur\") for the code: [388911329](https://codeforces.com/contest/2258/submission/388911329 \"Submission 388911329 by ItsNotMeItsYou\") \n\n**Answer**\n\n With using faster BigIntegers, you can solve in $$$\\\\mathcal{O}(n \\\\log n)$$$. For more info: <https://codeforces.com/problemset/problem/464/E>\n\n[2258E - DivMEX](https://codeforces.com/contest/2258/problem/E \"Codeforces Round 1118 (Div. 2)\") Idea: [ItsNotMeItsYou](https://codeforces.com/profile/ItsNotMeItsYou \"Master ItsNotMeItsYou\"), [carcinisation](https://codeforces.com/profile/carcinisation \"International Master carcinisation\") (This problem was actually designed differently. Thanks to [anpaio](https://codeforces.com/profile/anpaio \"International Grandmaster anpaio\") for this version.)\n\n**Hint1**\n\nWhat kind of integers can exist as an $$$f(l,r)=x$$$ value?\n\n**Hint2**\n\nHow to check a value $$$x$$$ exists as a $$$f(l,r)$$$?\n\n**Solution**\n\nIt can be seen that each value of $$$f$$$ must satisfy $$$f(l,r) \\\\leq \\\\text{smallest prime power bigger than n}$$$; for each $$$x$$$ that satisfies this constraint, we will try to check whether this value can be obtained or not. \n\nA range that satisfies $$$f(l,r)=x$$$ can't contain a multiple of $$$x$$$; otherwise, the set of divisors would have $$$x$$$ in it. When considering multiples of $$$x$$$ as obstacles, we realize that it is optimal to greedily select between two consecutive obstacles. Hence, at each obstacle, we check whether every number $$$y$$$ that satisfies $$$1 \\\\leq y \\\\le x$$$ appears as a factor of any integer between this and the previous obstacle. This solution works in $$$\\\\mathcal{O}(( \\\\sum\\_{i=1}^n |S\\_i|) \\\\cdot n$$$. Where $$$S\\_x$$$ represents the prime powers of $$$x$$$. \n\nTo optimize this solution, we iterate over all $$$a\\_i$$$ and check every factor of $$$a\\_i$$$, considering the index $$$i$$$ as an obstacle. In order to verify that every element smaller than $$$x$$$ is present between this and the previous obstacle, use a minimum segment tree. This solution works in $$$\\\\mathcal{O}(( \\\\sum\\_{i=1}^n |S\\_i|) \\\\log n) \\\\le \\\\mathcal{O}(n \\\\log^2 n)$$$.\n\n**Code (anpaio)**\n\n[388911266](https://codeforces.com/contest/2258/submission/388911266 \"Submission 388911266 by ItsNotMeItsYou\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") Good [ ](https://codeforces.com/data/like?action=like \"I like this\") Ok [ ](https://codeforces.com/data/like?action=like \"I like this\") Bad [ ](https://codeforces.com/data/like?action=like \"I like this\") Terrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2258F - Plus Minus Tree](https://codeforces.com/contest/2258/problem/F \"Codeforces Round 1118 (Div. 2)\") Idea: [ItsNotMeItsYou](https://codeforces.com/profile/ItsNotMeItsYou \"Master ItsNotMeItsYou\")\n\n**Solution**\n\nConsider $$$\\\\mathcal{O}(n^2)$$$ $$$dp\\[v\\]\\[\\\\text{#1's count in subtree of v}\\]$$$ using the merging subtrees technique. The dp is convex because we are adding only two convex functions, resulting in a convex function. Then, using small to large, you can achieve $$$\\\\mathcal{O}(n \\\\log^2 n)$$$ complexity.\n\nTo elaborate: \n\nFor $$$\\\\mathcal{O}(n^2)$$$, you can keep dp as pairs for simplicity: $$$(\\\\text{score of v}, \\\\text{minimum score})$$$. \n\nAs can be seen in the vector, scores are increasing by $$$2$$$. So you can keep dp with the vector's first element and then with the differences. When you merge two children's subtrees, you are basically doing a Minkowski sum, which is just updating the first element as $$$\\\\text{(lhs\\[0\\].first + rhs\\[0\\].first, lhs\\[0\\].second + rhs\\[0\\].second})$$$, and merging the differences. \n\nThen, for each point added to $$$|x|$$$, you must add $$$+2$$$ or $$$-2$$$ ($$$2$$$ due to parity) for all differences split by $$$0$$$ (be careful not to add any difference to $$$\\[-1, +1\\]$$$). \n\nYou can maintain two sets for each vertex, resulting in $$$\\\\mathcal{O}(n \\\\log^2 n)$$$ complexity from small to large. \n\n**Code**\n\n[388911297](https://codeforces.com/contest/2258/submission/388911297 \"Submission 388911297 by ItsNotMeItsYou\")\n\n**Alternative Solution**\n\nFirst, identify all nodes in the tree with an initial value of zero, set them to one, and store them in a candidate list. Then, until the list is completely empty, select the most advantageous node at each step. For each node, we trace the path up to the root, counting how many nodes along that path have a subtree sum greater than one ($$$\\\\text{pos}$$$) and how many have a subtree sum less than one ($$$\\\\text{neg}$$$), thereby calculating a score based on the difference $$$\\\\text{pos} - \\\\text{neg}$$$.\n\nAfter finding the node with the highest score among the list, select it, change its value to minus one, and decrease the subtree sums of all ancestors along the path from the chosen node to the root by two. We repeat this process until all special nodes are processed and the list is empty. And you can fasten this algorithm by HLD. \n\n$$$\\\\mathcal{O}(n^3)$$$ code by [anpaio](https://codeforces.com/profile/anpaio \"International Grandmaster anpaio\"): [388911953](https://codeforces.com/contest/2258/submission/388911953 \"Submission 388911953 by ItsNotMeItsYou\") $$$\\\\mathcal{O}(n \\\\log^2 n)$$$ code by [anpaio](https://codeforces.com/profile/anpaio \"International Grandmaster anpaio\"): [388911806](https://codeforces.com/contest/2258/submission/388911806 \"Submission 388911806 by ItsNotMeItsYou\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") Good [ ](https://codeforces.com/data/like?action=like \"I like this\") Ok [ ](https://codeforces.com/data/like?action=like \"I like this\") Bad [ ](https://codeforces.com/data/like?action=like \"I like this\") Terrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n![](//codeforces.com/codeforces.org/s/85295/images/icons/paperclip-16x16.png) Tutorial of [Codeforces Round 1118 (Div. 2)](https://codeforces.com/contest/2258) \n\n* [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/voteup-gray.png \"Vote: I like it\")](#)\n* +44\n* [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/votedown-gray.png \"Vote: I do not like it\")](#)\n\n* [![Author](//codeforces.com/codeforces.org/s/85295/images/blog/user_16x16.png \"Author\")](https://codeforces.com/profile/ItsNotMeItsYou) [ItsNotMeItsYou ](https://codeforces.com/profile/ItsNotMeItsYou)\n* ![Publication date](//codeforces.com/codeforces.org/s/85295/images/blog/date_16x16.png \"Publication date\") 3 weeks ago\n* [![Comments](//codeforces.com/codeforces.org/s/85295/images/blog/comments_16x16.png \"Comments\")](https://codeforces.com/blog/entry/156316#comments) [76 ](https://codeforces.com/blog/entry/156316#comments)\n\n  \n![Comments](//codeforces.com/codeforces.org/s/85295/images/icons/comments-48x48.png \"Comments\") Comments (76) \n\n[Write comment?](#) \n\n| » [MaxBlazeIceInk](https://codeforces.com/profile/MaxBlazeIceInk \"Candidate Master MaxBlazeIceInk\") | 3 weeks ago, [show](#) # |\n| --------------------------------------------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/2334734/avatar/fb176e49ae034a0a.jpg) ](https://codeforces.com/profile/MaxBlazeIceInk) [MaxBlazeIceInk](https://codeforces.com/profile/MaxBlazeIceInk \"Candidate Master MaxBlazeIceInk\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389109 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +34 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) On clist.by we can check the hardest div2D since 2026:wonderful 2500 difficulty! [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [rajakrishna](https://codeforces.com/profile/rajakrishna \"Pupil rajakrishna\") | 3 weeks ago, [show](#) # |\n| ------------------------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/5072937/avatar/d8cce84f7422827b.jpg) ](https://codeforces.com/profile/rajakrishna) [rajakrishna](https://codeforces.com/profile/rajakrishna \"Pupil rajakrishna\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389110 \"Link to comment\") \\| [←](#)Rev. 2 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) \\-17 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) . [→](#) [Reply](#) |\n| ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [Daniel-Alp](https://codeforces.com/profile/Daniel-Alp \"Pupil Daniel-Alp\") | 3 weeks ago, [show (+1)](#) # |\n| ---------------------------------------------------------------------------- | ----------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/Daniel-Alp) [Daniel-Alp](https://codeforces.com/profile/Daniel-Alp \"Pupil Daniel-Alp\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389112 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I enjoyed the contest! (despite being one of the goobers who killed all their time on B2 :P) [→](#) [Reply](#) |\n| -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [lmjsjg](https://codeforces.com/profile/lmjsjg \"Specialist lmjsjg\") | 3 weeks ago, [show](#) # |\n| ----------------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/2917220/avatar/dee6d6ea3380cb56.jpg) ](https://codeforces.com/profile/lmjsjg) [lmjsjg](https://codeforces.com/profile/lmjsjg \"Specialist lmjsjg\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389207 \"Link to comment\") [^](#comment-1389112 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) me too... [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n* | » » [superl61](https://codeforces.com/profile/superl61 \"Newbie superl61\") | 2 weeks ago, [show](#) # |\n| ------------------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/superl61) [superl61](https://codeforces.com/profile/superl61 \"Newbie superl61\") | 2 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389741 \"Link to comment\") [^](#comment-1389112 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) so do I [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [allcaps](https://codeforces.com/profile/allcaps \"Expert allcaps\") | 3 weeks ago, [show (+1)](#) # |\n| -------------------------------------------------------------------- | ----------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/3726532/avatar/64912144a3b5065f.jpg) ](https://codeforces.com/profile/allcaps) [allcaps](https://codeforces.com/profile/allcaps \"Expert allcaps\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389117 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +41 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) editorial: \"Thanks to robert9524 for providing stronger tests on cutting random solutions without an adaptive grader\"me: AC'd with a randomised solution and feels like the 0.01% of bacteria that don't get killed by soap [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [Hyder1102](https://codeforces.com/profile/Hyder1102 \"Expert Hyder1102\") | 3 weeks ago, [show (+1)](#) # |\n| ---------------------------------------------------------------------------- | ----------------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/4156580/avatar/3c7005d8f3eefc09.jpg) ](https://codeforces.com/profile/Hyder1102) [Hyder1102](https://codeforces.com/profile/Hyder1102 \"Expert Hyder1102\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389285 \"Link to comment\") [^](#comment-1389117 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +4 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) what was your randomized appraoch? [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n  * | » » » [allcaps](https://codeforces.com/profile/allcaps \"Expert allcaps\") | 3 weeks ago, [show (+1)](#) # |\n| ------------------------------------------------------------------------ | ----------------------------- |\n\n| » » » [ ![](https://codeforces.com/userpic.codeforces.org/3726532/avatar/64912144a3b5065f.jpg) ](https://codeforces.com/profile/allcaps) [allcaps](https://codeforces.com/profile/allcaps \"Expert allcaps\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389297 \"Link to comment\") [^](#comment-1389285 \"Parent comment\") \\| [←](#)Rev. 4 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +9 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) basically because so many interactive problems have binary search, my brain went straight to binary searching for the exact distance when you find a new farthest node, instead of realising that you can just increment it by one and that will happen at most $$$n - 1$$$ times.here's what i ended up doing: same general idea of the two DFS passes in each pass, keep track of furthest node $$$u$$$ and its distance $$$d$$$ maintain a variable $$$h$$$ as the maximum length a path in this graph can have (initially $$$n - 1$$$) when considering a new node, first check if it beats the distance by at least 1\\. if not, stop and go to the next node. otherwise, binary search between $$$\\[d + 1, h\\]$$$ to find distance of this node. then update $$$h$$$ to be $$$MIN(h, \\\\text{new distance + number of nodes left})$$$. the randomised part of the solution is that instead of starting at vertex $$$1$$$, and then querying $$$2, 3, 4, ... n$$$ in order, I shuffle the vertices so that I should only need to do $$$O(log(n))$$$ binary searches, so on average, I should have $$$2 n - 2 + O(log^2(n))$$$ queries. The worst case is still just $$$n log(n)$$$. intuitively, the worst case for such a solution should just be a path, so i wrote a program locally to simulate the process on a bunch of randomised paths, and it never exceeded $$$3n$$$ queries there. submission: [388829880](https://codeforces.com/contest/2258/submission/388829880 \"Submission 388829880 by allcaps\") [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n    * | » » » » [sergioRG](https://codeforces.com/profile/sergioRG \"Expert sergioRG\") | 3 weeks ago, [show (+1)](#) # |\n| ----------------------------------------------------------------------------- | ----------------------------- |\n\n| » » » » [ ![](https://codeforces.com/userpic.codeforces.org/82000/avatar/6a2b34040eea0d59.jpg) ](https://codeforces.com/profile/sergioRG) [sergioRG](https://codeforces.com/profile/sergioRG \"Expert sergioRG\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389300 \"Link to comment\") [^](#comment-1389297 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +3 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I wouldnt call that a \"0.01% bacteria that survives\" solution, the approach is also good even if it can exceed 3n bound in principle, also I admire anyone solving problems in Rust, I wouldnt be able to lol [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n      * | » » » » » [allcaps](https://codeforces.com/profile/allcaps \"Expert allcaps\") | 3 weeks ago, [show](#) # |\n| ---------------------------------------------------------------------------- | ------------------------ |\n\n| » » » » » [ ![](https://codeforces.com/userpic.codeforces.org/3726532/avatar/64912144a3b5065f.jpg) ](https://codeforces.com/profile/allcaps) [allcaps](https://codeforces.com/profile/allcaps \"Expert allcaps\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389302 \"Link to comment\") [^](#comment-1389300 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Thanks lol. Maybe the type of randomised solution they're talking about cutting is actually something different (tho I can't really think of what it would be). [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n    * | » » » » [Hyder1102](https://codeforces.com/profile/Hyder1102 \"Expert Hyder1102\") | 3 weeks ago, [show](#) # |\n| -------------------------------------------------------------------------------- | ------------------------ |\n\n| » » » » [ ![](https://codeforces.com/userpic.codeforces.org/4156580/avatar/3c7005d8f3eefc09.jpg) ](https://codeforces.com/profile/Hyder1102) [Hyder1102](https://codeforces.com/profile/Hyder1102 \"Expert Hyder1102\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389426 \"Link to comment\") [^](#comment-1389297 \"Parent comment\") \\| [←](#)Rev. 2 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +3 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) cool man, apart from randomized u had some nice cost cutting alsothe check +1 possible, if not skip and the upper limit for d [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n    * | » » » » [abcd1145](https://codeforces.com/profile/abcd1145 \"Expert abcd1145\") | 2 weeks ago, [show](#) # |\n| ----------------------------------------------------------------------------- | ------------------------ |\n\n| » » » » [ ![](https://codeforces.com/userpic.codeforces.org/3337986/avatar/44bb8552ff61c77d.jpg) ](https://codeforces.com/profile/abcd1145) [abcd1145](https://codeforces.com/profile/abcd1145 \"Expert abcd1145\") | 2 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389460 \"Link to comment\") [^](#comment-1389297 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) An amazing solution [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [Protorianes](https://codeforces.com/profile/Protorianes \"Specialist Protorianes\") | 3 weeks ago, [show](#) # |\n| ------------------------------------------------------------------------------------ | ------------------------ |\n\n\n...127483 bytes truncated...\n\nThe tool call succeeded but the output was truncated. Full output saved to: /Users/zayd/.local/share/opencode/tool-output/tool_0adfab628001X80z1G54TcXaRX\nUse the Task tool to have explore agent process this file with Grep and Read (with offset/limit). Do NOT read the full file yourself - delegate to save context.",
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              "command": "python3 - <<'PY'\nfrom itertools import product\nfrom math import lcm\n\ndef solve(a):\n n=len(a); pp=[]\n for x in range(2,2*n+1):\n  p=next(d for d in range(2,x+1) if x%d==0)\n  y=x\n  while y%p==0:y//=p\n  if y==1:\n   pp.append(x)\n   if x>n:break\n last=[0]*len(pp); ans=set()\n for i,v in enumerate(a,1):\n  ids=[j for j,q in enumerate(pp) if v%q==0]\n  for j in ids:\n   if last[j]+1<i and min(last[:j],default=n+1)>last[j]:ans.add(pp[j])\n  for j in ids:last[j]=i\n for j,q in enumerate(pp):\n  if last[j]<n and min(last[:j],default=n+1)>last[j]:ans.add(q)\n return ans\n\ndef brute(a):\n ans=set()\n for l in range(len(a)):\n  v=1\n  for r in range(l,len(a)):\n   v=lcm(v,a[r]);q=1\n   while v%q==0:q+=1\n   ans.add(q)\n return ans\ncount=0\nfor n in range(1,7):\n for a in product(range(1,n+1),repeat=n):\n  assert solve(a)==brute(a),(a,solve(a),brute(a))\n  count+=1\nprint('Exhaustive checks passed:',count)\nPY",
              "timeout": 120000
            },
            "output": "Exhaustive checks passed: 50069\n",
            "metadata": {
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              "exit": 0,
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            "title": "python3 - <<'PY'\nfrom itertools import product\nfrom math import lcm\n\ndef solve(a):\n n=len(a); pp=[]\n for x in range(2,2*n+1):\n  p=next(d for d in range(2,x+1) if x%d==0)\n  y=x\n  while y%p==0:y//=p\n  if y==1:\n   pp.append(x)\n   if x>n:break\n last=[0]*len(pp); ans=set()\n for i,v in enumerate(a,1):\n  ids=[j for j,q in enumerate(pp) if v%q==0]\n  for j in ids:\n   if last[j]+1<i and min(last[:j],default=n+1)>last[j]:ans.add(pp[j])\n  for j in ids:last[j]=i\n for j,q in enumerate(pp):\n  if last[j]<n and min(last[:j],default=n+1)>last[j]:ans.add(q)\n return ans\n\ndef brute(a):\n ans=set()\n for l in range(len(a)):\n  v=1\n  for r in range(l,len(a)):\n   v=lcm(v,a[r]);q=1\n   while v%q==0:q+=1\n   ans.add(q)\n return ans\ncount=0\nfor n in range(1,7):\n for a in product(range(1,n+1),repeat=n):\n  assert solve(a)==brute(a),(a,solve(a),brute(a))\n  count+=1\nprint('Exhaustive checks passed:',count)\nPY",
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          "hints": [
            {
              "order": 1,
              "content": "Don't compute the LCM itself—it can get enormous. Think about its prime exponents instead. If $x$ is the smallest positive integer that does not divide an LCM, can $x$ have two different prime factors?"
            },
            {
              "order": 2,
              "content": "Every answer is a prime power. For a prime power $q=p^k$, $q$ divides a subarray's LCM exactly when at least one element is divisible by $q$. So the problem becomes: which prime powers can be the smallest one missing from a subarray?"
            },
            {
              "order": 3,
              "content": "Fix a candidate prime power $x$. Positions with $x\\mid a_i$ are barriers: a valid subarray cannot cross or include them. It is enough to test the maximal nonempty gaps between consecutive barriers, including the prefix and suffix gaps. Enlarging a subarray inside one gap cannot lose any smaller prime power."
            },
            {
              "order": 4,
              "content": "Sweep from left to right and maintain $\\operatorname{last}[q]$, the latest processed position whose value is divisible by prime power $q$, initially $0$. Just before a barrier at position $i$ for $x$, let $b=\\operatorname{last}[x]$. The gap $[b+1,i-1]$ contains every smaller prime power exactly when $\\min_{q<x}\\operatorname{last}[q]>b$. Also check that the gap is nonempty."
            },
            {
              "order": 5,
              "content": "Store the latest positions in a minimum segment tree ordered by prime-power value. At position $i$, enumerate every prime-power divisor of $a_i$, query its preceding gap, and only then update all those latest positions to $i$. Finally test the suffix gap for every candidate. There are at most $\\log_2 n$ prime-power divisors per element, giving $O(n\\log^2 n)$ time. Include the smallest prime power greater than $n$ among the candidates."
            }
          ],
          "editorial": "The LCM is a distraction. We only need to know which **prime powers** occur as divisors of at least one element in a subarray.\n\n### 1. Only prime powers can be answers\n\nLet $L$ be a subarray's LCM and let $x$ be its smallest missing divisor. Since $1$ always divides $L$, we have $x\\ge 2$.\n\nSuppose $x$ is not a prime power. Write its prime factorization as\n$$x=p_1^{e_1}p_2^{e_2}\\cdots p_k^{e_k},\\qquad k\\ge 2.$$\nEach factor $p_j^{e_j}$ is smaller than $x$, so each must divide $L$. They are pairwise coprime, so their product also divides $L$. Contradiction.\n\nTherefore, **every answer is a prime power**.\n\nFor a prime power $q=p^e$,\n$$q\\mid\\operatorname{lcm}(a_l,\\ldots,a_r)\n\\iff \\exists i\\in[l,r]:q\\mid a_i.$$\nThis follows because an LCM takes the maximum exponent of each prime across its elements.\n\nConsequently, $f(l,r)=x$ precisely when:\n- no element in $[l,r]$ is divisible by $x$;\n- for every prime power $q<x$, some element in $[l,r]$ is divisible by $q$.\n\nChecking prime powers is enough: every positive integer smaller than $x$ is a product of its prime-power factors, all of which are also smaller than $x$.\n\nLet $B$ be the smallest prime power greater than $n$. No element can be divisible by $B$, so every answer is at most $B$. We consider all prime powers up to and including $B$.\n\n### 2. Test maximal gaps, not all subarrays\n\nFix a candidate $x$. Call every position with $x\\mid a_i$ a **barrier**.\n\nAny subarray missing $x$ lies entirely inside a gap between barriers. If a smaller subarray inside that gap contains every required prime power, the whole gap contains them too. Extending within the gap cannot introduce a multiple of $x$.\n\nThus, $x$ is achievable if and only if one of its maximal nonempty barrier-free gaps contains every prime power smaller than $x$.\n\nThat gets rid of the quadratic number of subarrays. Now we need to check these gaps efficiently across all candidates.\n\n### 3. Turn gap checks into prefix-minimum queries\n\nSweep the array from left to right. Maintain\n$$\\operatorname{last}[q]=\\text{latest processed index }j\\text{ such that }q\\mid a_j,$$\nwith $\\operatorname{last}[q]=0$ if there is no such index.\n\nJust before processing position $i$, suppose $x\\mid a_i$. Let\n$$b=\\operatorname{last}[x].$$\nThen the gap immediately before this barrier is $[b+1,i-1]$.\n\nFor any smaller prime power $q$, all processed positions are at most $i-1$. Therefore,\n$$q\\text{ occurs in this gap}\\iff\\operatorname{last}[q]>b.$$\nAll smaller prime powers occur exactly when\n$$\\min_{\\substack{q<x\\\\q\\text{ is a prime power}}}\\operatorname{last}[q]>b.$$\n\nStore the latest positions in a **minimum segment tree**, with leaves ordered by increasing prime-power value. This condition becomes a prefix-minimum query.\n\nFor $x=2$, the prefix is empty. Define its minimum as $+\\infty$: there are no smaller prime powers to require. We must still explicitly check that the gap is nonempty, meaning $b+1<i$. An empty subarray doesn't get a free pass.\n\n### 4. Sweep only the relevant events\n\nPosition $i$ is a barrier only for prime-power divisors of $a_i$.\n\nFactor $a_i$ using a smallest-prime-factor sieve. If a prime occurs with exponent $e$, enumerate\n$$p,p^2,\\ldots,p^e.$$\nInclude **all** these powers, not just the largest one.\n\nAt each position $i$:\n1. Enumerate its prime-power divisors.\n2. For each such divisor $x$, test the preceding gap using the current latest positions.\n3. After all queries, set $\\operatorname{last}[x]=i$ for every enumerated divisor and update the segment tree.\n\n**All queries must happen before any updates at that position.** The tested gaps end at $i-1$, so using position $i$ as evidence would be wrong.\n\nAfter the sweep, each candidate still has a final gap\n$$[\\operatorname{last}[x]+1,n].$$\nTest it with the same condition, requiring $\\operatorname{last}[x]<n$. These final checks can use a running prefix minimum over the latest-position array, taking linear time.\n\n### 5. Implementation details\n\n- Sieve up to $2n$. This always covers $B$, because the smallest power of two greater than $n$ is at most $2n$.\n- Generate prime powers, scan them in increasing order, and stop after the first one greater than $n$.\n- Assign each candidate a compressed index. Segment-tree query $[0,j)$ checks every candidate smaller than the one at index $j$.\n- Initialize all latest positions to zero and use $n+1$ as infinity for an empty minimum.\n- Mark successful candidates and output them in increasing order.\n\nNo LCM arithmetic is needed, so overflow never enters the picture.\n\n### Correctness proof\n\n**Lemma 1.** Every possible value of $f$ is a prime power at most $B$.\n\nIf the smallest missing divisor had multiple distinct prime factors, its smaller prime-power factors would all divide the LCM, forcing it to divide the LCM too. Also, $B$ cannot divide any array element, so it cannot divide any subarray LCM. Hence the smallest missing divisor is at most $B$.\n\n**Lemma 2.** A candidate $x$ is achievable if and only if some maximal nonempty gap between its barriers contains every smaller prime power.\n\nAny subarray with value $x$ avoids all barriers and lies inside such a gap. Enlarging it to the whole gap preserves every smaller prime power while still excluding $x$. Conversely, a gap containing every smaller prime power and excluding $x$ has smallest missing divisor exactly $x$.\n\n**Lemma 3.** Every gap check performed by the algorithm succeeds exactly when that gap has value $x$.\n\nBefore processing a barrier at $i$, the maintained latest positions describe precisely the prefix ending at $i-1$. A smaller prime power $q$ occurs after the previous barrier $b$ exactly when $\\operatorname{last}[q]>b$. The prefix minimum tests this simultaneously for all $q<x$. The explicit length condition excludes empty gaps. The same argument applies to suffix gaps after the sweep.\n\n**Theorem.** The algorithm outputs exactly all achievable values.\n\nBy Lemma 1, every achievable value is considered. For each candidate, the sweep checks every gap ending before a barrier, and the final pass checks its suffix gap. By Lemmas 2 and 3, the candidate is marked exactly when a subarray with that value exists.\n\n### Complexity\n\nIf\n$$a_i=\\prod_j p_j^{e_j},$$\nthen the number of enumerated prime-power divisors is $\\sum_j e_j$. Since\n$$2^{\\sum_j e_j}\\le a_i\\le n,$$\nthere are at most $\\log_2 n$ such divisors per element.\n\nEach causes one $O(\\log n)$ query and one $O(\\log n)$ update. Sieve construction, candidate generation, and the final pass fit within the resulting bound.\n\n- **Time:** $O(n\\log^2 n)$ per test case, with constant work for $n=1$.\n- **Space:** $O(n)$.",
          "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\nstruct SegTree {\n    int size = 1, inf;\n    vector<int> tree;\n\n    SegTree(int n, int inf) : inf(inf) {\n        while (size < n) size <<= 1;\n        tree.assign(2 * size, 0);\n    }\n\n    void update(int p, int value) {\n        p += size;\n        tree[p] = value;\n        while (p > 1) {\n            p >>= 1;\n            tree[p] = min(tree[2 * p], tree[2 * p + 1]);\n        }\n    }\n\n    int prefixMin(int r) const {\n        int result = inf;\n        for (int l = size, right = size + r; l < right;\n             l >>= 1, right >>= 1) {\n            if (l & 1) result = min(result, tree[l++]);\n            if (right & 1) result = min(result, tree[--right]);\n        }\n        return result;\n    }\n};\n\nint main() {\n    setIO();\n\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        int limit = 2 * n;\n        vector<int> spf(limit + 1);\n        vector<char> isPower(limit + 1, false);\n\n        for (int p = 2; p <= limit; ++p) {\n            if (spf[p]) continue;\n            spf[p] = p;\n            for (ll v = 1LL * p * p; v <= limit; v += p) {\n                if (!spf[v]) spf[v] = p;\n            }\n            for (ll q = p; q <= limit; q *= p) {\n                isPower[q] = true;\n            }\n        }\n\n        vector<int> powers, id(limit + 1, -1);\n        for (int q = 2; q <= limit; ++q) {\n            if (!isPower[q]) continue;\n            id[q] = (int)powers.size();\n            powers.push_back(q);\n            if (q > n) break;\n        }\n\n        int m = (int)powers.size();\n        vector<int> last(m, 0);\n        vector<char> possible(m, false);\n        SegTree seg(m, n + 1);\n        vector<int> divisors;\n        divisors.reserve(20);\n\n        for (int i = 1; i <= n; ++i) {\n            int value;\n            cin >> value;\n            divisors.clear();\n            while (value > 1) {\n                int p = spf[value], q = 1;\n                while (value % p == 0) {\n                    value /= p;\n                    q *= p;\n                    divisors.push_back(id[q]);\n                }\n            }\n\n            // Check gaps ending at i - 1 before inserting position i.\n            for (int j : divisors) {\n                if (!possible[j] && last[j] + 1 < i &&\n                    seg.prefixMin(j) > last[j]) {\n                    possible[j] = true;\n                }\n            }\n            for (int j : divisors) {\n                last[j] = i;\n                seg.update(j, i);\n            }\n        }\n\n        int prefix = n + 1;\n        vector<int> answer;\n        for (int j = 0; j < m; ++j) {\n            if (last[j] < n && prefix > last[j]) {\n                possible[j] = true;\n            }\n            prefix = min(prefix, last[j]);\n            if (possible[j]) answer.push_back(powers[j]);\n        }\n\n        cout << answer.size() << '\\n';\n        for (int i = 0; i < (int)answer.size(); ++i) {\n            if (i) cout << ' ';\n            cout << answer[i];\n        }\n        cout << '\\n';\n    }\n}\n"
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              "hints": [
                {
                  "order": 1,
                  "content": "Don't compute the LCM itself—it can get enormous. Think about its prime exponents instead. If $x$ is the smallest positive integer that does not divide an LCM, can $x$ have two different prime factors?"
                },
                {
                  "order": 2,
                  "content": "Every answer is a prime power. For a prime power $q=p^k$, $q$ divides a subarray's LCM exactly when at least one element is divisible by $q$. So the problem becomes: which prime powers can be the smallest one missing from a subarray?"
                },
                {
                  "order": 3,
                  "content": "Fix a candidate prime power $x$. Positions with $x\\mid a_i$ are barriers: a valid subarray cannot cross or include them. It is enough to test the maximal nonempty gaps between consecutive barriers, including the prefix and suffix gaps. Enlarging a subarray inside one gap cannot lose any smaller prime power."
                },
                {
                  "order": 4,
                  "content": "Sweep from left to right and maintain $\\operatorname{last}[q]$, the latest processed position whose value is divisible by prime power $q$, initially $0$. Just before a barrier at position $i$ for $x$, let $b=\\operatorname{last}[x]$. The gap $[b+1,i-1]$ contains every smaller prime power exactly when $\\min_{q<x}\\operatorname{last}[q]>b$. Also check that the gap is nonempty."
                },
                {
                  "order": 5,
                  "content": "Store the latest positions in a minimum segment tree ordered by prime-power value. At position $i$, enumerate every prime-power divisor of $a_i$, query its preceding gap, and only then update all those latest positions to $i$. Finally test the suffix gap for every candidate. There are at most $\\log_2 n$ prime-power divisors per element, giving $O(n\\log^2 n)$ time. Include the smallest prime power greater than $n$ among the candidates."
                }
              ],
              "editorial": "The LCM is a distraction. We only need to know which **prime powers** occur as divisors of at least one element in a subarray.\n\n### 1. Only prime powers can be answers\n\nLet $L$ be a subarray's LCM and let $x$ be its smallest missing divisor. Since $1$ always divides $L$, we have $x\\ge 2$.\n\nSuppose $x$ is not a prime power. Write its prime factorization as\n$$x=p_1^{e_1}p_2^{e_2}\\cdots p_k^{e_k},\\qquad k\\ge 2.$$\nEach factor $p_j^{e_j}$ is smaller than $x$, so each must divide $L$. They are pairwise coprime, so their product also divides $L$. Contradiction.\n\nTherefore, **every answer is a prime power**.\n\nFor a prime power $q=p^e$,\n$$q\\mid\\operatorname{lcm}(a_l,\\ldots,a_r)\n\\iff \\exists i\\in[l,r]:q\\mid a_i.$$\nThis follows because an LCM takes the maximum exponent of each prime across its elements.\n\nConsequently, $f(l,r)=x$ precisely when:\n- no element in $[l,r]$ is divisible by $x$;\n- for every prime power $q<x$, some element in $[l,r]$ is divisible by $q$.\n\nChecking prime powers is enough: every positive integer smaller than $x$ is a product of its prime-power factors, all of which are also smaller than $x$.\n\nLet $B$ be the smallest prime power greater than $n$. No element can be divisible by $B$, so every answer is at most $B$. We consider all prime powers up to and including $B$.\n\n### 2. Test maximal gaps, not all subarrays\n\nFix a candidate $x$. Call every position with $x\\mid a_i$ a **barrier**.\n\nAny subarray missing $x$ lies entirely inside a gap between barriers. If a smaller subarray inside that gap contains every required prime power, the whole gap contains them too. Extending within the gap cannot introduce a multiple of $x$.\n\nThus, $x$ is achievable if and only if one of its maximal nonempty barrier-free gaps contains every prime power smaller than $x$.\n\nThat gets rid of the quadratic number of subarrays. Now we need to check these gaps efficiently across all candidates.\n\n### 3. Turn gap checks into prefix-minimum queries\n\nSweep the array from left to right. Maintain\n$$\\operatorname{last}[q]=\\text{latest processed index }j\\text{ such that }q\\mid a_j,$$\nwith $\\operatorname{last}[q]=0$ if there is no such index.\n\nJust before processing position $i$, suppose $x\\mid a_i$. Let\n$$b=\\operatorname{last}[x].$$\nThen the gap immediately before this barrier is $[b+1,i-1]$.\n\nFor any smaller prime power $q$, all processed positions are at most $i-1$. Therefore,\n$$q\\text{ occurs in this gap}\\iff\\operatorname{last}[q]>b.$$\nAll smaller prime powers occur exactly when\n$$\\min_{\\substack{q<x\\\\q\\text{ is a prime power}}}\\operatorname{last}[q]>b.$$\n\nStore the latest positions in a **minimum segment tree**, with leaves ordered by increasing prime-power value. This condition becomes a prefix-minimum query.\n\nFor $x=2$, the prefix is empty. Define its minimum as $+\\infty$: there are no smaller prime powers to require. We must still explicitly check that the gap is nonempty, meaning $b+1<i$. An empty subarray doesn't get a free pass.\n\n### 4. Sweep only the relevant events\n\nPosition $i$ is a barrier only for prime-power divisors of $a_i$.\n\nFactor $a_i$ using a smallest-prime-factor sieve. If a prime occurs with exponent $e$, enumerate\n$$p,p^2,\\ldots,p^e.$$\nInclude **all** these powers, not just the largest one.\n\nAt each position $i$:\n1. Enumerate its prime-power divisors.\n2. For each such divisor $x$, test the preceding gap using the current latest positions.\n3. After all queries, set $\\operatorname{last}[x]=i$ for every enumerated divisor and update the segment tree.\n\n**All queries must happen before any updates at that position.** The tested gaps end at $i-1$, so using position $i$ as evidence would be wrong.\n\nAfter the sweep, each candidate still has a final gap\n$$[\\operatorname{last}[x]+1,n].$$\nTest it with the same condition, requiring $\\operatorname{last}[x]<n$. These final checks can use a running prefix minimum over the latest-position array, taking linear time.\n\n### 5. Implementation details\n\n- Sieve up to $2n$. This always covers $B$, because the smallest power of two greater than $n$ is at most $2n$.\n- Generate prime powers, scan them in increasing order, and stop after the first one greater than $n$.\n- Assign each candidate a compressed index. Segment-tree query $[0,j)$ checks every candidate smaller than the one at index $j$.\n- Initialize all latest positions to zero and use $n+1$ as infinity for an empty minimum.\n- Mark successful candidates and output them in increasing order.\n\nNo LCM arithmetic is needed, so overflow never enters the picture.\n\n### Correctness proof\n\n**Lemma 1.** Every possible value of $f$ is a prime power at most $B$.\n\nIf the smallest missing divisor had multiple distinct prime factors, its smaller prime-power factors would all divide the LCM, forcing it to divide the LCM too. Also, $B$ cannot divide any array element, so it cannot divide any subarray LCM. Hence the smallest missing divisor is at most $B$.\n\n**Lemma 2.** A candidate $x$ is achievable if and only if some maximal nonempty gap between its barriers contains every smaller prime power.\n\nAny subarray with value $x$ avoids all barriers and lies inside such a gap. Enlarging it to the whole gap preserves every smaller prime power while still excluding $x$. Conversely, a gap containing every smaller prime power and excluding $x$ has smallest missing divisor exactly $x$.\n\n**Lemma 3.** Every gap check performed by the algorithm succeeds exactly when that gap has value $x$.\n\nBefore processing a barrier at $i$, the maintained latest positions describe precisely the prefix ending at $i-1$. A smaller prime power $q$ occurs after the previous barrier $b$ exactly when $\\operatorname{last}[q]>b$. The prefix minimum tests this simultaneously for all $q<x$. The explicit length condition excludes empty gaps. The same argument applies to suffix gaps after the sweep.\n\n**Theorem.** The algorithm outputs exactly all achievable values.\n\nBy Lemma 1, every achievable value is considered. For each candidate, the sweep checks every gap ending before a barrier, and the final pass checks its suffix gap. By Lemmas 2 and 3, the candidate is marked exactly when a subarray with that value exists.\n\n### Complexity\n\nIf\n$$a_i=\\prod_j p_j^{e_j},$$\nthen the number of enumerated prime-power divisors is $\\sum_j e_j$. Since\n$$2^{\\sum_j e_j}\\le a_i\\le n,$$\nthere are at most $\\log_2 n$ such divisors per element.\n\nEach causes one $O(\\log n)$ query and one $O(\\log n)$ update. Sieve construction, candidate generation, and the final pass fit within the resulting bound.\n\n- **Time:** $O(n\\log^2 n)$ per test case, with constant work for $n=1$.\n- **Space:** $O(n)$.",
              "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\nstruct SegTree {\n    int size = 1, inf;\n    vector<int> tree;\n\n    SegTree(int n, int inf) : inf(inf) {\n        while (size < n) size <<= 1;\n        tree.assign(2 * size, 0);\n    }\n\n    void update(int p, int value) {\n        p += size;\n        tree[p] = value;\n        while (p > 1) {\n            p >>= 1;\n            tree[p] = min(tree[2 * p], tree[2 * p + 1]);\n        }\n    }\n\n    int prefixMin(int r) const {\n        int result = inf;\n        for (int l = size, right = size + r; l < right;\n             l >>= 1, right >>= 1) {\n            if (l & 1) result = min(result, tree[l++]);\n            if (right & 1) result = min(result, tree[--right]);\n        }\n        return result;\n    }\n};\n\nint main() {\n    setIO();\n\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        int limit = 2 * n;\n        vector<int> spf(limit + 1);\n        vector<char> isPower(limit + 1, false);\n\n        for (int p = 2; p <= limit; ++p) {\n            if (spf[p]) continue;\n            spf[p] = p;\n            for (ll v = 1LL * p * p; v <= limit; v += p) {\n                if (!spf[v]) spf[v] = p;\n            }\n            for (ll q = p; q <= limit; q *= p) {\n                isPower[q] = true;\n            }\n        }\n\n        vector<int> powers, id(limit + 1, -1);\n        for (int q = 2; q <= limit; ++q) {\n            if (!isPower[q]) continue;\n            id[q] = (int)powers.size();\n            powers.push_back(q);\n            if (q > n) break;\n        }\n\n        int m = (int)powers.size();\n        vector<int> last(m, 0);\n        vector<char> possible(m, false);\n        SegTree seg(m, n + 1);\n        vector<int> divisors;\n        divisors.reserve(20);\n\n        for (int i = 1; i <= n; ++i) {\n            int value;\n            cin >> value;\n            divisors.clear();\n            while (value > 1) {\n                int p = spf[value], q = 1;\n                while (value % p == 0) {\n                    value /= p;\n                    q *= p;\n                    divisors.push_back(id[q]);\n                }\n            }\n\n            // Check gaps ending at i - 1 before inserting position i.\n            for (int j : divisors) {\n                if (!possible[j] && last[j] + 1 < i &&\n                    seg.prefixMin(j) > last[j]) {\n                    possible[j] = true;\n                }\n            }\n            for (int j : divisors) {\n                last[j] = i;\n                seg.update(j, i);\n            }\n        }\n\n        int prefix = n + 1;\n        vector<int> answer;\n        for (int j = 0; j < m; ++j) {\n            if (last[j] < n && prefix > last[j]) {\n                possible[j] = true;\n            }\n            prefix = min(prefix, last[j]);\n            if (possible[j]) answer.push_back(powers[j]);\n        }\n\n        cout << answer.size() << '\\n';\n        for (int i = 0; i < (int)answer.size(); ++i) {\n            if (i) cout << ' ';\n            cout << answer[i];\n        }\n        cout << '\\n';\n    }\n}\n"
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