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+[{"task_id": "mmlu-5882", "model": "openai/gpt-5.6-luna", "prompt": "[MMLU] This question refers to the following information.\nBoth authors below were speaking of the French Revolution.\nIt was the best of times, it was the worst of times, it was the age of wisdom, it was the age of foolishness, it was the epoch of belief, it was the epoch of incredulity, it was the season of Light, it was the season of Darkness, it was the spring of hope, it was the winter of despair, we had everything before us, we had nothing before us, we were all going direct to Heaven, we were all going direct the other way\u2014in short, the period was so far like the present period, that some of its noisiest authorities insisted on its being received, for good or for evil, in the superlative degree of comparison only.\nThere were a king with a large jaw and a queen with a plain face, on the throne of England; there were a king with a large jaw and a queen with a fair face, on the throne of France. In both countries it was clearer than crystal . . . that things in general were settled for ever.\n\u2014A Tale of Two Cities, by Charles Dickens\nBliss was it in that dawn to be alive. But to be young was very heaven.\n\u2014The Prelude, by William Wordsworth\nThe intellectual foundations of the eighteenth-century political revolutions were based on\nA) the Reformation\nB) mercantilism\nC) the Enlightenment\nD) the Reconquista\n\nAnswer with the letter only.\n\nThink step by step about each option, then end with exactly: ANSWER: <letter>", "response": "ANSWER: C", "cost_usd": 8.18e-05, "tokens_in": 338, "tokens_out": 80, "latency_s": 3.589}, {"task_id": "mmlu-8151", "model": "openai/gpt-5.6-luna", "prompt": "[MMLU] What is the name of the Newtonian law that states 'An object in Motion tends to stay in motion'?\nA) Law of Inertia\nB) Law of Thermodynamics\nC) Law of Relativity\nD) Law of Gravitation\n\nAnswer with the letter only.\n\nThink step by step about each option, then end with exactly: ANSWER: <letter>", "response": "ANSWER: A", "cost_usd": 7.460000000000001e-05, "tokens_in": 86, "tokens_out": 110, "latency_s": 3.569}, {"task_id": "gsm8k-507", "model": "openai/gpt-5.6-luna", "prompt": "Janet takes two bus trips five days a week. If each bus trip costs her $2.20, how much would she save by buying a weekly bus pass for $20?\n\nReason step by step, then give the final answer as a single number.\n\nReason step by step, then end with the final answer as a single number.\n\n(ref 407875)", "response": "2 trips/day \u00d7 5 days = 10 trips  \n10 \u00d7 $2.20 = $22  \n$22 \u2212 $20 = $2  \n\n2", "cost_usd": 6.91e-05, "tokens_in": 79, "tokens_out": 102, "latency_s": 3.042}, {"task_id": "gsm8k-191", "model": "openai/gpt-5.6-luna", "prompt": "Ducks need to eat 3.5 pounds of insects each week to survive. If there is a flock of ten ducks, how many pounds of insects do they need per day?\n\nReason step by step, then give the final answer as a single number.\n\nReason step by step, then end with the final answer as a single number.\n\n(ref 471531)", "response": "Each duck needs 3.5 pounds per week.\n\nFor 10 ducks:  \n3.5 \u00d7 10 = 35 pounds per week.\n\nSince 1 week has 7 days:  \n35 \u00f7 7 = 5 pounds per day.\n\n5", "cost_usd": 6.849999999999998e-05, "tokens_in": 79, "tokens_out": 101, "latency_s": 2.491}, {"task_id": "lcb-abc389_b", "model": "openai/gpt-5.6-luna", "prompt": "You are given an integer X not less than 2.\nFind the positive integer N such that N! = X.\nHere, N! denotes the factorial of N, and it is guaranteed that there is exactly one such N.\n\nInput\n\nThe input is given from Standard Input in the following format:\nX\n\nOutput\n\nPrint the answer.\n\nConstraints\n\n\n- 2 \\leq X \\leq 3 \\times 10^{18}\n- There is exactly one positive integer N such that N!=X.\n- All input values are integers.\n\nSample Input 1\n\n6\n\nSample Output 1\n\n3\r\n\nFrom 3!=3\\times2\\times1=6, print 3.\n\nSample Input 2\n\n2432902008176640000\n\nSample Output 2\n\n20\r\n\nFrom 20!=2432902008176640000, print 20.\n\nWrite a complete Python 3 program that reads from standard input and writes the answer to standard output. Output ONLY the program source, no prose and no code fences.\n\nSolve this competitive-programming problem in Python 3. Read all of stdin at once with sys.stdin.buffer.read().split() and index across tokens; build output in a list and emit it with one write at the end. Check the constraint block and pick an algorithm whose running time stays under ~1e8 steps at the largest allowed input. Return ONLY raw complete Python 3 source \u2014 no Markdown fences, no prose, no comments.\n\nOUTPUT FORMAT: this judge compares printed output EXACTLY, token for token, even when the statement promises a numeric tolerance, and the hidden tests may not use the same precision as the statement's worked examples. Apply BOTH: (1) if an input exactly matches one of the statement's worked examples, print that example's expected output byte-for-byte exactly as the statement shows it; (2) for every other input print floats at a fixed width with format(x, '.12f'). Never pin an input that is not a worked example; compute those normally.", "response": "import sys\n\ntokens = sys.stdin.buffer.read().split()\nx = int(tokens[0])\nfactorial = 1\nn = 1\n\nwhile factorial < x:\n    n += 1\n    factorial *= n\n\nsys.stdout.write(str(n))", "cost_usd": 0.0001203, "tokens_in": 417, "tokens_out": 131, "latency_s": 2.875}, {"task_id": "lcb-abc394_d", "model": "openai/gpt-5.6-luna", "prompt": "You are given a string S consisting of six types of characters: (, ), [, ], <, >.\nA string T is called a colorful bracket sequence if it satisfies the following condition:\n\nIt is possible to turn T into an empty string by repeating the following operation any number of times (possibly zero):\n\n- If there exists a contiguous substring of T that is one of (), [], or <>, choose one such substring and delete it.\n- If the deleted substring was at the beginning or end of T, the remainder becomes the new T.\n- Otherwise, concatenate the part before the deleted substring and the part after the deleted substring, and that becomes the new T.\n\n\nDetermine whether S is a colorful bracket sequence.\n\nInput\n\nThe input is given from Standard Input in the following format:\nS\n\nOutput\n\nIf S is a colorful bracket sequence, print Yes; otherwise, print No.\n\nConstraints\n\n\n- S is a string of length between 1 and 2\\times 10^5, inclusive.\n- S consists of (, ), [, ], <, >.\n\nSample Input 1\n\n([])<>()\n\nSample Output 1\n\nYes\n\nFor S=([])<>(), it is possible to turn it into an empty string by repeating the operation as follows:\n\n- Delete the substring [] from the 2nd to the 3rd character in ([])<>(), then concatenate the parts before and after it. The string becomes ()<>().\n- Delete the substring () from the 1st to the 2nd character in ()<>(). The string becomes <>().\n- Delete the substring <> from the 1st to the 2nd character in <>(). The string becomes ().\n- Delete the substring () from the 1st to the 2nd character in (). The string becomes empty.\n\nThus, S=([])<>() is a colorful bracket sequence, so print Yes.\n\nSample Input 2\n\n([<)]>\n\nSample Output 2\n\nNo\n\nSince S=([<)]> does not contain (), [], or <> as a contiguous substring, we cannot perform the 1st operation, and in particular S is not a colorful bracket sequence. Therefore, print No.\n\nSample Input 3\n\n())\n\nSample Output 3\n\nNo\n\nIt is impossible to turn S into an empty string by repeating the operations.\nTherefore, S is not a colorful bracket sequence, so print No.\n\nWrite a complete Python 3 program that reads from standard input and writes the answer to standard output. Output ONLY the program source, no prose and no code fences.\n\nSolve this competitive-programming problem in Python 3. Read all of stdin at once with sys.stdin.buffer.read().split() and index across tokens; build output in a list and emit it with one write at the end. Check the constraint block and pick an algorithm whose running time stays under ~1e8 steps at the largest allowed input. Return ONLY raw complete Python 3 source \u2014 no Markdown fences, no prose, no comments.\n\nOUTPUT FORMAT: this judge compares printed output EXACTLY, token for token, even when the statement promises a numeric tolerance, and the hidden tests may not use the same precision as the statement's worked examples. Apply BOTH: (1) if an input exactly matches one of the statement's worked examples, print that example's expected output byte-for-byte exactly as the statement shows it; (2) for every other input print floats at a fixed width with format(x, '.12f'). Never pin an input that is not a worked example; compute those normally.", "response": "import sys\n\ndata = sys.stdin.buffer.read().split()\ns = data[0].decode()\n\npairs = {')': '(', ']': '[', '>': '<'}\nstack = []\nok = True\n\nfor ch in s:\n    if ch in '([{<':\n        stack.append(ch)\n    elif not stack or stack.pop() != pairs[ch]:\n        ok = False\n        break\n\nif stack:\n    ok = False\n\nout = ['Yes\\n' if ok else 'No\\n']\nsys.stdout.write(''.join(out))\n", "cost_usd": 0.00020560000000000006, "tokens_in": 718, "tokens_out": 223, "latency_s": 2.325}, {"task_id": "lcb-abc394_d", "model": "openai/gpt-5.6-luna", "prompt": "You are given a string S consisting of six types of characters: (, ), [, ], <, >.\nA string T is called a colorful bracket sequence if it satisfies the following condition:\n\nIt is possible to turn T into an empty string by repeating the following operation any number of times (possibly zero):\n\n- If there exists a contiguous substring of T that is one of (), [], or <>, choose one such substring and delete it.\n- If the deleted substring was at the beginning or end of T, the remainder becomes the new T.\n- Otherwise, concatenate the part before the deleted substring and the part after the deleted substring, and that becomes the new T.\n\n\nDetermine whether S is a colorful bracket sequence.\n\nInput\n\nThe input is given from Standard Input in the following format:\nS\n\nOutput\n\nIf S is a colorful bracket sequence, print Yes; otherwise, print No.\n\nConstraints\n\n\n- S is a string of length between 1 and 2\\times 10^5, inclusive.\n- S consists of (, ), [, ], <, >.\n\nSample Input 1\n\n([])<>()\n\nSample Output 1\n\nYes\n\nFor S=([])<>(), it is possible to turn it into an empty string by repeating the operation as follows:\n\n- Delete the substring [] from the 2nd to the 3rd character in ([])<>(), then concatenate the parts before and after it. The string becomes ()<>().\n- Delete the substring () from the 1st to the 2nd character in ()<>(). The string becomes <>().\n- Delete the substring <> from the 1st to the 2nd character in <>(). The string becomes ().\n- Delete the substring () from the 1st to the 2nd character in (). The string becomes empty.\n\nThus, S=([])<>() is a colorful bracket sequence, so print Yes.\n\nSample Input 2\n\n([<)]>\n\nSample Output 2\n\nNo\n\nSince S=([<)]> does not contain (), [], or <> as a contiguous substring, we cannot perform the 1st operation, and in particular S is not a colorful bracket sequence. Therefore, print No.\n\nSample Input 3\n\n())\n\nSample Output 3\n\nNo\n\nIt is impossible to turn S into an empty string by repeating the operations.\nTherefore, S is not a colorful bracket sequence, so print No.\n\nWrite a complete Python 3 program that reads from standard input and writes the answer to standard output. Output ONLY the program source, no prose and no code fences.\n\nNow write TWO programs and nothing else. FIRST a REFERENCE solution that is correct by construction and may be arbitrarily slow (brute force / full simulation); do NOT reuse the idea from the fast solution \u2014 it must be able to disagree. SECOND a GENERATOR taking two argv ints (seed, size): call random.seed(seed) and print ONE valid random input in the exact stdin format, obeying every constraint; at size 2 emit the smallest legal input, larger sizes proportionally bigger, with repeats and extreme values. Output exactly two fenced blocks:\n```reference\n<prog>\n```\n```generator\n<prog>\n```", "response": "```reference\nimport sys\nfrom functools import lru_cache\n\nsys.setrecursionlimit(1_000_000)\n\ns = sys.stdin.readline().strip()\npairs = {\")\": \"(\", \"]\": \"[\", \">\": \"<\"}\n\n@lru_cache(maxsize=None)\ndef can_reduce(t):\n    if not t:\n        return True\n\n    for i in range(len(t) - 1):\n        if t[i + 1] in pairs and pairs[t[i + 1]] == t[i]:\n            u = t[:i] + t[i + 2:]\n            if can_reduce(u):\n                return True\n\n    return False\n\nprint(\"Yes\" if can_reduce(s) else \"No\")\n```\n\n```generator\nimport random\nimport sys\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\nif size == 2:\n    n = 1\nelse:\n    n = max(1, min(200000, size - 1))\n\nchars = \"()[]<>\"\npatterns = [\n    \"()\",\n    \"[]\",\n    \"<>\",\n    \"([<>])\",\n    \"([])<>\",\n    \"([<)]>\",\n    \"())\",\n    \"(((())))\",\n]\n\nresult = []\nwhile len(result) < n:\n    if random.random() < 0.35:\n        block = random.choice(patterns)\n        result.extend(block[:min(len(block), n - len(result))])\n    else:\n        result.append(random.choice(chars))\n\nprint(\"\".join(result))\n```", "cost_usd": 0.0005241, "tokens_in": 645, "tokens_out": 766, "latency_s": 7.415}]