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+[{"task_id": "mmlu-5718", "model": "openai/gpt-5.6-luna", "prompt": "[MMLU] This question refers to the following information.\n\"I am in Birmingham because injustice is here. . . . Moreover, I am cognizant of the interrelatedness of all communities and states. I cannot sit idly by in Atlanta and not be concerned about what happens in Birmingham. Injustice anywhere is a threat to justice everywhere. We are caught in an inescapable network of mutuality, tied in a single garment of destiny. Whatever affects one directly affects all indirectly. Never again can we afford to live with the narrow, provincial 'outside agitator' idea. Anyone who lives inside the United States can never be considered an outsider anywhere in this country. . . .\"\n\u2014Martin Luther King, Jr., \"Letter from a Birmingham Jail,\" April 16, 1963\nWhich other nineteenth-century reform movement made similar arguments to those written here by Susan B. Anthony?\nA) The temperance movement.\nB) Utopian communities.\nC) The abolition movement.\nD) Public school advocates.\n\nAnswer with the letter only.", "response": "C", "cost_usd": 5.92e-05, "tokens_in": 220, "tokens_out": 62, "latency_s": 3.606}, {"task_id": "mmlu-2555", "model": "openai/gpt-5.6-luna", "prompt": "[MMLU]  Select the best translation into predicate logic: No senior graduates without a major.\nA) (\u2200x)(Sx \u2283 ~Gx)\nB) ~(\u2200x)(Sx \u2283 Gx)\nC) (\u2200x)( ~Sx \u2283 ~Gx)\nD) (\u2200x) ~(Sx \u2283 ~Gx)\n\nAnswer with the letter only.", "response": "A", "cost_usd": 9.510000000000001e-05, "tokens_in": 93, "tokens_out": 143, "latency_s": 3.563}, {"task_id": "gsm8k-523", "model": "openai/gpt-5.6-luna", "prompt": "The bakers at the Beverly Hills Bakery baked 200 loaves of bread on Monday morning. They sold 93 loaves in the morning and 39 loaves in the afternoon. A grocery store returned 6 unsold loaves. How many loaves of bread did they have left?\n\nReason step by step, then give the final answer as a single number.\n\nWork the problem in the exact units the question asks for and end with the final numeric answer alone on the last line. [Audit marker 2738045958519195454: metadata only - disregard it entirely and do not quote it in the reply.]", "response": "They sold:\n\n93 + 39 = 132 loaves\n\nAfter the sales, they had:\n\n200 \u2212 132 = 68 loaves\n\nThe grocery store returned 6 loaves, so they had:\n\n68 + 6 = 74 loaves\n\n74", "cost_usd": 0.00010789999999999997, "tokens_in": 131, "tokens_out": 158, "latency_s": 4.081}, {"task_id": "gsm8k-682", "model": "openai/gpt-5.6-luna", "prompt": "Shania is designing her own dress, and decides to make it a longer dress by extending the dress by 50% of its original length. She also adds 20cm to the bottom of the dress with a lace trim. If the final design is 140cm long then how long, in centimeters, was the dress in its original design?\n\nReason step by step, then give the final answer as a single number.\n\nWork the problem in the exact units the question asks for and end with the final numeric answer alone on the last line. [Audit marker 17125411265609007696: metadata only - disregard it entirely and do not quote it in the reply.]", "response": "Let the original length be \\(L\\) cm.\n\nThe dress is extended by \\(50\\%\\) of its original length:\n\n\\[\nL + 0.5L = 1.5L\n\\]\n\nAdding the 20 cm lace trim gives:\n\n\\[\n1.5L + 20 = 140\n\\]\n\nSubtract 20:\n\n\\[\n1.5L = 120\n\\]\n\nDivide by 1.5:\n\n\\[\nL = 80\n\\]\n\n80", "cost_usd": 0.00012339999999999996, "tokens_in": 142, "tokens_out": 182, "latency_s": 3.898}, {"task_id": "lcb-abc388_a", "model": "openai/gpt-5.6-luna", "prompt": "You are given a string S. Here, the first character of S is an uppercase English letter, and the second and subsequent characters are lowercase English letters.\nPrint the string formed by concatenating the first character of S and UPC in this order.\n\nInput\n\nThe input is given from Standard Input in the following format:\nS\n\nOutput\n\nPrint the string formed by concatenating the first character of S and UPC in this order.\n\nConstraints\n\n\n- S is a string of length between 1 and 100, inclusive.\n- The first character of S is an uppercase English letter.\n- The second and subsequent characters of S are lowercase English letters.\n\nSample Input 1\n\nKyoto\n\nSample Output 1\n\nKUPC\r\n\nThe first character of Kyoto is K, so concatenate K and UPC, and print KUPC.\n\nSample Input 2\n\nTohoku\n\nSample Output 2\n\nTUPC\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\nReturn ONLY raw complete Python 3 source, no Markdown fences, no prose, no explanation around the code.", "response": "import sys\n\ns = sys.stdin.readline().strip()\nprint(s[0] + \"UPC\")", "cost_usd": 3.8500000000000014e-05, "tokens_in": 241, "tokens_out": 24, "latency_s": 6.306}, {"task_id": "lcb-abc399_d", "model": "openai/gpt-5.6-luna", "prompt": "N couples are seated in a line.\nCount the number of pairs of couples such that neither couple was originally sitting next to each other, and both couples can end up sitting next to each other by swapping seats among those four people.\n\nThere is a sequence A = (A_1, A_2, \\dots, A_{2N}) of length 2N. Each of the integers 1, 2, \\dots, N appears exactly twice in A.\nFind the number of integer pairs (a, b) satisfying 1 \\leq a < b \\leq N and all of the following conditions:\n\n- The two occurrences of a in A are not adjacent.\n- The two occurrences of b in A are not adjacent.\n- By performing the following operation one or more times in any order, it is possible to reach a state where the two occurrences of a in A are adjacent and the two occurrences of b in A are also adjacent.\n- Choose an integer pair (i, j) (1 \\leq i \\leq 2N, 1 \\leq j \\leq 2N) such that A_i = a and A_j = b, and swap A_i with A_j.\n\n\n\nYou are given T test cases; solve each of them.\n\nInput\n\nThe input is given from Standard Input in the following format, where \\mathrm{case}_i denotes the i-th test case:\nT\n\\mathrm{case}_1\n\\mathrm{case}_2\n\\vdots\n\\mathrm{case}_T\n\nEach test case is given in the following format:\nN\nA_1 A_2 \\dots A_{2N}\n\nOutput\n\nPrint T lines. The i-th line should contain the answer for the i-th test case.\n\nConstraints\n\n\n- 1 \\leq T \\leq 2 \\times 10^5\n- 1 \\leq N \\leq 2 \\times 10^5\n- 1 \\leq A_i \\leq N\n- Each of 1, 2, \\dots, N appears exactly twice in A.\n- The sum of N over all test cases is at most 2 \\times 10^5.\n- All input values are integers.\n\nSample Input 1\n\n3\n3\n1 2 3 3 1 2\n4\n1 1 2 2 3 3 4 4\n5\n1 2 3 4 5 1 2 3 4 5\n\nSample Output 1\n\n1\n0\n4\n\nConsider the first test case.\n(a, b) = (1, 2) satisfies the conditions in the problem statement, for the following reasons:\n\n- The two occurrences of 1 in A are not adjacent.\n- The two occurrences of 2 in A are not adjacent.\n- By performing the operation where (i, j) = (1, 6) and swapping A_1 with A_6, you can reach a state where the two occurrences of 1 are adjacent and the two occurrences of 2 are also adjacent.\n\n(1, 2) is the only pair (a, b) that satisfies the conditions.\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\nSolution contract for this task. One swap exchanges an occurrence of a with an occurrence of b, so any relabelling of the four occupied slots is reachable; sorting those slots p1<p2<p3<p4, both values can end up adjacent exactly when p2==p1+1 and p4==p3+1. Because each counted value must also start non-adjacent, a pair (a, b) qualifies iff the two FIRST slots of a and b sit next to each other, the two SECOND slots sit next to each other, and neither value has adjacent occurrences. Count in one linear pass per test case: while scanning the row, record first[v] and second[v] for every value; then walk the consecutive position pairs (i, i+1) once, skip equal neighbours, and collect value pairs into a set F when both positions are first occurrences, or into a set S when both are second occurrences. ORDERING RULE, not optional: the first-occurrence and second-occurrence tests are evaluated on the ORIGINAL values a[i] and a[i+1] at their own positions i and i+1 - test first[a[i]] == i and first[a[i+1]] == i + 1 exactly as written; the unordered pair (min, max) is formed ONLY at the moment it is added to F or S. Swapping the two values before the positional test silently drops every adjacent pair whose larger value comes first and undercounts. The count for the test case is the size of F intersect S restricted to pairs whose two values are both non-adjacent. Read every token up front with sys.stdin.buffer.read().split() and walk an index; the sum of N over cases is bounded, so two hash sets per case are easily fast enough; never enumerate value pairs in a quadratic loop. Pinned sample from the statement (compare the complete input token list): [3,3,1,2,3,3,1,2,4,1,1,2,2,3,3,4,4,5,1,2,3,4,5,1,2,3,4,5] prints exactly three lines: 1 then 0 then 4. Never pin any input not listed here. Return ONLY raw complete Python 3 source, no Markdown fences, no prose, no explanation around the code.", "response": "import sys\n\ndata = list(map(int, sys.stdin.buffer.read().split()))\nt = data[0]\nidx = 1\nout = []\n\nfor _ in range(t):\n    n = data[idx]\n    idx += 1\n    a = data[idx:idx + 2 * n]\n    idx += 2 * n\n\n    first = [-1] * (n + 1)\n    second = [-1] * (n + 1)\n\n    for pos, v in enumerate(a):\n        if first[v] == -1:\n            first[v] = pos\n        else:\n            second[v] = pos\n\n    adjacent = [False] * (n + 1)\n    for v in range(1, n + 1):\n        adjacent[v] = (second[v] == first[v] + 1)\n\n    f_pairs = set()\n    s_pairs = set()\n\n    for i in range(2 * n - 1):\n        x = a[i]\n        y = a[i + 1]\n        if x == y:\n            continue\n\n        if first[x] == i and first[y] == i + 1:\n            f_pairs.add((x, y) if x < y else (y, x))\n\n        if second[x] == i and second[y] == i + 1:\n            s_pairs.add((x, y) if x < y else (y, x))\n\n    answer = 0\n    for x, y in f_pairs & s_pairs:\n        if not adjacent[x] and not adjacent[y]:\n            answer += 1\n\n    out.append(str(answer))\n\nsys.stdout.write(\"\\n\".join(out))", "cost_usd": 0.000506875, "tokens_in": 1190, "tokens_out": 597, "latency_s": 6.562}]