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+[{"task_id": "mmlu-2694", "model": "openai/gpt-5.6-luna", "prompt": "[MMLU] Which of the following statements best describes what a Barr body is and its significance?\nA) It is an inactivated Y chromosome and results in a man being sterile.\nB) It is an inactivated Y chromosome, and the person who has it appears female.\nC) It is an inactivated X chromosome and results in females with half their cells having one X inactivated and the other half of their cells having the other X inactivated.\nD) It is an inactivated X chromosome and results in females who are sterile.\n\nAnswer with the letter only.", "response": "C", "cost_usd": 1.51e-05, "tokens_in": 121, "tokens_out": 5, "latency_s": 1.673}, {"task_id": "mmlu-1521", "model": "openai/gpt-5.6-luna", "prompt": "[MMLU] Applications developed by programming languages like ____ and ______ have this common buffer-overflow error.\nA) C, Ruby\nB) Python, Ruby\nC) C, C++\nD) Tcl, C#\n\nAnswer with the letter only.", "response": "C", "cost_usd": 8.7e-06, "tokens_in": 57, "tokens_out": 5, "latency_s": 0.945}, {"task_id": "gsm8k-195", "model": "openai/gpt-5.6-luna", "prompt": "A pen costs as much as a pencil and eraser combined. A pencil costs $1.20 and an eraser costs $0.30. How much will 8 pens cost?\n\nReason step by step, then give the final answer as a single number.\n\nSolve in the requested units and privately verify arithmetic, signs, rounding, and boundary assumptions. Put only the final numeric result on the last line.", "response": "A pen costs the same as a pencil and eraser combined:\n\n$1.20 + $0.30 = $1.50\n\nFor 8 pens:\n\n8 \u00d7 $1.50 = $12.00\n\n12.00", "cost_usd": 6.94e-05, "tokens_in": 88, "tokens_out": 101, "latency_s": 2.369}, {"task_id": "gsm8k-863", "model": "openai/gpt-5.6-luna", "prompt": "Gretchen has some coins. There are 30 more gold coins than silver coins. If she had 70 gold coins, how many coins did Gretchen have in total?\n\nReason step by step, then give the final answer as a single number.\n\nSolve in the requested units and privately verify arithmetic, signs, rounding, and boundary assumptions. Put only the final numeric result on the last line.", "response": "Let the number of silver coins be \\(70 - 30 = 40\\).\n\nTotal coins: \\(70 + 40 = 110\\).\n\n110", "cost_usd": 5.3e-05, "tokens_in": 86, "tokens_out": 74, "latency_s": 2.827}, {"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\nDerive the algorithm from the complete specification and maximum constraints. Privately construct a small direct specification and try to falsify the proposed algorithm. Check ordering, multiplicity, repeated values, boundaries, state changes, and complexity. Trace every published example. The execution harness compares output tokens exactly even when the problem prose grants mathematical tolerance. Match the required canonical representation, precision, rounding, ordering, and separators shown by the statement and examples. Do not print extra precision merely because it is available. Return only one complete raw Python 3 program without Markdown, fences, or explanation.", "response": "import sys\n\ns = sys.stdin.readline().strip()\nprint(s[0] + \"UPC\")", "cost_usd": 4.7799999999999996e-05, "tokens_in": 334, "tokens_out": 24, "latency_s": 1.465}, {"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\nWork independently from the statement and do not assume any candidate implementation. Write a correctness-first oracle for small legal inputs using direct simulation or exhaustive search. Also write a generator accepting seed and size arguments, seeding Python random, and printing one varied legal input. Systematically include legal ties and equalities, repeated values, minimum counts, endpoints, mandatory final transitions, and choices where an equal local action changes a later opportunity. Vary these boundary families from the seed instead of emitting a fixed example. The oracle must not reuse the efficient algorithm requested by the statement. The execution harness compares output tokens exactly even when the prose permits numerical tolerance. Reproduce every published output token exactly and infer one canonical representation for unseen outputs from the statement and examples without inventing extra precision. Return exactly two fenced blocks named oracle and generator, with no other text.", "response": "```oracle\nimport sys\n\ns = sys.stdin.readline().strip()\nprint(s[0] + \"UPC\")\n```\n\n```generator\nimport random\nimport string\nimport sys\n\nseed = int(sys.argv[1]) if len(sys.argv) > 1 else 0\nsize = int(sys.argv[2]) if len(sys.argv) > 2 else 1\n\nrandom.seed(seed)\nsize = max(1, min(100, size))\n\nfirst = random.choice(string.ascii_uppercase)\n\nif size == 1:\n    print(first)\n    sys.exit()\n\nfamilies = [\n    lambda n: \"a\" * n,\n    lambda n: \"z\" * n,\n    lambda n: (\"ab\" * ((n + 1) // 2))[:n],\n    lambda n: (\"zy\" * ((n + 1) // 2))[:n],\n    lambda n: \"\".join(random.choice(string.ascii_lowercase) for _ in range(n)),\n]\n\nfamily = families[seed % len(families)]\nprint(first + family(size - 1))\n```", "cost_usd": 0.00033750000000000007, "tokens_in": 387, "tokens_out": 498, "latency_s": 6.227}, {"task_id": "lcb-abc399_c", "model": "openai/gpt-5.6-luna", "prompt": "You are given a simple undirected graph with N vertices and M edges, where the vertices are labeled 1 to N. The i-th edge connects vertices u_i and v_i.\r\nWhat is the minimum number of edges that need to be deleted from this graph so that the graph becomes a forest?\n\nWhat is a forest?\r\n\r\nA simple undirected graph F is called a forest if and only if F does not contain any cycle.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN M\r\nu_1 v_1\r\nu_2 v_2\r\n\\vdots\r\nu_M v_M\n\nOutput\n\nPrint the answer.\n\nConstraints\n\n\n- 1 \\leq N \\leq 2 \\times 10^5\n- 0 \\leq M \\leq \\min \\left( \\frac{N(N-1)}{2}, 2 \\times 10^5\\right)\n- 1 \\leq u_i < v_i \\leq N\n- The given graph is simple.\n- All input values are integers.\n\nSample Input 1\n\n4 4\r\n1 2\r\n1 3\r\n2 4\r\n3 4\n\nSample Output 1\n\n1\r\n\nFor example, if you delete the first edge, the graph becomes a forest.\n\nSample Input 2\n\n5 0\n\nSample Output 2\n\n0\n\nSample Input 3\n\n10 10\r\n7 9\r\n4 6\r\n6 10\r\n2 5\r\n5 6\r\n5 9\r\n6 8\r\n4 8\r\n1 5\r\n1 4\n\nSample Output 3\n\n2\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\nDerive the algorithm from the complete specification and maximum constraints. Privately construct a small direct specification and try to falsify the proposed algorithm. Check ordering, multiplicity, repeated values, boundaries, state changes, and complexity. Trace every published example. The execution harness compares output tokens exactly even when the problem prose grants mathematical tolerance. Match the required canonical representation, precision, rounding, ordering, and separators shown by the statement and examples. Do not print extra precision merely because it is available. Return only one complete raw Python 3 program without Markdown, fences, or explanation.", "response": "import sys\n\ndef main():\n    data = list(map(int, sys.stdin.buffer.read().split()))\n    if not data:\n        return\n\n    n, m = data[0], data[1]\n    parent = list(range(n + 1))\n    size = [1] * (n + 1)\n\n    def find(x):\n        while parent[x] != x:\n            parent[x] = parent[parent[x]]\n            x = parent[x]\n        return x\n\n    used = 0\n    idx = 2\n    for _ in range(m):\n        u = data[idx]\n        v = data[idx + 1]\n        idx += 2\n\n        ru = find(u)\n        rv = find(v)\n        if ru != rv:\n            if size[ru] < size[rv]:\n                ru, rv = rv, ru\n            parent[rv] = ru\n            size[ru] += size[rv]\n            used += 1\n\n    print(m - used)\n\nif __name__ == \"__main__\":\n    main()\n", "cost_usd": 0.00024030000000000004, "tokens_in": 495, "tokens_out": 318, "latency_s": 3.503}, {"task_id": "lcb-abc399_c", "model": "openai/gpt-5.6-luna", "prompt": "You are given a simple undirected graph with N vertices and M edges, where the vertices are labeled 1 to N. The i-th edge connects vertices u_i and v_i.\r\nWhat is the minimum number of edges that need to be deleted from this graph so that the graph becomes a forest?\n\nWhat is a forest?\r\n\r\nA simple undirected graph F is called a forest if and only if F does not contain any cycle.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN M\r\nu_1 v_1\r\nu_2 v_2\r\n\\vdots\r\nu_M v_M\n\nOutput\n\nPrint the answer.\n\nConstraints\n\n\n- 1 \\leq N \\leq 2 \\times 10^5\n- 0 \\leq M \\leq \\min \\left( \\frac{N(N-1)}{2}, 2 \\times 10^5\\right)\n- 1 \\leq u_i < v_i \\leq N\n- The given graph is simple.\n- All input values are integers.\n\nSample Input 1\n\n4 4\r\n1 2\r\n1 3\r\n2 4\r\n3 4\n\nSample Output 1\n\n1\r\n\nFor example, if you delete the first edge, the graph becomes a forest.\n\nSample Input 2\n\n5 0\n\nSample Output 2\n\n0\n\nSample Input 3\n\n10 10\r\n7 9\r\n4 6\r\n6 10\r\n2 5\r\n5 6\r\n5 9\r\n6 8\r\n4 8\r\n1 5\r\n1 4\n\nSample Output 3\n\n2\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\nWork independently from the statement and do not assume any candidate implementation. Write a correctness-first oracle for small legal inputs using direct simulation or exhaustive search. Also write a generator accepting seed and size arguments, seeding Python random, and printing one varied legal input. Systematically include legal ties and equalities, repeated values, minimum counts, endpoints, mandatory final transitions, and choices where an equal local action changes a later opportunity. Vary these boundary families from the seed instead of emitting a fixed example. The oracle must not reuse the efficient algorithm requested by the statement. The execution harness compares output tokens exactly even when the prose permits numerical tolerance. Reproduce every published output token exactly and infer one canonical representation for unseen outputs from the statement and examples without inventing extra precision. Return exactly two fenced blocks named oracle and generator, with no other text.", "response": "```oracle\nimport sys\nfrom itertools import combinations\n\ndef is_forest(n, edges, mask):\n    parent = list(range(n))\n    size = [1] * n\n\n    def find(x):\n        while parent[x] != x:\n            parent[x] = parent[parent[x]]\n            x = parent[x]\n        return x\n\n    for i, (u, v) in enumerate(edges):\n        if not (mask >> i) & 1:\n            continue\n        ru, rv = find(u), find(v)\n        if ru == rv:\n            return False\n        if size[ru] < size[rv]:\n            ru, rv = rv, ru\n        parent[rv] = ru\n        size[ru] += size[rv]\n    return True\n\ndef solve():\n    data = list(map(int, sys.stdin.buffer.read().split()))\n    if not data:\n        return\n    n, m = data[0], data[1]\n    edges = []\n    p = 2\n    for _ in range(m):\n        edges.append((data[p] - 1, data[p + 1] - 1))\n        p += 2\n\n    best_kept = 0\n    for mask in range(1 << m):\n        kept = mask.bit_count()\n        if kept <= best_kept:\n            continue\n        if is_forest(n, edges, mask):\n            best_kept = kept\n\n    print(m - best_kept)\n\nif __name__ == \"__main__\":\n    solve()\n```\n\n```generator\nimport sys\nimport random\n\ndef main():\n    if len(sys.argv) < 3:\n        raise SystemExit(\"usage: generator SEED SIZE\")\n\n    seed = int(sys.argv[1])\n    size = max(1, int(sys.argv[2]))\n    random.seed(seed)\n\n    n = size\n    max_m = min(200000, n * (n - 1) // 2)\n\n    family = random.randrange(7)\n    all_edges = [(u, v) for u in range(1, n + 1) for v in range(u + 1, n + 1)]\n\n    if family == 0 or n == 1:\n        m = 0\n        edges = []\n    elif family == 1:\n        m = min(max_m, max(0, n - 1))\n        edges = [(i, i + 1) for i in range(1, n)]\n        random.shuffle(edges)\n        edges = edges[:m]\n    elif family == 2:\n        m = min(max_m, n)\n        edges = [(i, i + 1) for i in range(1, n)]\n        if n >= 3:\n            edges.append((1, n))\n        random.shuffle(edges)\n        edges = edges[:m]\n    elif family == 3:\n        m = min(max_m, max(0, 2 * n - 1))\n        edges = random.sample(all_edges, m)\n    elif family == 4:\n        m = max_m\n        edges = all_edges\n    elif family == 5:\n        m = random.randrange(max_m + 1)\n        edges = random.sample(all_edges, m)\n    else:\n        if max_m == 0:\n            m = 0\n            edges = []\n        else:\n            targets = [0, 1, max_m, max(0, n - 1), min(max_m, n),\n                       random.randrange(max_m + 1)]\n            m = targets[random.randrange(len(targets))]\n            edges = random.sample(all_edges, m)\n\n    print(n, len(edges))\n    for u, v in edges:\n        print(u, v)\n\nif __name__ == \"__main__\":\n    main()\n```", "cost_usd": 0.0007592000000000001, "tokens_in": 548, "tokens_out": 1174, "latency_s": 195.703}]