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+[{"task_id": "mmlu-4925", "model": "openai/gpt-5.6-luna", "prompt": "[MMLU] Which of the following is primarily a chemical sense?\nA) Touch\nB) Vision\nC) Taste\nD) Hearing\n\nAnswer with the letter only.", "response": "C", "cost_usd": 7.2e-06, "tokens_in": 42, "tokens_out": 5, "latency_s": 1.406}, {"task_id": "mmlu-10279", "model": "openai/gpt-5.6-luna", "prompt": "[MMLU] When did the Inca begin their military expansion across South America?\nA) 200 A.D.\nB) 900 A.D.\nC) 1450 A.D.\nD) 1650 A.D.\n\nAnswer with the letter only.", "response": "B", "cost_usd": 8.9e-06, "tokens_in": 59, "tokens_out": 5, "latency_s": 0.751}, {"task_id": "gsm8k-808", "model": "openai/gpt-5.6-luna", "prompt": "There are 10 quarts of tea left from the party. Four students each drank 1.5 quarts of tea and 16 students each drank 2 quarts of tea. How many gallons of tea were at the beginning of the party?\n\nReason step by step, then give the final answer as a single number.\n\nSolve in the requested units and put only the final numeric result on the last line. Audit marker 9208378043942665589 is answer-independent metadata; do not reproduce it.", "response": "Four students drank:\n\\(4 \\times 1.5 = 6\\) quarts\n\nSixteen students drank:\n\\(16 \\times 2 = 32\\) quarts\n\nTotal consumed:\n\\(6 + 32 = 38\\) quarts\n\nTea at the beginning:\n\\(38 + 10 = 48\\) quarts\n\nConvert to gallons:\n\\(48 \\div 4 = 12\\) gallons\n\n12", "cost_usd": 0.0001045, "tokens_in": 109, "tokens_out": 156, "latency_s": 3.161}, {"task_id": "gsm8k-1291", "model": "openai/gpt-5.6-luna", "prompt": "Timothy has three times the number of pens that George owns. Sarah has half as many pens as Timothy. If George has 18 pens, how many pens does Sarah have?\n\nReason step by step, then give the final answer as a single number.\n\nSolve in the requested units and put only the final numeric result on the last line. Audit marker 3704499846150824406 is answer-independent metadata; do not reproduce it.", "response": "George has 18 pens.\n\nTimothy has three times as many:\n\\(3 \\times 18 = 54\\)\n\nSarah has half as many as Timothy:\n\\(54 \\div 2 = 27\\)\n\n27", "cost_usd": 6.520000000000001e-05, "tokens_in": 94, "tokens_out": 93, "latency_s": 2.688}, {"task_id": "lcb-abc398_a", "model": "openai/gpt-5.6-luna", "prompt": "Find a length-N string that satisfies all of the following conditions:\n\n- Each character is - or =.\n- It is a palindrome.\n- It contains exactly one or exactly two =s. If it contains two =s, they are adjacent.\n\nSuch a string is unique.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\n\nOutput\n\nPrint the answer.\n\nConstraints\n\n\n- 1 \\leq N \\leq 100\n- N is an integer.\n\nSample Input 1\n\n4\n\nSample Output 1\n\n-==-\n\nSample Input 2\n\n7\n\nSample Output 2\n\n---=---\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 a complete solution from the full statement and constraints. Privately create a small-state specification and use it to challenge your algorithm. Check boundaries, ordering, multiplicity, repeated values, overflow, and complexity. Trace every published example and return only complete raw Python 3 source with no Markdown or prose.", "response": "n = int(input())\nk = 1 if n % 2 else 2\nd = (n - k) // 2\nprint(\"-\" * d + \"=\" * k + \"-\" * d)\n", "cost_usd": 0.0001339, "tokens_in": 229, "tokens_out": 185, "latency_s": 2.258}, {"task_id": "lcb-abc390_d", "model": "openai/gpt-5.6-luna", "prompt": "There are N bags, labeled bag 1, bag 2, \\ldots, bag N.\r\nBag i (1 \\leq i \\leq N) contains A_i stones.\nTakahashi can perform the following operation any number of times, possibly zero:\n\nChoose two bags A and B, and move all stones from bag A into bag B.\n\nFind the number of different possible values for the following after repeating the operation.\n\n- B_1 \\oplus B_2 \\oplus \\cdots \\oplus B_N, where B_i is the final number of stones in bag i.\r\nHere, \\oplus denotes bitwise XOR.\n\n\nAbout bitwise XOR\r\nFor non-negative integers a and b, the bitwise XOR a \\oplus b is defined as follows:\r\n\r\n\r\nIn the binary representation of a \\oplus b, the digit in the 2^k place (k \\ge 0) is 1 if and only if exactly one of the digits in the 2^k place of a and b is 1; otherwise, it is 0.\r\n\r\n\r\nFor example, 3 \\oplus 5 = 6 (in binary, 011 \\oplus 101 = 110).\r\nIn general, for k non-negative integers x_1, x_2, \\ldots, x_k, their bitwise XOR x_1 \\oplus x_2 \\oplus \\cdots \\oplus x_k is defined as (\\cdots((x_1 \\oplus x_2) \\oplus x_3) \\oplus \\cdots) \\oplus x_k, which does not depend on the order of x_1, x_2, \\ldots, x_k.\r\n\nIt can be proved that under the constraints of this problem, the number of possible values is finite.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\r\nA_1 A_2 \\ldots A_N\n\nOutput\n\nPrint the number of different possible values for B_1 \\oplus B_2 \\oplus \\cdots \\oplus B_N after repeating the operation.\n\nConstraints\n\n\n- 2 \\leq N \\leq 12\n- 1 \\leq A_i \\leq 10^{17}\n- All input values are integers.\n\nSample Input 1\n\n3\r\n2 5 7\n\nSample Output 1\n\n3\r\n\nFor example, if Takahashi chooses bags 1 and 3 for the operation, then the numbers of stones in bags 1, 2, 3 become 0, 5, 9.\r\nIf he stops at this point, the XOR is 0 \\oplus 5 \\oplus 9 = 12.\nThe other possible XOR values after repeating the operation are 0 and 14.\r\nTherefore, the possible values are 0, 12, 14; there are three values, so the output is 3.\n\nSample Input 2\n\n2\r\n100000000000000000 100000000000000000\n\nSample Output 2\n\n2\n\nSample Input 3\n\n6\r\n71 74 45 34 31 60\n\nSample Output 3\n\n84\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 a complete solution from the full statement and constraints. Privately create a small-state specification and use it to challenge your algorithm. Check boundaries, ordering, multiplicity, repeated values, overflow, and complexity. Trace every published example and return only complete raw Python 3 source with no Markdown or prose.", "response": "import sys\nfrom array import array\n\ndata = list(map(int, sys.stdin.buffer.read().split()))\nn = data[0]\nvalues = data[1:]\n\nbell = [1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597]\nrequired = bell[n] * 17 // 10 + 1\ntable_bits = required.bit_length()\ntable_size = 1 << table_bits\ntable_mask = table_size - 1\n\ntable = array('Q', [0]) * table_size\nblocks = [0] * n\ncount = [0]\n\nmul = 11400714819323198485\nfull_mask = (1 << 64) - 1\nshift = 64 - table_bits\n\ndef dfs(i, block_count, current_xor,\n        values=values, blocks=blocks, table=table, count=count,\n        n=n, mul=mul, full_mask=full_mask, shift=shift,\n        table_mask=table_mask):\n    if i == n:\n        key = current_xor + 1\n        index = ((current_xor * mul) & full_mask) >> shift\n\n        while True:\n            current = table[index]\n            if current == 0:\n                table[index] = key\n                count[0] += 1\n                return\n            if current == key:\n                return\n            index = (index + 1) & table_mask\n\n    value = values[i]\n    next_i = i + 1\n\n    for j in range(block_count):\n        old_sum = blocks[j]\n        new_sum = old_sum + value\n        blocks[j] = new_sum\n        dfs(next_i, block_count, current_xor ^ old_sum ^ new_sum)\n        blocks[j] = old_sum\n\n    blocks[block_count] = value\n    dfs(next_i, block_count + 1, current_xor ^ value)\n\nblocks[0] = values[0]\ndfs(1, 1, values[0])\n\nprint(count[0])", "cost_usd": 0.0068728, "tokens_in": 766, "tokens_out": 11327, "latency_s": 101.596}, {"task_id": "lcb-abc390_d", "model": "google/gemini-3.6-flash", "prompt": "There are N bags, labeled bag 1, bag 2, \\ldots, bag N.\r\nBag i (1 \\leq i \\leq N) contains A_i stones.\nTakahashi can perform the following operation any number of times, possibly zero:\n\nChoose two bags A and B, and move all stones from bag A into bag B.\n\nFind the number of different possible values for the following after repeating the operation.\n\n- B_1 \\oplus B_2 \\oplus \\cdots \\oplus B_N, where B_i is the final number of stones in bag i.\r\nHere, \\oplus denotes bitwise XOR.\n\n\nAbout bitwise XOR\r\nFor non-negative integers a and b, the bitwise XOR a \\oplus b is defined as follows:\r\n\r\n\r\nIn the binary representation of a \\oplus b, the digit in the 2^k place (k \\ge 0) is 1 if and only if exactly one of the digits in the 2^k place of a and b is 1; otherwise, it is 0.\r\n\r\n\r\nFor example, 3 \\oplus 5 = 6 (in binary, 011 \\oplus 101 = 110).\r\nIn general, for k non-negative integers x_1, x_2, \\ldots, x_k, their bitwise XOR x_1 \\oplus x_2 \\oplus \\cdots \\oplus x_k is defined as (\\cdots((x_1 \\oplus x_2) \\oplus x_3) \\oplus \\cdots) \\oplus x_k, which does not depend on the order of x_1, x_2, \\ldots, x_k.\r\n\nIt can be proved that under the constraints of this problem, the number of possible values is finite.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\r\nA_1 A_2 \\ldots A_N\n\nOutput\n\nPrint the number of different possible values for B_1 \\oplus B_2 \\oplus \\cdots \\oplus B_N after repeating the operation.\n\nConstraints\n\n\n- 2 \\leq N \\leq 12\n- 1 \\leq A_i \\leq 10^{17}\n- All input values are integers.\n\nSample Input 1\n\n3\r\n2 5 7\n\nSample Output 1\n\n3\r\n\nFor example, if Takahashi chooses bags 1 and 3 for the operation, then the numbers of stones in bags 1, 2, 3 become 0, 5, 9.\r\nIf he stops at this point, the XOR is 0 \\oplus 5 \\oplus 9 = 12.\nThe other possible XOR values after repeating the operation are 0 and 14.\r\nTherefore, the possible values are 0, 12, 14; there are three values, so the output is 3.\n\nSample Input 2\n\n2\r\n100000000000000000 100000000000000000\n\nSample Output 2\n\n2\n\nSample Input 3\n\n6\r\n71 74 45 34 31 60\n\nSample Output 3\n\n84\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 without seeing another solution. Return one raw executable Python program derived by a structurally different route. Oracle must be a simple correctness-first program for small legal inputs and must not reuse the candidate algorithm. Generator must accept seed and size arguments, seed Python random, and print one varied legal input. Append oracle lines between comments `# ORACLE-BEGIN` and `# ORACLE-END`, prefixing every line with `#|`. Append generator lines the same way between `# GENERATOR-BEGIN` and `# GENERATOR-END`. Those comment sections must be inert when the complete response is executed as the candidate.", "response": "import sys\n\ndef main():\n    input = sys.stdin.read\n    data = input().split()\n    if not data:\n        return\n    N = int(data[0])\n    A = [int(x) for x in data[1:N+1]]\n\n    sum_mask = [0] * (1 << N)\n    for i in range(N):\n        bit = 1 << i\n        val = A[i]\n        for mask in range(bit):\n            sum_mask[mask | bit] = sum_mask[mask] + val\n\n    f = [None] * (1 << N)\n    f[0] = (0,)\n\n    for S in range(1, 1 << N):\n        b = S & -S\n        R = S ^ b\n        rem = R\n        f_S = set()\n        add = f_S.add\n        while True:\n            val = sum_mask[S ^ rem]\n            for x in f[rem]:\n                add(val ^ x)\n            if not rem:\n                break\n            rem = (rem - 1) & R\n        f[S] = tuple(f_S)\n\n    print(len(f[(1 << N) - 1]))\n\nif __name__ == '__main__':\n    main()\n\n# ORACLE-BEGIN\n#| import sys\n#|\n#| def get_partitions(elements):\n#|     if not elements:\n#|         yield []\n#|         return\n#|     first = elements[0]\n#|     for p in get_partitions(elements[1:]):\n#|         yield [[first]] + p\n#|         for i in range(len(p)):\n#|             yield p[:i] + [p[i] + [first]] + p[i+1:]\n#|\n#| def main():\n#|     input = sys.stdin.read\n#|     data = input().split()\n#|     if not data:\n#|         return\n#|     N = int(data[0])\n#|     A = [int(x) for x in data[1:N+1]]\n#|     \n#|     seen = set()\n#|     for partition in get_partitions(A):\n#|         xor_sum = 0\n#|         for block in partition:\n#|             xor_sum ^= sum(block)\n#|         seen.add(xor_sum)\n#|     print(len(seen))\n#|\n#| if __name__ == '__main__':\n#|     main()\n# ORACLE-END\n\n# GENERATOR-BEGIN\n#| import sys\n#| import random\n#|\n#| def main():\n#|     seed = int(sys.argv[1]) if len(sys.argv) > 1 else 42\n#|     size = int(sys.argv[2]) if len(sys.argv) > 2 else 5\n#|     random.seed(seed)\n#|     \n#|     N = max(2, min(12, size))\n#|     A = [random.randint(1, 10**17) for _ in range(N)]\n#|     \n#|     print(N)\n#|     print(*(A))\n#|\n#| if __name__ == '__main__':\n#|     main()\n# GENERATOR-END", "cost_usd": 0.039576, "tokens_in": 858, "tokens_out": 10382, "latency_s": 56.035}]