| {"schema": 2, "epoch": 87594, "nonce": "856ff8f4f21e3678", "hotkey": "5CaXH581GtSjxFaFaJzSN35CyKzvFMQhRAxrNy6qmoztNiPz", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "19aafcfeb694bb06e7fbbedbfd7fce6c50bd3389a57eeefd1b5d2e5475afdd62", "model_id": "router", "total_cost_usd": 0.00296986, "n_calls": 6, "call_log_hash": "17fd1e32cdd50b83f2eb3418c90a2c49688b9914e3fb291896b5c14294ca61c1", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 153.273, "tokens_in": 1254, "tokens_out": 20552, "results": [{"benchmark": "mmlu", "task_id": "mmlu-6570", "answer": "D", "cost_usd": 4.63e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.00046, 2.3e-05, 2.2e-05, 3.1e-05, 0.999384, 6.3e-05, 1.8e-05]}, {"benchmark": "mmlu", "task_id": "mmlu-8674", "answer": "D", "cost_usd": 4.477000000000001e-05, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.905042, 0.00043, 0.000323, 0.000481, 0.092483, 0.000897, 0.000343]}, {"benchmark": "math", "task_id": "gsm8k-954", "answer": "He has 24 measuring cups.\n\nMeasuring spoons: \\( \\frac{2}{3} \\times 24 = 16\\).\n\nTotal utensils initially: \\(24 + 16 = 40\\).\n\nAfter gifting 6 spoons: \\(40 - 6 = 34\\).\n\n34", "cost_usd": 8.320000000000002e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.089758, 0.000185, 0.00014, 0.000232, 0.909051, 0.000504, 0.00013]}, {"benchmark": "math", "task_id": "gsm8k-505", "answer": "1. A 2 \u00d7 4 costs $4 and makes 200 sticks: \n \\(200 \\div 4 = 50\\) sticks per dollar.\n\n2. A 2 \u00d7 8 costs $6 and makes 400 sticks: \n \\(400 \\div 6 \\approx 66.7\\) sticks per dollar.\n\n3. The 2 \u00d7 8 gives more sticks per dollar, so buy as many 2 \u00d7 8 pieces as possible.\n\n4. With $24: \n \\(24 \\div 6 = 4\\) pieces.\n\n5. Total sticks: \n \\(4 \\times 400 = 1600\\)\n\n1600", "cost_usd": 0.0001603, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.001819, 2.5e-05, 2e-05, 3.6e-05, 0.99801, 7.2e-05, 1.8e-05]}, {"benchmark": "code", "task_id": "lcb-abc389_b", "answer": "import sys\n\nx = int(sys.stdin.readline())\nn = 1\nfactorial = 1\n\nwhile factorial < x:\n n += 1\n factorial *= n\n\nprint(n)", "cost_usd": 4.8099999999999977e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.035223, 7.3e-05, 5.7e-05, 9.3e-05, 0.96432, 0.000186, 4.8e-05]}, {"benchmark": "code", "task_id": "lcb-abc388_c", "answer": "```python\nimport sys\n\ndef main():\n # Read all input from standard input efficiently\n # sys.stdin.read().split() reads the entire input and splits by whitespace\n input_data = sys.stdin.read().split()\n \n if not input_data:\n return\n\n # Create an iterator to process tokens sequentially\n iterator = iter(input_data)\n \n try:\n # The first token is N\n N = int(next(iterator))\n except StopIteration:\n return\n\n # The remaining tokens are the sizes of the mochi A_1, ..., A_N\n # Convert them to integers. The problem guarantees A is sorted in non-decreasing order.\n A = list(map(int, iterator))\n \n # Free memory used by the input string list\n del input_data\n \n # Two pointers approach to count valid pairs\n # 'right' pointer will track the first index that can serve as a bottom mochi\n # for the current top mochi at 'left'.\n right = 0\n count = 0\n n_len = len(A)\n \n # Iterate through each mochi, treating it as the top mochi\n for left in range(n_len):\n # The condition for placing mochi A (size a) on top of mochi B (size b) is a <= b / 2.\n # This is equivalent to b >= 2 * a.\n # Here, A[left] is the size of the top mochi.\n target = 2 * A[left]\n \n # We need to find the first index 'right' such that A[right] >= target.\n # Since the array A is sorted, as 'left' increases, 'target' increases (or stays same),\n # so the valid 'right' index will also move to the right (increase).\n while right < n_len and A[right] < target:\n right += 1\n \n # All mochi from index 'right' to 'n_len - 1' satisfy the condition A[i] >= target.\n # The number of such mochi is n_len - right.\n count += (n_len - right)\n \n print(count)\n\nif __name__ == '__main__':\n main()\n```", "cost_usd": 0.00258719, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.996716, 2.2e-05, 1.6e-05, 2.1e-05, 0.003166, 4.6e-05, 1.3e-05]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "767279dbefd2c2c0ceda8d0b7fc7a9d6d24063f6e3b8819f081e8fd3440638c1", "platform_sig": 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