{"schema": 2, "epoch": 87561, "nonce": "69fa48bfb4c3a8ca", "hotkey": "5CaXH581GtSjxFaFaJzSN35CyKzvFMQhRAxrNy6qmoztNiPz", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "19aafcfeb694bb06e7fbbedbfd7fce6c50bd3389a57eeefd1b5d2e5475afdd62", "model_id": "router", "total_cost_usd": 0.00380885, "n_calls": 6, "call_log_hash": "4a55bf5fb42159fe500d145d44a8874cb5a2a1779f0355598d2f7c6625af576b", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 174.452, "tokens_in": 1514, "tokens_out": 27511, "results": [{"benchmark": "mmlu", "task_id": "mmlu-12371", "answer": "D", "cost_usd": 4.4800000000000005e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.000559, 3.3e-05, 3.3e-05, 4.3e-05, 0.999218, 8.6e-05, 2.7e-05]}, {"benchmark": "mmlu", "task_id": "mmlu-1429", "answer": "A", "cost_usd": 1.4399999999999998e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.267798, 0.000689, 0.000603, 0.000724, 0.728089, 0.001587, 0.000511]}, {"benchmark": "math", "task_id": "gsm8k-1172", "answer": "Monday: 3  \nTuesday: 6  \nWednesday: 12  \nThursday: 24  \nFriday: 48  \n\nTotal: \\(3+6+12+24+48=93\\)\n\n93", "cost_usd": 7.320000000000002e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.000394, 2.1e-05, 1.9e-05, 2.7e-05, 0.999467, 5.6e-05, 1.5e-05]}, {"benchmark": "math", "task_id": "gsm8k-325", "answer": "Step 1: Determine the number of junior programmers. They make up 2/5 of the total 100 employees:  \n(2/5) \u00d7 100 = 40 junior programmers.\n\nStep 2: Determine the number of senior programmers. Subtract the number of junior programmers from the total:  \n100 \u2212 40 = 60 senior programmers.\n\nStep 3: Find the monthly salary for each group. Junior programmers earn $2000 per month. Senior programmers earn $400 more:  \n$2000 + $400 = $2400 per month for senior programmers.\n\nStep 4: Calculate the total monthly payroll for all programmers. Multiply the number of each type by their respective salary and add the results:  \n(40 \u00d7 $2000) + (60 \u00d7 $2400) = $80,000 + $144,000 = $224,000.\n\n224000", "cost_usd": 0.00022883000000000002, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.840255, 0.000293, 0.000235, 0.000319, 0.157927, 0.000739, 0.000232]}, {"benchmark": "code", "task_id": "lcb-abc388_b", "answer": "import sys\n\ndef main():\n    data = list(map(int, sys.stdin.buffer.read().split()))\n    n, d = data[0], data[1]\n    snakes = [(data[i], data[i + 1]) for i in range(2, 2 + 2 * n, 2)]\n\n    results = []\n    for k in range(1, d + 1):\n        results.append(str(max(t * (l + k) for t, l in snakes)))\n\n    sys.stdout.write(\"\\n\".join(results))\n\nif __name__ == \"__main__\":\n    main()\n", "cost_usd": 0.00011510000000000003, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.085031, 8e-05, 6.4e-05, 9.7e-05, 0.914445, 0.000229, 5.4e-05]}, {"benchmark": "code", "task_id": "lcb-abc388_d", "answer": "```python\nimport sys\n\ndef solve():\n    # Read all input from stdin\n    input_data = sys.stdin.read().split()\n    \n    if not input_data:\n        return\n    \n    iterator = iter(input_data)\n    try:\n        N = int(next(iterator))\n    except StopIteration:\n        return\n\n    # diff array for difference array technique\n    # Size N+2 to handle indices up to N+1 safely\n    # Indices used: 1 to N+1\n    diff = [0] * (N + 2)\n    \n    current_received = 0\n    results = []\n    \n    # Process each alien from 1 to N\n    # Alien i corresponds to index i in 1-based logic\n    for i in range(1, N + 1):\n        try:\n            A_i = int(next(iterator))\n        except StopIteration:\n            break\n        \n        # Update the number of stones received from previous aliens\n        # diff[i] contains the net change in received stones at index i\n        current_received += diff[i]\n        \n        # Total stones alien i has upon becoming an adult\n        # Initial stones + stones received\n        X = A_i + current_received\n        \n        # Alien i will give 1 stone to each subsequent alien k (i < k <= N)\n        # as long as he has stones.\n        # He can give to at most X aliens.\n        # The range of aliens he gives to is [i+1, min(N, i + X)].\n        \n        # Calculate the end of the range\n        # If i + X >= N, he gives to everyone up to N\n        if i + X >= N:\n            end = N\n        else:\n            end = i + X\n        \n        # Apply the difference array update for the range [i+1, end]\n        # This marks that aliens in this range will receive 1 stone from alien i\n        if end >= i + 1:\n            diff[i+1] += 1\n            diff[end+1] -= 1\n        \n        # Calculate remaining stones\n        # He gives min(X, N - i) stones to future aliens\n        # Remaining = X - min(X, N - i) = max(0, X - (N - i))\n        remaining = X - (N - i)\n        if remaining < 0:\n            remaining = 0\n        \n        results.append(str(remaining))\n    \n    sys.stdout.write(' '.join(results) + '\\n')\n\nif __name__ == '__main__':\n    solve()\n```", "cost_usd": 0.00333252, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.998965, 1.2e-05, 8e-06, 1.2e-05, 0.000968, 2.7e-05, 8e-06]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "017710f930891fdc1c729cddb0e35a5cdd3e8bb515df1aa8b059b0e2fb7e5e72", "platform_sig": 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