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+ {"schema": 2, "epoch": 87864, "nonce": "6eba5cc6fd54a0dd", "hotkey": "5FRE1kqMJz8ddModcXUsCquuWhB8QHnjYib2iK2VfiFetkC5", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "2e4d6a88063177a0a7a564f24e10b347ef48e87c9595a209162369a4fa4d4477", "model_id": "router", "total_cost_usd": 0.0336591654, "n_calls": 6, "call_log_hash": "79c5eed3a2317e380b0dc1786dc3736f520cc1427c30d1e028a06329a0c013c6", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 42.193, "tokens_in": 2223, "tokens_out": 5831, "results": [{"benchmark": "mmlu", "task_id": "mmlu-1820", "answer": "B", "cost_usd": 6.725e-05, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.999992, 2e-06, 0.0, 0.0, 3e-06, 2e-06, 0.0]}, {"benchmark": "mmlu", "task_id": "mmlu-3982", "answer": "D", "cost_usd": 4.7130000000000004e-05, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.999941, 1.6e-05, 0.0, 0.0, 4.3e-05, 0.0, 0.0]}, {"benchmark": "math", "task_id": "gsm8k-780", "answer": "The difference between their scores is 25 points, with Emma having the higher score.\n\n\\boxed{25}", "cost_usd": 0.00010064739999999999, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.007545, 0.990194, 0.0, 0.0, 0.002052, 0.00021, 0.0]}, {"benchmark": "math", "task_id": "gsm8k-80", "answer": "The two girls each received 1/6 of 24 liters, which is 4 liters each, so together they took 8 liters. The boy then took 6 liters, making a total of 14 liters taken. The remaining water is 24 - 14 = 10 liters.\n\n10", "cost_usd": 4.0338000000000006e-05, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.001124, 0.998796, 0.0, 0.0, 5.5e-05, 2.5e-05, 0.0]}, {"benchmark": "code", "task_id": "lcb-abc399_b", "answer": "import sys\n\ndef main():\n input = sys.stdin.readline\n n = int(input())\n scores = list(map(int, input().split()))\n\n rank = [0] * n\n current_rank = 1\n\n for score in sorted(set(scores), reverse=True):\n indices = [i for i, value in enumerate(scores) if value == score]\n for i in indices:\n rank[i] = current_rank\n current_rank += len(indices)\n\n print(\"\\n\".join(map(str, rank)))\n\nif __name__ == \"__main__\":\n main()\n", "cost_usd": 0.0001383, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.000174, 0.000382, 0.0, 0.0, 0.993096, 0.006347, 0.0]}, {"benchmark": "code", "task_id": "lcb-abc400_d", "answer": "import sys\nfrom collections import deque\n\ndef solve():\n input = sys.stdin.read\n data = input().split()\n if not data:\n return\n \n H = int(data[0])\n W = int(data[1])\n grid = data[2:2+H]\n A = int(data[2+H]) - 1\n B = int(data[2+H+1]) - 1\n C = int(data[2+H+2]) - 1\n D = int(data[2+H+3]) - 1\n\n dist = [-1] * (H * W)\n start_idx = A * W + B\n target_idx = C * W + D\n\n dist[start_idx] = 0\n dq = deque([start_idx])\n\n dr_dc_0 = [(-1, 0), (1, 0), (0, -1), (0, 1)]\n dr_dc_1 = [\n (-1, 0), (-2, 0),\n (1, 0), (2, 0),\n (0, -1), (0, -2),\n (0, 1), (0, 2)\n ]\n\n while dq:\n u = dq.popleft()\n d = dist[u]\n\n if u == target_idx:\n print(d)\n return\n\n r = u // W\n c = u % W\n\n # 0-cost transitions: adjacent cells that are roads '.'\n for dr, dc in dr_dc_0:\n nr, nc = r + dr, c + dc\n if 0 <= nr < H and 0 <= nc < W:\n if grid[nr][nc] == '.':\n v = nr * W + nc\n if dist[v] == -1 or dist[v] > d:\n dist[v] = d\n dq.appendleft(v)\n\n # 1-cost transitions: kick up to 2 steps in 4 directions\n for dr, dc in dr_dc_1:\n nr, nc = r + dr, c + dc\n if 0 <= nr < H and 0 <= nc < W:\n v = nr * W + nc\n if dist[v] == -1 or dist[v] > d + 1:\n dist[v] = d + 1\n dq.append(v)\n\nif __name__ == '__main__':\n solve()", "cost_usd": 0.0332655, "chosen_rung": 5, "rungs_used": [5], "distribution": [0.000115, 7.7e-05, 0.0, 0.0, 0.000713, 0.999095, 0.0]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "2b7df261b4279e6736b5d669e7882c888b511062c80fca3a2ae44d1c364f572b", "platform_sig": 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