| {"schema": 2, "epoch": 87569, "nonce": "9a38438598412408", "hotkey": "5CaXH581GtSjxFaFaJzSN35CyKzvFMQhRAxrNy6qmoztNiPz", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "19aafcfeb694bb06e7fbbedbfd7fce6c50bd3389a57eeefd1b5d2e5475afdd62", "model_id": "router", "total_cost_usd": 0.00627808, "n_calls": 6, "call_log_hash": "ea22ab6503804cf704714d77d4d26676e7bdc5de57cc1507356fe2de2b9ad6e0", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 293.916, "tokens_in": 2364, "tokens_out": 45161, "results": [{"benchmark": "mmlu", "task_id": "mmlu-7563", "answer": "B", "cost_usd": 7.5e-06, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.093242, 0.000381, 0.000339, 0.000488, 0.904383, 0.000898, 0.00027]}, {"benchmark": "mmlu", "task_id": "mmlu-9954", "answer": "A", "cost_usd": 2.0300000000000002e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.034478, 0.000227, 0.000218, 0.000296, 0.964039, 0.000549, 0.000192]}, {"benchmark": "math", "task_id": "gsm8k-642", "answer": "10 suits: \\(10 \\times 750 = 7{,}500\\)\n\n10 dress pants: \\(10 \\times (750 \\div 5) = 10 \\times 150 = 1{,}500\\)\n\nDress shirts: \\(10 \\times 3 = 30\\) shirts, and \\(30 \\times 60 = 1{,}800\\)\n\nTotal: \\(7{,}500 + 1{,}500 + 1{,}800 = 10{,}800\\)\n\n10800", "cost_usd": 0.00012550000000000001, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.001296, 2.5e-05, 2.2e-05, 3.3e-05, 0.998529, 7.5e-05, 2e-05]}, {"benchmark": "math", "task_id": "gsm8k-419", "answer": "1. The comb fell from 10,000 feet to \\(10{,}000 - 4{,}000 = 6{,}000\\) feet.\n2. Oliver is at 3,000 feet, so he must climb \\(6{,}000 - 3{,}000 = 3{,}000\\) feet.\n\n3000", "cost_usd": 9.92e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.014986, 9.9e-05, 8.7e-05, 0.000123, 0.984356, 0.000271, 7.7e-05]}, {"benchmark": "code", "task_id": "lcb-abc395_b", "answer": "import sys\n\nN = int(sys.stdin.readline())\nfor i in range(1, N + 1):\n row = []\n for j in range(1, N + 1):\n depth = min(i, j, N + 1 - i, N + 1 - j)\n row.append('#' if depth % 2 == 1 else '.')\n print(''.join(row))", "cost_usd": 0.00018840000000000003, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.010037, 3.2e-05, 2.4e-05, 3.6e-05, 0.989772, 7.9e-05, 2.1e-05]}, {"benchmark": "code", "task_id": "lcb-abc400_d", "answer": "```python\nimport sys\nfrom collections import deque\n\ndef solve():\n # Read all input from stdin efficiently\n input_data = sys.stdin.read().split()\n if not input_data:\n return\n\n iterator = iter(input_data)\n \n try:\n H = int(next(iterator))\n W = int(next(iterator))\n \n S = []\n for _ in range(H):\n S.append(next(iterator))\n \n A = int(next(iterator))\n B = int(next(iterator))\n C = int(next(iterator))\n D = int(next(iterator))\n except StopIteration:\n return\n\n # Convert 1-based coordinates to 0-based\n start_r, start_c = A - 1, B - 1\n target_r, target_c = C - 1, D - 1\n\n # Distance matrix initialized to infinity\n INF = 10**9\n dist = [[INF] * W for _ in range(H)]\n \n # Deque for 0-1 BFS\n dq = deque()\n \n # Initial state\n dist[start_r][start_c] = 0\n dq.append((start_r, start_c))\n \n # Directions: Up, Down, Left, Right\n directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]\n \n while dq:\n r, c = dq.popleft()\n d = dist[r][c]\n \n # If we reached the target, print and exit\n # Since it's 0-1 BFS, the first time we pop the target, it's the shortest path\n if r == target_r and c == target_c:\n print(d)\n return\n\n # 1. Move to adjacent road cells (Cost 0)\n # We can only move to cells that are currently roads ('.')\n # Note: If a cell was a wall but cleared by a kick, its distance would have been updated,\n # so we don't strictly need to move into it to discover it, but we might need to move\n # to explore further. However, since cleared walls have low distance, they are in the queue.\n for dr, dc in directions:\n nr, nc = r + dr, c + dc\n if 0 <= nr < H and 0 <= nc < W:\n if S[nr][nc] == '.':\n if dist[nr][nc] > d:\n dist[nr][nc] = d\n dq.appendleft((nr, nc))\n \n # 2. Perform front kick (Cost 1)\n # A kick at (r, c) in a direction clears cells at distance 1 and 2 in that direction.\n # These cells become roads, so we can reach them with cost d + 1.\n for dr, dc in directions:\n # Cell at distance 1\n tr1, tc1 = r + dr, c + dc\n if 0 <= tr1 < H and 0 <= tc1 < W:\n # This cell becomes a road.\n # We can reach it with cost d + 1.\n if dist[tr1][tc1] > d + 1:\n dist[tr1][tc1] = d + 1\n dq.append((tr1, tc1))\n \n # Cell at distance 2\n tr2, tc2 = r + 2*dr, c + 2*dc\n if 0 <= tr2 < H and 0 <= tc2 < W:\n # This cell becomes a road.\n if dist[tr2][tc2] > d + 1:\n dist[tr2][tc2] = d + 1\n dq.append((tr2, tc2))\n\n # If target is unreachable (should not happen based on problem statement)\n print(dist[target_r][target_c])\n\nif __name__ == '__main__':\n solve()\n```", "cost_usd": 0.00583718, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.997247, 2.1e-05, 1.4e-05, 2e-05, 0.002637, 4.7e-05, 1.4e-05]}], "quote": {"measurement": 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