{"schema": 2, "epoch": 87572, "nonce": "f03d68fcf248033e", "hotkey": "5CaXH581GtSjxFaFaJzSN35CyKzvFMQhRAxrNy6qmoztNiPz", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "19aafcfeb694bb06e7fbbedbfd7fce6c50bd3389a57eeefd1b5d2e5475afdd62", "model_id": "router", "total_cost_usd": 0.00639035, "n_calls": 6, "call_log_hash": "d043943cfb311928f6876bce103a5237e11fc6ff1e7be0ae54c322b14a1eba2f", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 329.723, "tokens_in": 1769, "tokens_out": 46880, "results": [{"benchmark": "mmlu", "task_id": "mmlu-4392", "answer": "B", "cost_usd": 3.8e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.066216, 0.000192, 0.000157, 0.000231, 0.932608, 0.000454, 0.000144]}, {"benchmark": "mmlu", "task_id": "mmlu-454", "answer": "A", "cost_usd": 1.5100000000000001e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.01085, 0.000149, 0.000131, 0.0002, 0.988183, 0.000368, 0.000119]}, {"benchmark": "math", "task_id": "gsm8k-1171", "answer": "Each bedroom: \\(20 \\times 12 = 240\\) square feet.\n\nFour bedrooms: \\(4 \\times 240 = 960\\) square feet.\n\nLiving room: \\(5 \\times 240 = 1200\\) square feet.\n\nTotal area: \\(960 + 1200 + 1000 = 3160\\) square feet.\n\n3160", "cost_usd": 0.00010420000000000003, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.264273, 0.00029, 0.000234, 0.000327, 0.73388, 0.00078, 0.000216]}, {"benchmark": "math", "task_id": "gsm8k-447", "answer": "Joey\u2019s score: 214 + 26 = 240  \nMarcy\u2019s score: 225 + 10 = 235  \nDifference: 240 \u2212 235 = 5  \n\n5", "cost_usd": 6.16e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.008402, 4.4e-05, 3.4e-05, 5.5e-05, 0.991313, 0.000119, 3.1e-05]}, {"benchmark": "code", "task_id": "lcb-abc398_a", "answer": "n = int(input())\ns = ['-'] * n\n\nif n % 2:\n    s[n // 2] = '='\nelse:\n    s[n // 2 - 1] = '='\n    s[n // 2] = '='\n\nprint(''.join(s))", "cost_usd": 9.650000000000004e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.017201, 3.4e-05, 2.5e-05, 4.5e-05, 0.982587, 8.6e-05, 2.3e-05]}, {"benchmark": "code", "task_id": "lcb-abc400_d", "answer": "```python\nimport sys\nfrom collections import deque\n\n# Increase recursion depth just in case, though not strictly needed here\nsys.setrecursionlimit(2000)\n\ndef solve():\n    # Read all input from standard input\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)) - 1\n        B = int(next(iterator)) - 1\n        C = int(next(iterator)) - 1\n        D = int(next(iterator)) - 1\n    except StopIteration:\n        return\n\n    # Distance matrix initialized to infinity\n    # Using a large number for infinity\n    INF = 10**9\n    dist = [[INF] * W for _ in range(H)]\n    \n    # Deque for 0-1 BFS, storing (row, col, distance)\n    dq = deque()\n    dist[A][B] = 0\n    dq.append((A, B, 0))\n    \n    # Kick offsets: (dr, dc) for distance 1 and 2 in 4 directions\n    # Right: (0, 1), (0, 2)\n    # Left: (0, -1), (0, -2)\n    # Down: (1, 0), (2, 0)\n    # Up: (-1, 0), (-2, 0)\n    kick_offsets = [\n        (0, 1), (0, 2),\n        (0, -1), (0, -2),\n        (1, 0), (2, 0),\n        (-1, 0), (-2, 0)\n    ]\n    \n    # Directions for moving to adjacent cells\n    move_dirs = [(-1, 0), (1, 0), (0, -1), (0, 1)]\n    \n    while dq:\n        r, c, d = dq.popleft()\n        \n        # If we found a shorter path to this cell already, skip processing\n        if d > dist[r][c]:\n            continue\n            \n        # If we reached the target cell\n        if r == C and c == D:\n            print(d)\n            return\n        \n        # 1. Move to adjacent roads (weight 0)\n        # We can move to any adjacent cell that is a road ('.').\n        # Note: Even if the current cell (r,c) was originally a wall but reached via a kick,\n        # it is now considered a road, so we can move to adjacent roads.\n        for dr, dc in move_dirs:\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, d))\n        \n        # 2. Perform front kick (weight 1)\n        # Kicking allows reaching cells at distance 1 and 2 in cardinal directions.\n        # The cost increases by 1.\n        for dr, dc in kick_offsets:\n            tr, tc = r + dr, c + dc\n            if 0 <= tr < H and 0 <= tc < W:\n                if dist[tr][tc] > d + 1:\n                    dist[tr][tc] = d + 1\n                    dq.append((tr, tc, d + 1))\n\n    # If the target is not reachable (though problem constraints imply it should be)\n    print(-1)\n\nif __name__ == '__main__':\n    solve()\n```", "cost_usd": 0.00607495, "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": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "5d81219f742e48218b6a6921019a0321ec7c10914e8aaa94bc6716932198216a", "platform_sig": 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