{"schema": 2, "epoch": 87622, "nonce": "c534dd314e25dbfc", "hotkey": "5GuVFWnG62s4r1AGAfrhhHBMnsMPMXKNxrmBpeJdFGyCQEaX", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "095ffbac36e314358948a86370be2e0be5384b2278d5ee9bc924d7f155c7d474", "model_id": "router", "total_cost_usd": 0.0357786, "n_calls": 6, "call_log_hash": "b3058f3482ed1e26cd2956d5d45a1c326ab7102ea35b01182ecefbf75f449afe", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 37.307, "tokens_in": 1835, "tokens_out": 4852, "results": [{"benchmark": "mmlu", "task_id": "mmlu-8333", "answer": "D", "cost_usd": 4.11e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.019127, 0.005942, 0.005888, 0.006041, 0.683066, 0.150664, 0.129273]}, {"benchmark": "mmlu", "task_id": "mmlu-13551", "answer": "D", "cost_usd": 9.300000000000002e-06, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.017402, 0.005379, 0.005187, 0.005317, 0.787952, 0.116721, 0.062042]}, {"benchmark": "math", "task_id": "gsm8k-766", "answer": "Marissa had $80 and spent $20 on the ball:\n\n$80 - $20 = $60\n\nEach candy bar cost $5:\n\n$60 \u00f7 $5 = 12\n\n12", "cost_usd": 6.42e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.008964, 0.002419, 0.002291, 0.002285, 0.917993, 0.03578, 0.030267]}, {"benchmark": "math", "task_id": "gsm8k-923", "answer": "There are \\(27 - 15 = 12\\) cups for chamomile and mint tea.\n\nDividing these equally among 3 rows gives \\(12 \\div 3 = 4\\) cups per row.\n\nEach row has equal amounts of chamomile and mint tea, so \\(4 \\div 2 = 2\\) mint tea cups per row.\n\n2", "cost_usd": 0.00011840000000000002, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.003536, 0.00085, 0.000767, 0.00076, 0.978156, 0.0092, 0.006732]}, {"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": [0.004621, 0.001175, 0.001046, 0.001044, 0.898875, 0.009311, 0.083928]}, {"benchmark": "code", "task_id": "lcb-abc400_d", "answer": "To find the minimum number of front kicks Takahashi needs to reach the fish shop, we can model this as a shortest path problem on a grid using **0-1 BFS (Deque)**.\n\n### Key Idea\n1. **0-cost moves**: Takahashi can move to any orthogonally adjacent cell $(r', c')$ if it is already a road (`.`).\n2. **1-cost moves (Front Kick)**: From a cell $(r, c)$, Takahashi can perform a front kick in any of the 4 cardinal directions (up, down, left, right). This turns the walls in the next 1 or 2 steps in that direction into roads. \n\nBy kicking in a direction, he creates a 2-cell bridge of roads. From these newly created road cells, he can step into any adjacent existing road cells. Across all 4 kick directions, any cell $(r', c')$ within a $5 \\times 5$ bounding box centered at $(r, c)$ (i.e., $|r - r'| \\le 2$ and $|c - c'| \\le 2$) can be reached with **1 kick**, provided that $(r', c')$ is inside the grid bounds.\n\nThus:\n- Moving to an adjacent road cell costs `0`.\n- Moving to any cell in the $5 \\times 5$ box around the current cell costs `1` (which acts as performing 1 kick).\n\n### Python 3 Source Code\n\n```python\nimport sys\nfrom collections import deque\n\ndef solve():\n    input = sys.stdin.read\n    data = input().split()\n    \n    H = int(data[0])\n    W = int(data[1])\n    \n    grid = data[2:2+H]\n    \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 = [[float('inf')] * W for _ in range(H)]\n    dist[A][B] = 0\n    \n    q = deque([(A, B)])\n    \n    while q:\n        r, c = q.popleft()\n        d = dist[r][c]\n        \n        if r == C and c == D:\n            print(d)\n            return\n\n        # 0-cost transitions: adjacent road cells\n        for dr, dc in [(-1, 0), (1, 0), (0, -1), (0, 1)]:\n            nr, nc = r + dr, c + dc\n            if 0 <= nr < H and 0 <= nc < W:\n                if grid[nr][nc] == '.' and dist[nr][nc] > d:\n                    dist[nr][nc] = d\n                    q.appendleft((nr, nc))\n\n        # 1-cost transitions: any cell in the 5x5 area centered at (r, c)\n        for dr in range(-2, 3):\n            for dc in range(-2, 3):\n                if dr == 0 and dc == 0:\n                    continue\n                nr, nc = r + dr, c + dc\n                if 0 <= nr < H and 0 <= nc < W:\n                    if dist[nr][nc] > d + 1:\n                        dist[nr][nc] = d + 1\n                        q.append((nr, nc))\n\nif __name__ == '__main__':\n    solve()\n```", "cost_usd": 0.0354975, "chosen_rung": 5, "rungs_used": [5], "distribution": [0.003914, 0.001721, 0.002135, 0.002377, 0.003852, 0.981493, 0.004508]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "5e72f2ca7549a849e80aab2713780a913268f90f8c24d2775bbd37128135609a", "platform_sig": 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