{"schema": 2, "epoch": 87623, "nonce": "00ff08d189d71f7e", "hotkey": "5Hg4fML2RQ86kiurzDJsHXBMwJuCDzW6Jv4MU8zTeLjonNxH", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "f105aad63d7b96494c90b5ac1f2a4e3fa0b94f9a6a7430370730cc1f32438ded", "model_id": "router", "total_cost_usd": 0.00483457, "n_calls": 6, "call_log_hash": "54f8598f29eafea1b146e4f46141f13bdc4b6cc52778c4b4b5cdfe2da69447e2", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 239.504, "tokens_in": 1387, "tokens_out": 34990, "results": [{"benchmark": "mmlu", "task_id": "mmlu-3982", "answer": "D", "cost_usd": 8.8e-06, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.06173, 0.070776, 0.061335, 0.061335, 0.625729, 0.062406, 0.056689]}, {"benchmark": "mmlu", "task_id": "mmlu-9027", "answer": "D", "cost_usd": 1.3999999999999998e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.107778, 0.108964, 0.067883, 0.067883, 0.521267, 0.04476, 0.081465]}, {"benchmark": "math", "task_id": "gsm8k-420", "answer": "First bank: $4,000  \nSecond bank: $4,000 \u00d7 2 = $8,000  \nTotal funding: $4,000 + $8,000 = $12,000  \nCapital now: $5,000 + $12,000 = $17,000  \n\n17000", "cost_usd": 7.88e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.089636, 0.081632, 0.070127, 0.070127, 0.498264, 0.1023, 0.087914]}, {"benchmark": "math", "task_id": "gsm8k-415", "answer": "1. Joanne gathers 30 apples from the tallest trees.\n2. She gathers half as many from the shortest trees: \\(30 \\div 2 = 15\\).\n3. Her sister gathers twice as many from the tallest trees: \\(2 \\times 30 = 60\\).\n4. Her sister gathers three times as many from the shortest trees: \\(3 \\times 15 = 45\\).\n5. Together, they gather \\(30 + 15 + 60 + 45 = 150\\) apples from the tallest and shortest trees.\n6. Therefore, Joanne gathers \\(500 - 150 = 350\\) apples from the average trees.\n\n350", "cost_usd": 0.0001458, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.093493, 0.080014, 0.067274, 0.067274, 0.5435, 0.068216, 0.080229]}, {"benchmark": "code", "task_id": "lcb-abc400_b", "answer": "n, m = map(int, input().split())\n\nlimit = 10**9\ntotal = 0\npower = 1\n\nfor _ in range(m + 1):\n    total += power\n    if total > limit:\n        print(\"inf\")\n        break\n    power *= n\nelse:\n    print(total)\n", "cost_usd": 7.14e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.104071, 0.108677, 0.074204, 0.074204, 0.345218, 0.145565, 0.148061]}, {"benchmark": "code", "task_id": "lcb-abc399_d", "answer": "```python\nimport sys\n\n# Increase recursion depth just in case, though not strictly needed here\nsys.setrecursionlimit(2000)\n\ndef solve():\n    # Read all input from stdin efficiently\n    # split() handles whitespace (spaces, newlines) automatically\n    input_data = sys.stdin.read().split()\n    \n    if not input_data:\n        return\n\n    iterator = iter(input_data)\n    \n    try:\n        t_str = next(iterator)\n    except StopIteration:\n        return\n    T = int(t_str)\n    \n    results = []\n    \n    for _ in range(T):\n        try:\n            n_str = next(iterator)\n        except StopIteration:\n            break\n        N = int(n_str)\n        \n        # Arrays to store positions of each number\n        # 1-based indexing for values 1..N\n        # first_pos[x] stores the first occurrence index of x\n        # second_pos[x] stores the second occurrence index of x\n        # Size N + 1 to accommodate 1-based values\n        first_pos = [0] * (N + 1)\n        second_pos = [0] * (N + 1)\n        \n        current_pos = 1\n        # Read 2*N integers representing the sequence A\n        for _ in range(2 * N):\n            val = int(next(iterator))\n            if first_pos[val] == 0:\n                first_pos[val] = current_pos\n            else:\n                second_pos[val] = current_pos\n            current_pos += 1\n            \n        # Collect valid points (l, r) where r - l > 1\n        # A point (l, r) represents a number x whose first occurrence is at l\n        # and second occurrence is at r.\n        # The condition \"occurrences not adjacent\" means r - l > 1.\n        valid_points = set()\n        for x in range(1, N + 1):\n            l = first_pos[x]\n            r = second_pos[x]\n            if r - l > 1:\n                valid_points.add((l, r))\n        \n        ans = 0\n        # We need to count pairs of numbers {a, b} such that:\n        # 1. Neither a nor b has adjacent occurrences (handled by valid_points filter).\n        # 2. It is possible to make both a's adjacent and both b's adjacent by swapping a's and b's.\n        #\n        # Analysis shows that this is possible if and only if the set of 4 positions\n        # {pos(a,1), pos(a,2), pos(b,1), pos(b,2)} can be partitioned into two pairs of adjacent integers.\n        # Let La, Ra be positions of a (La < Ra) and Lb, Rb be positions of b (Lb < Rb).\n        # The condition simplifies to: |La - Lb| = 1 AND |Ra - Rb| = 1.\n        #\n        # Assuming La < Lb, this implies Lb = La + 1.\n        # Then |Ra - Rb| = 1 implies Rb = Ra + 1 or Rb = Ra - 1.\n        # So for a point (l, r), we look for points (l+1, r+1) or (l+1, r-1).\n        # By only checking neighbors with a larger first coordinate (l+1), we count each pair exactly once.\n        \n        for l, r in valid_points:\n            # Check neighbor (l+1, r+1)\n            if (l + 1, r + 1) in valid_points:\n                ans += 1\n            # Check neighbor (l+1, r-1)\n            if (l + 1, r - 1) in valid_points:\n                ans += 1\n        \n        results.append(str(ans))\n        \n    print('\\n'.join(results))\n\nif __name__ == '__main__':\n    solve()\n```", "cost_usd": 0.00451577, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.31854, 0.119812, 0.06536, 0.06536, 0.150592, 0.15006, 0.130277]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "bc86aa591eac9d9f756db1151f895ce259efb53005006427dc399ea78929ebbb", "platform_sig": 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