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+[{"task_id": "mmlu-8265", "model": "deepseek/deepseek-v4-flash", "prompt": "[MMLU]  In accordance with the theory of justice proposed by John Rawls, Stephen Macedo argues,\"We must consider the justifiability of policies from the standpoint of the _____ well-off among our fellow citizens.\"\nA) most\nB) least\nC) historically\nD) permanently\n\nAnswer with the letter only.\n\nEnd with one final line of exactly the form 'Answer: X' and nothing afterwards \u2014 X is the option letter when options are listed, otherwise the final value.", "response": "B", "cost_usd": 2.862e-05, "tokens_in": 104, "tokens_out": 107, "latency_s": 3.513}, {"task_id": "mmlu-4111", "model": "deepseek/deepseek-v4-flash", "prompt": "[MMLU] Imagine an economy that produces only two goods cheese and crackers. Calculate GDP for this economy if cheese retails for $3 a pound and 10 pounds are produced while crackers sell for $2 a pound and 20 pounds are produced.\nA) $35 \nB) $1,200 \nC) $70 \nD) $150 \n\nAnswer with the letter only.\n\nEnd with one final line of exactly the form 'Answer: X' and nothing afterwards \u2014 X is the option letter when options are listed, otherwise the final value.", "response": "Answer: C", "cost_usd": 3.4920000000000004e-05, "tokens_in": 116, "tokens_out": 136, "latency_s": 4.443}, {"task_id": "gsm8k-860", "model": "openai/gpt-5.6-luna", "prompt": "Jennifer bought 12 oranges from the market, she gave her three daughters 2 oranges each, and her only boy got 3 oranges. How many oranges did she remain with?\n\nReason step by step, then give the final answer as a single number.\n\nEnd with one final line of exactly the form 'Answer: X' and nothing afterwards \u2014 X is the option letter when options are listed, otherwise the final value.", "response": "Jennifer gave her daughters \\(3 \\times 2 = 6\\) oranges.  \nShe gave her son 3 oranges, for a total of \\(6 + 3 = 9\\) oranges.  \nShe had \\(12 - 9 = 3\\) oranges remaining.\n\nAnswer: 3", "cost_usd": 0.000144, "tokens_in": 90, "tokens_out": 105, "latency_s": 2.496}, {"task_id": "gsm8k-918", "model": "openai/gpt-5.6-luna", "prompt": "There are 6 students playing tennis and twice that number playing volleyball. There are 16 boys and 22 girls playing soccer. If each student only participates in one group, how many students are there in total?\n\nReason step by step, then give the final answer as a single number.\n\nEnd with one final line of exactly the form 'Answer: X' and nothing afterwards \u2014 X is the option letter when options are listed, otherwise the final value.", "response": "Tennis: 6 students  \nVolleyball: \\(2 \\times 6 = 12\\) students  \nSoccer: \\(16 + 22 = 38\\) students  \n\nTotal: \\(6 + 12 + 38 = 56\\)\n\nAnswer: 56", "cost_usd": 0.0001418, "tokens_in": 97, "tokens_out": 102, "latency_s": 2.908}, {"task_id": "lcb-abc398_a", "model": "openai/gpt-5.6-luna", "prompt": "Find a length-N string that satisfies all of the following conditions:\n\n- Each character is - or =.\n- It is a palindrome.\n- It contains exactly one or exactly two =s. If it contains two =s, they are adjacent.\n\nSuch a string is unique.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\n\nOutput\n\nPrint the answer.\n\nConstraints\n\n\n- 1 \\leq N \\leq 100\n- N is an integer.\n\nSample Input 1\n\n4\n\nSample Output 1\n\n-==-\n\nSample Input 2\n\n7\n\nSample Output 2\n\n---=---\n\nWrite a complete Python 3 program that reads from standard input and writes the answer to standard output. Output ONLY the program source, no prose and no code fences.\n\nThe checker compares stdout token by token; anything but an exact match scores zero. Mirror the published sample outputs literally \u2014 identical lines, and identical decimal digit counts wherever the samples print decimals.\n\nUse Python 3. Read the whole of stdin with one buffered read before parsing; assemble the entire output first and emit it with a single write. Prefer exact integer or Fraction arithmetic over floats for counting and comparison. Reply with the program alone: no commentary, no code fences.\n\nBefore finishing: when a loop compares or swaps two picked items, store them in fresh local names instead of rebinding the loop variables, or later iterations read corrupted state. Duplicated elements count once per occurrence \u2014 never deduplicate through a set unless the statement says distinct.", "response": "import sys\n\nn = int(sys.stdin.buffer.read())\nk = 1 if n % 2 else 2\nresult = \"-\" * ((n - k) // 2) + \"=\" * k + \"-\" * ((n - k) // 2)\nsys.stdout.write(result + \"\\n\")", "cost_usd": 0.00037860000000000004, "tokens_in": 315, "tokens_out": 263, "latency_s": 4.54}, {"task_id": "lcb-abc398_a", "model": "openai/gpt-5.6-luna", "prompt": "Find a length-N string that satisfies all of the following conditions:\n\n- Each character is - or =.\n- It is a palindrome.\n- It contains exactly one or exactly two =s. If it contains two =s, they are adjacent.\n\nSuch a string is unique.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\n\nOutput\n\nPrint the answer.\n\nConstraints\n\n\n- 1 \\leq N \\leq 100\n- N is an integer.\n\nSample Input 1\n\n4\n\nSample Output 1\n\n-==-\n\nSample Input 2\n\n7\n\nSample Output 2\n\n---=---\n\nWrite a complete Python 3 program that reads from standard input and writes the answer to standard output. Output ONLY the program source, no prose and no code fences.\n\nThe checker compares stdout token by token; anything but an exact match scores zero. Mirror the published sample outputs literally \u2014 identical lines, and identical decimal digit counts wherever the samples print decimals.\n\nUse Python 3. Read the whole of stdin with one buffered read before parsing; assemble the entire output first and emit it with a single write. Prefer exact integer or Fraction arithmetic over floats for counting and comparison. Reply with the program alone: no commentary, no code fences.\n\nBefore finishing: when a loop compares or swaps two picked items, store them in fresh local names instead of rebinding the loop variables, or later iterations read corrupted state. Duplicated elements count once per occurrence \u2014 never deduplicate through a set unless the statement says distinct.\n\nNow build test tooling for that solution and output nothing else. First a REFERENCE program: obviously correct, allowed to be very slow \u2014 simulate directly or enumerate every possibility, and deliberately avoid the algorithmic idea used above so the two can disagree. It reads the same stdin format and prints the same output format; handling small inputs only is fine. Second a GENERATOR program: it takes two command line integers, seed and size, calls random.seed(seed), and prints one random valid input in exactly the statement's input format, size being a rough magnitude clamped to the constraints. At size 2 print the smallest legal input. At maximum size, every seed must produce genuinely independent high-entropy values, while smaller sizes should also cover repeats, range extremes and adversarial orderings. Never draw most maximum-size values from a tiny pool. Also write a second BOUNDARY GENERATOR with a different construction strategy. It has the same seed and size interface, but concentrates on valid maximum-size, high-diversity and worst-case-shaped inputs rather than ordinary random cases. Respond with exactly three fenced blocks:\n```reference\n<program>\n```\n```generator\n<program>\n```\n```boundary_generator\n<program>\n```", "response": "```reference\nimport sys\nfrom itertools import product\n\ndata = sys.stdin.buffer.read().split()\nn = int(data[0])\n\ndef valid(s):\n    if s != s[::-1]:\n        return False\n    eq = [i for i, c in enumerate(s) if c == \"=\"]\n    if len(eq) == 1:\n        return True\n    return len(eq) == 2 and eq[1] == eq[0] + 1\n\nif n <= 16:\n    answer = None\n    for chars in product(\"-=\", repeat=n):\n        candidate = \"\".join(chars)\n        if valid(candidate):\n            answer = candidate\n            break\nelse:\n    answer = None\n    for i in range(n):\n        chars = [\"-\"] * n\n        chars[i] = \"=\"\n        candidate = \"\".join(chars)\n        if valid(candidate):\n            answer = candidate\n            break\n    if answer is None:\n        for i in range(n - 1):\n            chars = [\"-\"] * n\n            chars[i] = chars[i + 1] = \"=\"\n            candidate = \"\".join(chars)\n            if valid(candidate):\n                answer = candidate\n                break\n\nsys.stdout.write(answer + \"\\n\")\n```\n\n```generator\nimport random\nimport sys\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\nsize = max(1, min(100, size))\nif size == 2:\n    n = 1\nelse:\n    n = random.randint(1, size)\n\nsys.stdout.write(str(n) + \"\\n\")\n```\n\n```boundary_generator\nimport random\nimport sys\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\nsize = max(1, min(100, size))\nif size == 2:\n    n = 1\nelif size == 1:\n    n = 1\nelse:\n    low = max(1, size - max(1, size // 5))\n    n = random.randint(low, size)\n\nsys.stdout.write(str(n) + \"\\n\")\n```", "cost_usd": 0.0012652, "tokens_in": 560, "tokens_out": 961, "latency_s": 9.245}, {"task_id": "lcb-abc399_d", "model": "openai/gpt-5.6-luna", "prompt": "N couples are seated in a line.\nCount the number of pairs of couples such that neither couple was originally sitting next to each other, and both couples can end up sitting next to each other by swapping seats among those four people.\n\nThere is a sequence A = (A_1, A_2, \\dots, A_{2N}) of length 2N. Each of the integers 1, 2, \\dots, N appears exactly twice in A.\nFind the number of integer pairs (a, b) satisfying 1 \\leq a < b \\leq N and all of the following conditions:\n\n- The two occurrences of a in A are not adjacent.\n- The two occurrences of b in A are not adjacent.\n- By performing the following operation one or more times in any order, it is possible to reach a state where the two occurrences of a in A are adjacent and the two occurrences of b in A are also adjacent.\n- Choose an integer pair (i, j) (1 \\leq i \\leq 2N, 1 \\leq j \\leq 2N) such that A_i = a and A_j = b, and swap A_i with A_j.\n\n\n\nYou are given T test cases; solve each of them.\n\nInput\n\nThe input is given from Standard Input in the following format, where \\mathrm{case}_i denotes the i-th test case:\nT\n\\mathrm{case}_1\n\\mathrm{case}_2\n\\vdots\n\\mathrm{case}_T\n\nEach test case is given in the following format:\nN\nA_1 A_2 \\dots A_{2N}\n\nOutput\n\nPrint T lines. The i-th line should contain the answer for the i-th test case.\n\nConstraints\n\n\n- 1 \\leq T \\leq 2 \\times 10^5\n- 1 \\leq N \\leq 2 \\times 10^5\n- 1 \\leq A_i \\leq N\n- Each of 1, 2, \\dots, N appears exactly twice in A.\n- The sum of N over all test cases is at most 2 \\times 10^5.\n- All input values are integers.\n\nSample Input 1\n\n3\n3\n1 2 3 3 1 2\n4\n1 1 2 2 3 3 4 4\n5\n1 2 3 4 5 1 2 3 4 5\n\nSample Output 1\n\n1\n0\n4\n\nConsider the first test case.\n(a, b) = (1, 2) satisfies the conditions in the problem statement, for the following reasons:\n\n- The two occurrences of 1 in A are not adjacent.\n- The two occurrences of 2 in A are not adjacent.\n- By performing the operation where (i, j) = (1, 6) and swapping A_1 with A_6, you can reach a state where the two occurrences of 1 are adjacent and the two occurrences of 2 are also adjacent.\n\n(1, 2) is the only pair (a, b) that satisfies the conditions.\n\nWrite a complete Python 3 program that reads from standard input and writes the answer to standard output. Output ONLY the program source, no prose and no code fences.\n\nThe checker compares stdout token by token; anything but an exact match scores zero. Mirror the published sample outputs literally \u2014 identical lines, and identical decimal digit counts wherever the samples print decimals.\n\nUse Python 3. Read the whole of stdin with one buffered read before parsing; assemble the entire output first and emit it with a single write. Prefer exact integer or Fraction arithmetic over floats for counting and comparison. Reply with the program alone: no commentary, no code fences.\n\nBefore finishing: when a loop compares or swaps two picked items, store them in fresh local names instead of rebinding the loop variables, or later iterations read corrupted state. Duplicated elements count once per occurrence \u2014 never deduplicate through a set unless the statement says distinct.", "response": "import sys\n\ndata = list(map(int, sys.stdin.buffer.read().split()))\nit = iter(data)\nt = next(it)\nout = []\n\nfor _ in range(t):\n    n = next(it)\n    m = 2 * n\n    arr = [next(it) for _ in range(m)]\n\n    bad = bytearray(n + 1)\n    for i in range(m - 1):\n        x = arr[i]\n        y = arr[i + 1]\n        if x == y:\n            bad[x] = 1\n\n    counts = {}\n    for i in range(m - 1):\n        x = arr[i]\n        y = arr[i + 1]\n        if x != y:\n            if x > y:\n                x, y = y, x\n            key = (x, y)\n            counts[key] = counts.get(key, 0) + 1\n\n    shared = {}\n    for i in range(m - 2):\n        x = arr[i]\n        y = arr[i + 1]\n        z = arr[i + 2]\n        if x == z and x != y:\n            if x > y:\n                x, y = y, x\n            shared[(x, y)] = True\n\n    ans = 0\n    for key, count in counts.items():\n        x, y = key\n        if count >= 2 and not bad[x] and not bad[y]:\n            if count != 2 or key not in shared:\n                ans += 1\n\n    out.append(str(ans))\n\nsys.stdout.write(\"\\n\".join(out))", "cost_usd": 0.0067856, "tokens_in": 850, "tokens_out": 5513, "latency_s": 51.622}, {"task_id": "lcb-abc399_d", "model": "openai/gpt-5.6-luna", "prompt": "N couples are seated in a line.\nCount the number of pairs of couples such that neither couple was originally sitting next to each other, and both couples can end up sitting next to each other by swapping seats among those four people.\n\nThere is a sequence A = (A_1, A_2, \\dots, A_{2N}) of length 2N. Each of the integers 1, 2, \\dots, N appears exactly twice in A.\nFind the number of integer pairs (a, b) satisfying 1 \\leq a < b \\leq N and all of the following conditions:\n\n- The two occurrences of a in A are not adjacent.\n- The two occurrences of b in A are not adjacent.\n- By performing the following operation one or more times in any order, it is possible to reach a state where the two occurrences of a in A are adjacent and the two occurrences of b in A are also adjacent.\n- Choose an integer pair (i, j) (1 \\leq i \\leq 2N, 1 \\leq j \\leq 2N) such that A_i = a and A_j = b, and swap A_i with A_j.\n\n\n\nYou are given T test cases; solve each of them.\n\nInput\n\nThe input is given from Standard Input in the following format, where \\mathrm{case}_i denotes the i-th test case:\nT\n\\mathrm{case}_1\n\\mathrm{case}_2\n\\vdots\n\\mathrm{case}_T\n\nEach test case is given in the following format:\nN\nA_1 A_2 \\dots A_{2N}\n\nOutput\n\nPrint T lines. The i-th line should contain the answer for the i-th test case.\n\nConstraints\n\n\n- 1 \\leq T \\leq 2 \\times 10^5\n- 1 \\leq N \\leq 2 \\times 10^5\n- 1 \\leq A_i \\leq N\n- Each of 1, 2, \\dots, N appears exactly twice in A.\n- The sum of N over all test cases is at most 2 \\times 10^5.\n- All input values are integers.\n\nSample Input 1\n\n3\n3\n1 2 3 3 1 2\n4\n1 1 2 2 3 3 4 4\n5\n1 2 3 4 5 1 2 3 4 5\n\nSample Output 1\n\n1\n0\n4\n\nConsider the first test case.\n(a, b) = (1, 2) satisfies the conditions in the problem statement, for the following reasons:\n\n- The two occurrences of 1 in A are not adjacent.\n- The two occurrences of 2 in A are not adjacent.\n- By performing the operation where (i, j) = (1, 6) and swapping A_1 with A_6, you can reach a state where the two occurrences of 1 are adjacent and the two occurrences of 2 are also adjacent.\n\n(1, 2) is the only pair (a, b) that satisfies the conditions.\n\nWrite a complete Python 3 program that reads from standard input and writes the answer to standard output. Output ONLY the program source, no prose and no code fences.\n\nThe checker compares stdout token by token; anything but an exact match scores zero. Mirror the published sample outputs literally \u2014 identical lines, and identical decimal digit counts wherever the samples print decimals.\n\nUse Python 3. Read the whole of stdin with one buffered read before parsing; assemble the entire output first and emit it with a single write. Prefer exact integer or Fraction arithmetic over floats for counting and comparison. Reply with the program alone: no commentary, no code fences.\n\nBefore finishing: when a loop compares or swaps two picked items, store them in fresh local names instead of rebinding the loop variables, or later iterations read corrupted state. Duplicated elements count once per occurrence \u2014 never deduplicate through a set unless the statement says distinct.\n\nNow build test tooling for that solution and output nothing else. First a REFERENCE program: obviously correct, allowed to be very slow \u2014 simulate directly or enumerate every possibility, and deliberately avoid the algorithmic idea used above so the two can disagree. It reads the same stdin format and prints the same output format; handling small inputs only is fine. Second a GENERATOR program: it takes two command line integers, seed and size, calls random.seed(seed), and prints one random valid input in exactly the statement's input format, size being a rough magnitude clamped to the constraints. At size 2 print the smallest legal input. At maximum size, every seed must produce genuinely independent high-entropy values, while smaller sizes should also cover repeats, range extremes and adversarial orderings. Never draw most maximum-size values from a tiny pool. Also write a second BOUNDARY GENERATOR with a different construction strategy. It has the same seed and size interface, but concentrates on valid maximum-size, high-diversity and worst-case-shaped inputs rather than ordinary random cases. Respond with exactly three fenced blocks:\n```reference\n<program>\n```\n```generator\n<program>\n```\n```boundary_generator\n<program>\n```", "response": "```reference\nimport sys\nfrom collections import deque\n\ndef solve():\n    data = list(map(int, sys.stdin.buffer.read().split()))\n    if not data:\n        return\n    t = data[0]\n    pos = 1\n    out = []\n\n    for _ in range(t):\n        n = data[pos]\n        pos += 1\n        arr = tuple(data[pos:pos + 2 * n])\n        pos += 2 * n\n\n        first = [[] for _ in range(n + 1)]\n        for i, x in enumerate(arr):\n            first[x].append(i)\n\n        ans = 0\n        for a in range(1, n + 1):\n            if first[a][1] == first[a][0] + 1:\n                continue\n            for b in range(a + 1, n + 1):\n                if first[b][1] == first[b][0] + 1:\n                    continue\n\n                q = deque([arr])\n                seen = {arr}\n                possible = False\n\n                while q and not possible:\n                    state = q.popleft()\n                    pa = []\n                    pb = []\n                    for i, x in enumerate(state):\n                        if x == a:\n                            pa.append(i)\n                        elif x == b:\n                            pb.append(i)\n\n                    if pa[1] == pa[0] + 1 and pb[1] == pb[0] + 1:\n                        possible = True\n                        break\n\n                    for ia in pa:\n                        for ib in pb:\n                            nxt = list(state)\n                            nxt[ia], nxt[ib] = nxt[ib], nxt[ia]\n                            nxt = tuple(nxt)\n                            if nxt not in seen:\n                                seen.add(nxt)\n                                q.append(nxt)\n\n                if possible:\n                    ans += 1\n\n        out.append(str(ans))\n\n    sys.stdout.write(\"\\n\".join(out))\n\nif __name__ == \"__main__\":\n    solve()\n```\n\n```generator\nimport sys\nimport random\n\ndef main():\n    seed = int(sys.argv[1])\n    size = int(sys.argv[2])\n    random.seed(seed)\n\n    if size == 2:\n        sys.stdout.write(\"1\\n1\\n1 1\\n\")\n        return\n\n    n = max(1, min(200000, size))\n    mode = random.randrange(5)\n\n    if mode == 0:\n        a = list(range(1, n + 1)) * 2\n        random.shuffle(a)\n    elif mode == 1:\n        p = list(range(1, n + 1))\n        random.shuffle(p)\n        a = []\n        for x in p:\n            a.extend((x, x))\n    elif mode == 2:\n        p = list(range(1, n + 1))\n        q = list(range(1, n + 1))\n        random.shuffle(p)\n        random.shuffle(q)\n        a = p + q\n    elif mode == 3:\n        p = list(range(1, n + 1))\n        random.shuffle(p)\n        q = p[:]\n        random.shuffle(q)\n        a = []\n        for i in range(n):\n            a.append(p[i])\n            a.append(q[i])\n    else:\n        p = list(range(1, n + 1))\n        random.shuffle(p)\n        a = p + p[:]\n        random.shuffle(a)\n\n    sys.stdout.write(\"1\\n\")\n    sys.stdout.write(str(n) + \"\\n\")\n    sys.stdout.write(\" \".join(map(str, a)) + \"\\n\")\n\nif __name__ == \"__main__\":\n    main()\n```\n\n```boundary_generator\nimport sys\nimport random\n\ndef main():\n    seed = int(sys.argv[1])\n    size = int(sys.argv[2])\n    random.seed(seed)\n\n    if size == 2:\n        sys.stdout.write(\"1\\n1\\n1 1\\n\")\n        return\n\n    n = max(1, min(200000, size))\n    p = list(range(1, n + 1))\n    random.shuffle(p)\n    mode = random.randrange(6)\n\n    if n == 1:\n        q = p[:]\n    elif mode == 0:\n        q = p[::-1]\n    elif mode == 1:\n        shift = random.randrange(1, n)\n        q = p[shift:] + p[:shift]\n    elif mode == 2:\n        q = p[:]\n        random.shuffle(q)\n    elif mode == 3:\n        q = sorted(p, key=lambda _: random.getrandbits(64))\n    elif mode == 4:\n        q = p[1:] + p[:1]\n        random.shuffle(q)\n    else:\n        q = list(range(1, n + 1))\n        random.shuffle(q)\n\n    if mode % 3 == 0:\n        a = p + q\n    elif mode % 3 == 1:\n        a = []\n        for i in range(n):\n            a.extend((p[i], q[i]))\n    else:\n        a = []\n        mid = n // 2\n        for i in range(mid):\n            a.extend((p[i], p[i + mid]))\n        if n % 2:\n            a.append(p[-1])\n            a.append(q[-1])\n        a.extend(q[:mid])\n\n    sys.stdout.write(\"1\\n\")\n    sys.stdout.write(str(n) + \"\\n\")\n    sys.stdout.write(\" \".join(map(str, a)) + \"\\n\")\n\nif __name__ == \"__main__\":\n    main()\n```", "cost_usd": 0.0033851999999999997, "tokens_in": 1095, "tokens_out": 2593, "latency_s": 21.873}]