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code_segments/segment_95.txt
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The secret behind Oscar's first magic trick has been revealed! Because he still wants to impress Lura, he comes up with a new idea: he still wants to sort a permutation $p_1, p_2, \ldots, p_n$ of $[1, 2, \ldots, n]$.
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This time, he chooses an integer $k$. He wants to sort the permutation in non-decreasing order using the following operation several times:
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1. Pick a continuous subarray of length $k$ and remove it from $p$. 2. Insert the continuous subarray back into $p$ at any position (perhaps, in the very front or the very back).
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To be as impressive as possible, Oscar would like to choose the maximal value of $k$ such that he can sort his permutation. Please help him find the maximal $k$ as well as a sequence of operations that will sort the permutation. You don't need to minimize the number of operations, but you are allowed to use at most $5n^2$ operations.
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We have a proof that, for the maximal $k$ such that you can sort the permutation in any number of operations, you can also sort it in at most $5n^2$ operations.
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The first line contains a single integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases.
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The first line of each test case contains a single integer $n$ ($5 \leq n \leq 10^3$) — the length of the permutation.
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The second line of each test case contains a permutation $p_1, p_2, \ldots, p_n$ of $[1, 2, \ldots, n]$.
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The sum of $n$ over all test cases does not exceed $2 \cdot 10^3$.
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For each test case, first output the chosen value of $k$ on a new line ($1 \leq k \leq n$).
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Then, output a single integer $m$ — the number of operations used ($0 \leq m \leq 5n^2$).
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Then, on each of the next $m$ lines, output the operations denoted by two integers $i$ and $j$ ($1 \leq i, j \leq n - k + 1$), representing an operation where you remove the subarray starting from index $i$ and replace it back into $p$ at index $j$.
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In the first test case, it is enough to move the last four numbers to the front.
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In the second test case, it can be shown that we cannot have $k = 4$ or $k = 5$. With $k =
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