CF1816B.Grid Reconstruction

Consider a $2 \times n$ grid, where $n$ is an even integer. You may place the integers $1, 2, \ldots, 2n$ on the grid, using each integer exactly once.

A path is a sequence of cells achieved by starting at $(1, 1)$, then repeatedly walking either downwards or to the right, and stopping when $(2, n)$ is reached. The path should not extend beyond the grid.

The cost of a path is the alternating sum of the numbers written on the cells in a path. That is, let the numbers written on the cells be $a_1, a_2, \ldots, a_k$ (in the order that it is visited), the cost of the path is $a_1 - a_2 + a_3 - a_4 + \ldots = \sum_{i=1}^k a_i \cdot (-1)^{i+1}$.

Construct a way to place the integers $1, 2, \ldots, 2n$ on the grid, such that the minimum cost over all paths from $(1, 1)$ to $(2, n)$ is maximized. If there are multiple such grids that result in the maximum value, output any of them.

考虑一个 $2 \times n$ 的网格，其中 $n$ 为偶数。你可以将整数 $1, 2, \ldots, 2n$ 恰好各使用一次填入该网格中。

一条路径是指从 $(1, 1)$ 出发，每次只能向下或向右移动，最终到达 $(2, n)$ 所经过的一系列格子（路径不能超出网格边界）。

一条路径的代价定义为该路径所经过格子上数字的交错和。即：设路径依次经过格子上的数字为 $a_1, a_2, \ldots, a_k$，则该路径的代价为

$$a_1 - a_2 + a_3 - a_4 + \ldots = \sum_{i=1}^k a_i \cdot (-1)^{i+1}.$$

请构造一种将整数 $1, 2, \ldots, 2n$ 填入网格的方式，使得所有从 $(1, 1)$ 到 $(2, n)$ 的路径中最小代价尽可能大。若存在多种方案能达到该最大值，输出任意一种即可。

The first line contains a single integer $t$ ($1 \leq t \leq 1000$) — the number of test cases. The description of test cases follows.

The first and the only line of each test case contains a single integer $n$ ($2 \leq n \leq 10^5$, $n$ is even) — the number of the columns in the grid.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

第一行包含一个整数 $t$（$1 \leq t \leq 1000$）—— 测试用例的数量。随后是各测试用例的描述。

每个测试用例仅有一行，包含一个整数 $n$（$2 \leq n \leq 10^5$，且 $n$ 为偶数）—— 网格中的列数。

保证所有测试用例中 $n$ 的总和不超过 $10^5$。

For each test case, output $2$ lines, each containing $n$ integers — the desired grid. If there are multiple solutions, output any of them.

对于每个测试用例，输出 $2$ 行，每行包含 $n$ 个整数——即所要求的网格。若存在多个解，输出任意一个即可。

In the first test case, there are only two paths from cell $(1, 1)$ to cell $(2, 2)$. Their costs are $3-1+4=6$ and $3-2+4=5$. Then the minimum cost is $5$, which is the maximum possible value.

In the second test case, there are four paths from cell $(1, 1)$ to cell $(2, 4)$. Their costs are $8-1+5-3+7=16$, $8-2+5-3+7=15$, $8-2+6-3+7=16$, and $8-2+6-4+7=15$. Then the minimum value is $15$, which is the maximum possible value.

在第一个测试用例中，从单元格 $(1, 1)$ 到单元格 $(2, 2)$ 的路径仅有两条。它们的代价分别为 $3-1+4=6$ 和 $3-2+4=5$。因此最小代价为 $5$，这也是可能的最大值。

在第二个测试用例中，从单元格 $(1, 1)$ 到单元格 $(2, 4)$ 的路径共有四条。它们的代价分别为 $8-1+5-3+7=16$、$8-2+5-3+7=15$、$8-2+6-3+7=16$ 和 $8-2+6-4+7=15$。因此最小值为 $15$，这也是可能的最大值。