C1. XOR Convenience (Easy Version)
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

This is the easy version of the problem. The difference between the versions is that in this version, $$$2 \le i \le n-1$$$. Note that a correct solution for the hard version is not necessarily a correct solution for the easy version.

Given a natural number $$$n$$$. Find a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$, such that for each $$$i$$$ ($$$\style{color:red}{2 \le i \le n-1}$$$) there exists a $$$j$$$ ($$$\style{color:red}{i \le j \le n}$$$) such that $$$p_i = p_j \oplus i$$$ $$$^{\text{†}}$$$.

It can be proven that under the constraints of the problem, there exists at least one suitable permutation $$$p$$$.

$$$^{\text{∗}}$$$A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

$$$^{\text{†}}$$$$$$\oplus$$$ denotes the bitwise XOR operation.

Input

Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The only line of each test case contains a single integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the length of the permutation.

It is guaranteed that the sum of $$$n$$$ across all test cases does not exceed $$$2\cdot 10^5$$$.

Output

For each test case, output $$$n$$$ integers $$$p_1,p_2,\ldots,p_n$$$ — the permutation $$$p$$$.

If there are multiple solutions, you may output any of them.

Example
Input
2
3
6
Output
2 1 3
3 6 2 5 1 4
Note

In the first test case, the permutation $$$p = [2,1,3]$$$ is suitable since $$$p_2 = 1$$$ and $$$p_3 \oplus 2 = 1$$$.

In the second test case, the permutation $$$p = [3,6,2,5,1,4]$$$ is suitable, as:

  • $$$p_2 = 6 = 4 \oplus 2 = p_6 \oplus 2$$$
  • $$$p_3 = 2 = 1 \oplus 3 = p_5 \oplus 3$$$
  • $$$p_4 = 5 = 1 \oplus 4 = p_5 \oplus 4$$$
  • $$$p_5 = 1 = 4 \oplus 5 = p_6 \oplus 5$$$