Let's define the value of the permutation $p$ of $n$ integers $1$, $2$, ..., $n$ (a permutation is an array where each element from $1$ to $n$ occurs exactly once) as follows:

initially, an integer variable $x$ is equal to $0$;

if $x < p_1$, then add $p_1$ to $x$ (set $x = x + p_1$), otherwise assign $0$ to $x$;

if $x < p_2$, then add $p_2$ to $x$ (set $x = x + p_2$), otherwise assign $0$ to $x$;

...

if $x < p_n$, then add $p_n$ to $x$ (set $x = x + p_n$), otherwise assign $0$ to $x$;

the value of the permutation is $x$ at the end of this process.

For example, for $p = [4, 5, 1, 2, 3, 6]$, the value of $x$ changes as follows: $0, 4, 9, 0, 2, 5, 11$, so the value of the permutation is $11$.

You are given an integer $n$. Find a permutation $p$ of size $n$ with the maximum possible value among all permutations of size $n$. If there are several such permutations, you can print any of them.


-----Input-----

The first line contains one integer $t$ ($1 \le t \le 97$) — the number of test cases.

The only line of each test case contains one integer $n$ ($4 \le n \le 100$).


-----Output-----

For each test case, print $n$ integers — the permutation $p$ of size $n$ with the maximum possible value among all permutations of size $n$.


-----Examples-----

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


-----Note-----

None

Difficulty: EASY

Solutions:
for _ in range(int(input())):
	a = []
	n = int(input())
	if 'erfjijferjr32r32r32' == 232434323235566:
		print('Tanmay')
	if n & 1:
		for i in range(n):
			if i % 2 == 1:
				a.append(i + 2)
				continue
			a.append(i)
		a[0] = 1
	else:
		for i in range(n):
			if i % 2 == 0:
				a.append(i + 2)
				continue
			a.append(i)
	a[n - 1] = n
	a[n - 2] = n - 1
	for i in a:
		print(i, end=' ')
	print()