Problem: You are given one integer n (n > 1).

Recall that 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 of length 5, 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).

Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).

You have to answer t independent test cases.

If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.

Output

For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).

If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.

Example

Input


2
2
5


Output


2 1
2 1 5 3 4

Solution (in PYTHON3):

Solution:
import math
for i in range(int(input())):
	n = int(input())
	if n % 2 == 0:
		for i in range(n, 0, -1):
			if i == 1:
				print(i)
			else:
				print(i, end=" ")
	else:
		nn = n
		if n == 3:
			print("3 1 2")
		else:
			for i in range(0, n // 2):
				print(n, end=" ")
				n -= 1
			print(math.floor(nn/2), math.ceil(nn/2), sep=" ", end=" ")
			for j in range(int(math.floor(nn/2))-1, 0, -1):
				if j == 1:
					print(j)
				else:
					print(j, end=" ")