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:
t = int(input())
x = []
for i in range(t):
	n = int(input())
	x.append(n)
for i in range(t):
	y = []
	if x[i] == 2:
		y.append(2)
		y.append(1)
		for j in range(2):
			print(y[j], end=" ")
		print()
	elif x[i] == 3:
		y.append(2)
		y.append(3)
		y.append(1)
		for j in range(3):
			print(y[j], end=" ")
		print()
	else:
		b = 3
		for j in range(x[i] - 2):
			y.append(b)
			b += 1
		y.append(1)
		y.append(2)	
		for k in range(x[i]):
			print(y[k], end=" ")
		print()