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 CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;
int a[105];
void biao()
{
	int i;
	for(i=1;i<=100;i++)
	{
		a[i]=i;
	}
}
int main() 
{
	int t,n,i,x,y,z;
	scanf("%d",&t);
	while(t--)
	{
		biao();
	    scanf("%d",&n);
		 if(n%2==0)
		{
		 	for(i=1;i<=n;i=i+2)
		 	{
		 		x=a[i];
		 		a[i]=a[i+1];
		 		a[i+1]=x;
			}
		}
		else
		{
			for(i=1;i<=n-3;i=i+2)
		 	{
		 		x=a[i];
		 		a[i]=a[i+1];
		 		a[i+1]=x;
			}
			y=a[n-2];
			z=a[n-1];
			a[n-2]=a[n];
			a[n-1]=y;
			a[n]=z;
		}
		for(i=1;i<=n-1;i++)
		{
			printf("%d ",a[i]);
		}
		printf("%d\n",a[n]);	
	} 
	return 0;
}