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 JAVA):

Solution:
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;
public class Nov22 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner obj= new Scanner(System.in);
		int t=obj.nextInt();
		while(t-->0)
		{
			int n=obj.nextInt();
			int[] arr=new int[n];
			int k=n;
			for(int i=0;i<n;i++)
			{
				arr[i]=k;
				k--;
			}
			if(n%2!=0)
			{
				int te=arr[0];
				arr[0]=arr[((n+1)/2)-1];
				arr[((n+1)/2)-1]=te;
			}
			for(int i=0;i<n;i++)
			{
				System.out.print(arr[i]+" ");
			}
			System.out.println();
		}


	}

}