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.*;
public class  test{

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