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 Main
{
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		while(T-->0)
		{
		    int n = sc.nextInt();
		    
		    Integer arr[]=new Integer[n];
		    for(int i=0;i<n;i++)
		    arr[i]=i+1;
		    
		    Arrays.sort(arr,Collections.reverseOrder());
		    
		    for(int i=0;i<n;i++)
		    {
		    if(i==n/2&&n>2&&n%2!=0)
		    {
		        int temp=arr[i];
		        arr[i]=arr[i-1];
		        arr[i-1]=temp;
		    }
		    }
		    for(int i=0;i<n;i++)
		    System.out.print(arr[i]+" ");
		    System.out.println();
		}
	}
}