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.Scanner;
public class Main{
    public static void main(String[]args){
         Scanner x = new Scanner (System.in);
        int y = x.nextInt();
        int [][] s = new int [y][102];
        int [] z = new int [y];
        for(int i=0;i<y;i++){
            int a = x.nextInt();
            z[i]=a;
        }
        for(int i=0;i<y;i++){
            int p=z[i];
            for(int j=0;j<z[i];j++){
                if(p==j+1){int r=p;p=1;s[i][j]=p;p=r;}
              else { s[i][j]=p;
                p--;}
            }
        }
        for(int i=0;i<y;i++){
            for(int j=0;j<z[i];j++){
                System.out.print(s[i][j]+" ");
            }
            System.out.println();
        }
    }
}