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 geek {

   /* public static int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }
    public static boolean prime(int x){
        if(x==0||x==1){
            return  false;
        }
        boolean b[]=new boolean[10000001];
        Arrays.fill(b,true);
        b[0]=false;
        b[1]=false;
        for(int i=2;i*i<=1000000;i++){
            if(b[i]){
                for(int j=i*i;j<=1000000;j+=i){
                    b[j]=false;
                }
            }
        }
        if(b[x])return true;
        return false;

    }

     */

/*public static boolean check(long x){
    for(long i=2;i<=Math.sqrt(x);i++){
        if(x%i==0){
            return false;
        }
    }
    return true;
}
public static boolean pow(long x){
    if(x==0)return false;
    if(x==1)return true;
    while(x!=1){
        if(x%2!=0){
            return false;
        }
        x=x/2;
    }
    return true;
}*/

    public static void main(String[] args) {
        try {
            Scanner s = new Scanner(System.in);

            int t=s.nextInt();
            while(t-->0){
                int n=s.nextInt();
               StringBuffer sb=new StringBuffer();
               int k=n;
               for(int i=0;i<n;i++){
                   if(k==(i+1)){

                       sb.append((k-1)+" ");

                           sb.append((k) + " ");
                        i++;
                       k-=2;
                   }else {
                       sb.append((k--) + " ");
                   }
               }
                System.out.println(sb);
            }






        } catch (Exception e) {
        }
    }
}