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.*;
import java.lang.*;
import java.io.*;  
import java.lang.Math;
public class commitment
{       
	static void fun(int inp )
	{  if(inp%2==0)
		{
		for(int a = inp ; a>0 ; a--)
		{
			System.out.print(a + " ") ;
		}
		System.out.println(); 
       }
       else
       {      int ans = inp/2 ; 
       	      int index = inp - ans  ;
          for(int x = inp ;x>index  ; x--)
          {
          	System.out.print(x + " ")  ;
          }
          for(int y = 1 ; y<=ans+1 ;y++)
          {
          	System.out.print(y + " ") ;
          }
          System.out.println() ;
       }





	}
public static void main(String[] args) {
   	    Scanner sc = new Scanner(System.in) ;
   	       	int n = sc.nextInt() ;
   	       	while(n>0)
   	       	{

                      int inp = sc.nextInt() ;
                      fun(inp) ;

   	       		n-- ;
   	       	}


          
   	   
   }   
   	}