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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;

public class Codechef {
	
	
	public static void main (String[] args) throws java.lang.Exception
	{
	    BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); 
  
        int t = Integer.parseInt(br.readLine());
        for(int i=0;i<t;i++) {
        	int n=Integer.parseInt(br.readLine());
        	int[] arr=new int[n];
        	for(int j=0;j<n;j++) {
        		if(j==0) {
        			arr[j]=n;
        		}
        		else {
        			arr[j]=j;
        		}
        	}
        	System.out.println(Arrays.toString(arr).replaceAll("[\\[|\\]|\\,]", ""));
        	
        }
	}

}