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.ArrayList;
import java.util.Scanner;

public class SpecialPermutation {

	public static void main(String[] args) {
		Scanner s=new Scanner(System.in);
		int t=s.nextInt();
		ArrayList<ArrayList<Integer>> last=new ArrayList<ArrayList<Integer>>();
		for(int i=0;i<t;i++) {
			int n=s.nextInt();
			last.add(permutation(n));
		}
		for(int i=0;i<last.size();i++) {
			for(int j=0;j<last.get(i).size();j++) {
				System.out.print(last.get(i).get(j)+" ");
			}
			System.out.println();
		}
		
		
	}

	private static ArrayList<Integer> permutation(int n) {
		ArrayList<Integer> temp=new ArrayList<Integer>(); 
		for(int i=n;i>0;i--){
			temp.add(i);
		}
		if(n%2!=0) {
			int tempno=temp.remove(n/2);
			temp.add(tempno);
			
		}
		return temp;
		
	}

}