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.IOException;
import java.util.Scanner;

public class Main {
	public static int[] f(int n){
		int[] arr = new int[n];
		for (int i = 0; i < arr.length; i++) {
			arr[i]=n-i;
		}
		if(arr.length>2){
		int a = arr[0];
		arr[0]=arr[arr.length/2];
		arr[arr.length/2]=a;}
		return arr;
	}
	
	public static void main(String[] args) throws IOException {
		Scanner sc  = new Scanner(System.in);
		int t = sc.nextInt();
		int[] n = new int[t];
		for (int i = 0; i < t; i++) {
			n[i]=sc.nextInt();
		}
		for (int i = 0; i < n.length; i++) {
			int[] arr = f(n[i]);
			for (int j = 0; j < arr.length; j++) {
				if(j==0)System.out.print(arr[0]);
				else{
				System.out.print(" "+arr[j]);
			}}
			System.out.println();
		}
	}

}