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.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Locale;
import java.util.Objects;
import java.util.Queue;
import java.util.Scanner;
import java.util.Stack;
import java.util.TreeSet;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;

public class Main {

    public static void fill(int[] permutaciones, int max) {
        for (int i = 1; i < max + 1; i++) {
            permutaciones[i - 1] = i;
        }
    }

    public static int[] reor(int[] permutaciones) {
        for (int i = 0; i < permutaciones.length; i++) {
            if (permutaciones[i] == i + 1 && i != 0) {
                int trash = permutaciones[i];
                permutaciones[i] = permutaciones[i - 1];
                permutaciones[i - 1] = trash;
                return reor(permutaciones);
            } else if (permutaciones[i] == i + 1 && i == 0) {
                int trash = permutaciones[i];
                permutaciones[i] = permutaciones[i + 1];
                permutaciones[i + 1] = trash;
                return reor(permutaciones);
            }
        }
        return permutaciones;
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int casos = Integer.parseInt(br.readLine());
        while (casos > 0) {
            int permutacion = Integer.parseInt(br.readLine());
            int[] permutaciones = new int[permutacion];
            fill(permutaciones, permutacion);
            reor(permutaciones);
            for (int i = 0; i < permutacion; i++) {
                if(i==permutacion-1){
                    System.out.println(permutaciones[i]);
                }else{
                    System.out.print(permutaciones[i] + " ");
                }
            }
            casos--;
        }
    }
}