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.*;
public class Main
{
  public static void main(String[] args)
  {
    Scanner read = new Scanner(System.in);
    
    int t = read.nextInt();
    for(int x = 1; x<=t; x++)
    {
      int n = read.nextInt();
      if(n % 2 == 0)
      {
        for(int i = n; i>=1; i--)
        {
          System.out.print(i+" ");
        }
      }     
      else if(n == 3)
      {
        System.out.print("3 1 2");
      }
      else
      {
        for(int i = n / 2 + 1; i<=n-1; i++)
        {
          System.out.print(i+" ");
        }
        System.out.print(n+" ");
        for(int i = n / 2; i>=1; i--)
        {
          System.out.print(i+" ");
        }
      }
      
      System.out.println();
    }
  }
}