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:
/**
 *
 * @author aditya
 */
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
 
public class SpecialPermutation {
 
	public static void main(String[] args) {
            FastScanner fs=new FastScanner();
            int T=fs.nextInt();
            PrintWriter out=new PrintWriter(System.out);
            for (int tt=0; tt<T; tt++) {
                int n = fs.nextInt();
                System.out.print(2+" ");
                for(int i=3;i<=n;i++){
                    System.out.print(i+" ");
                }
                System.out.println(1);
            }
            out.close();
	}
	
	static class FastScanner {
            BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
            StringTokenizer st=new StringTokenizer("");
            public String next() {
            while (!st.hasMoreElements())
		try {
                    st=new StringTokenizer(br.readLine());
		} catch (IOException e) {
                    e.printStackTrace();
		}
		return st.nextToken();
            }
		
            int[] readArray(int n) {
            int[] a=new int[n]; 
            for (int i=0; i<n; i++) a[i]=nextInt();
		return a;
            }
            int nextInt () {
                return Integer.parseInt(next());
            }
	}
 
}