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.*;
import java.util.*;
import java.math.*;

public class a{
	public static void main(String args[])throws Exception{
		FastScanner fs=new FastScanner();
		int t=fs.nextInt();
		while(t-->0){
			int n=fs.nextInt();
			System.out.print(n+" ");
			for(int i=1;i<n;i++){
				System.out.print(i+" ");
			}
			System.out.print("\n");
		}
		fs.close();
	}
	
	static int gcd(int a,int b){if(a==0)return b;return gcd(b%a,a);}
    static int lcm(int a,int b){return (a/gcd(a,b))*b;}
	
	static class FastScanner{
		InputStreamReader is;
		BufferedReader br;
		StringTokenizer st;
		
		public FastScanner(){ 
			is=new InputStreamReader(System.in);
			br=new BufferedReader(is);
		} 
		
		String next(){ 
			while (st == null || !st.hasMoreElements()){ 
				try{ 
					st = new StringTokenizer(br.readLine()); 
				} 
				catch (IOException  e){ 
					e.printStackTrace(); 
				} 
			} 
			return st.nextToken(); 
		} 
		
		int nextInt(){ 
			return Integer.parseInt(next()); 
		} 
		
		long nextLong(){ 
			return Long.parseLong(next()); 
		}
		
		int[] readArray(int n){
			int arr[]=new int[n];
			for(int i=0;i<n;i++){
				arr[i]=nextInt();
			}
			return arr;
		}
		
		int[][] readArrayTwo(int n,int m){
			int arr[][]=new int[n][m];
			for(int i=0;i<n;i++){
				for(int j=0;j<m;j++){
					arr[i][j]=nextInt();
				}
			}
			return arr;
		}
		
		int[][] readArrayTwo(int n){
			int arr[][]=new int[n][n];
			for(int i=0;i<n;i++){
				for(int j=0;j<n;j++){
					arr[i][j]=nextInt();
				}
			}
			return arr;
		}
		
		String nextLine(){ 
			String str = ""; 
			try{ 
				str = br.readLine(); 
			} 
			catch (IOException e){
				e.printStackTrace(); 
			}
			return str; 
		}
		
		void close() throws Exception{
			is.close();
			br.close();
		}
	}
}