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.*;
import java.io.*;
import java.lang.*;
public class Hello{
	
	public static void main(String[] args)throws IOException{
		MyScanner sc=new MyScanner(System.in);
		
		///* By Santosh.. 
		
	
		int t=sc.nextInt();
		
		while(t-->0) {
			StringBuilder sb=new StringBuilder();
			int n=sc.nextInt();
			int a[]=new int[n];
			
			for(int i=0;i<n-1;i++) {
				a[i]=i+2;
			}
			a[n-1]=1;
			for(int i=0;i<n;i++) {
				sb.append(a[i]+" ");
			}
			System.out.println(sb);
		}
		
		//*/
	}
	
	
	public static class MyScanner{ 
		InputStream inn;
		byte[] buffer = new byte[1 << 13];
		int p=0;
		public MyScanner(String fileName) throws IOException {
			inn= new FileInputStream(fileName);
		}
		public MyScanner(InputStream is) throws IOException {
			inn= is;
		}
		byte nextByte() throws IOException {
			if (p== buffer.length) {
				inn.read(buffer, 0, buffer.length);
				p = 0;
			}
			return buffer[p++];
		}
		BufferedReader read;
		StringTokenizer token;	
        public MyScanner(){     
        	read=new BufferedReader(new InputStreamReader(System.in));
        }
        String next() throws IOException {
			StringBuffer ret = new StringBuffer();
 
			byte b;
			do {
				b = nextByte();
			} while (b <= ' ');
			while (b > ' ') {
				ret.appendCodePoint(b);
				b = nextByte();
			}
 
			return ret.toString();
		}
        int nextInt() throws IOException {
			int ret = 0;
 
			byte b;
			do {
				b = nextByte();
			} while (b <= ' ');
			boolean negative = false;
			if (b == '-') {
				negative = true;
				b = nextByte();
			}
			while (b >= '0' && b <= '9') {
				ret = 10 * ret + b - '0';
				b = nextByte();
			}
 
			return (negative) ? -ret : ret;
		}    
        long nextLong() throws IOException{
            return Long.parseLong(next());
        }
        double nextDouble() throws IOException{
        	return Double.parseDouble(next());
        }
        float nextFloat() throws IOException{
        	return Float.parseFloat(next());
        }
        String nextLine(){
            String str="";
            try{
            	str=read.readLine();
            }
            catch(IOException e){
            	e.printStackTrace();
            }
            return str;
        }	
	}
}