Problem: You are given a positive integer n, it is guaranteed that n is even (i.e. divisible by 2).

You want to construct the array a of length n such that: 

  * The first n/2 elements of a are even (divisible by 2); 
  * the second n/2 elements of a are odd (not divisible by 2); 
  * all elements of a are distinct and positive; 
  * the sum of the first half equals to the sum of the second half (∑_{i=1}^{n/2} a_i = ∑_{i=n/2 + 1}^{n} a_i). 



If there are multiple answers, you can print any. It is not guaranteed that the answer exists.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of the array. It is guaranteed that that n is even (i.e. divisible by 2).

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — "NO" (without quotes), if there is no suitable answer for the given test case or "YES" in the first line and any suitable array a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) satisfying conditions from the problem statement on the second line.

Example

Input


5
2
4
6
8
10


Output


NO
YES
2 4 1 5
NO
YES
2 4 6 8 1 3 5 11
NO

Solution (in JAVA):

Solution:

import java.util.*;
public class Main
{
	public static void main(String[] args) {
	    Scanner s=new Scanner(System.in);
		int t=s.nextInt();
		while(t!=0){
		    int n=s.nextInt();
		    int k=n/2;
		    int v=0,f=0;
		    int e=2;
		    int o=1;
		    int sume=0,sumo=0;
		    int a[]=new int[k];
		    int b[]=new int[k];
		    int c[]=new int[n];
		    if(k%2!=0){
		        System.out.println("NO");
		    }
		    
		    else{
		        while(v<k){
		            a[v]=e;
		            sume=sume+a[v];
		            v++;
		            e=e+2;
		        }
		        
		        while(f<k){
		            b[f]=o;
		            sumo=sumo+b[f];
		            f++;
		            o=o+2;
		        }
		        
		        int x=sume-sumo;
		        b[k-1]=b[k-1]+x;
		        int j=0;
		        for(int i=0;i<k;i++){
		            c[j]=a[i];
		            j++;
		        }
		        
		        for(int i=0;i<k;i++){
		            c[j]=b[i];
		            j++;
		        }
		        
		        System.out.println("YES");
		        for(int i=0;i<n;i++){
		            System.out.print(c[i]+" ");
		        }
		        
		        System.out.println();
		        
		            
		        }
		        t--;
		        
		    }
		    
		}
	
}