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.io.*;
import java.util.*;
public class M
{
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int test=sc.nextInt();

        while(test>0)
        {
          int size=sc.nextInt();
        int[] m=new int[size/2];
       

 int[] n=new int[size/2];
        int p=0,q=0,a=0,r=1;
            
            if((size/2)%2==1)
            System.out.println("NO");
            else
            {System.out.println("YES");
                for(int j=0;j<(size/2)-1;j++)
                {
                    a=a+2;
                    m[p++]=a;
                    n[q++]=r;
                    r=r+2;
                    
                }
                m[p++]=(a+2);
    int s=0;
   for(int k=0;k<size/2;k++)
   {
       s=s+m[k];
   System.out.print(m[k]+" ");
   
            }
    int s2=0;
       for(int k=0;k<(size/2)-1;k++)
       {
           s2=s2+n[k];
           System.out.print(n[k]+" ");
       }
       System.out.print(s-s2);
     System.out.println();
    
       }
       test--;
        }
    }
}