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 MyClass {
    public static void main(String args[]) {
     Scanner in=new Scanner(System.in);int t=0,n=0;
     if(in.hasNext())
     t=in.nextInt();
     for(int j=0;j<t;j++)
     {
         if(in.hasNext())
         n=in.nextInt();
         
         
         long div=n/2;
         
         if(div%2==1)
         System.out.println("NO");
         else
         {
             System.out.println("YES");
             long a[]=new long[n];
             int half=n/2;
             long k=2;
             boolean b=true;
             for(int i=0;i<n/2;i++)
         {
             a[i]=k;
             if(b==true)
             {a[i+half]=k-1;b=false;}
             else
             {a[i+half]=k+1;b=true;}
             k+=4;
         }
            /*long totevensum=div*(div+1);
            long k=2;
            for(int i=1;i<=div;i++)
            {
                System.out.print(k+" ");
                k+=2;
            }
            k=1;
            long sum=0;*/
            for(int i=0;i<n;i++)
            {
                System.out.print(a[i]+" ");
            }
            System.out.println();
         }
         
     }
    }
}