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 Main
{
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int tes=sc.nextInt();
            for(int ii=0;ii<tes;ii++)
            {
                int n=sc.nextInt();
                int[] ar=new int[n/2];
                int[] br=new int[n/2];
                int p=0,q=0;
                int l=0,r=1;
                if((n/2)%2==1)
                System.out.println("NO");
                else
                {   System.out.println("YES");
                    for(int j=0;j<(n/2)-1;j++)
                    {
                        l=l+2;
                        ar[p++]=l;
                        br[q++]=r;
                        r=r+2;
                    }
                    ar[p++]=(l+2);
            int su=0,su2=0;
            for(int k=0;k<n/2;k++)
            {
                su=su+ar[k];
                System.out.print(ar[k]+" ");
            }
            for(int k=0;k<(n/2)-1;k++)
            {
            su2=su2+br[k];
            System.out.print(br[k]+" ");
            }
            System.out.print(su-su2);
            System.out.println();
       }
    }
    }
}