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 solution
{
    public static void main(String args[])
    {
        Scanner in=new Scanner(System.in);
        int t=in.nextInt();
        while(t-->0)
        {
            int n=in.nextInt();
            int arr[]=new int[n];
            int b[]=new int[n/2];
            int a[]=new int[n/2];
            int k=n/2,j=0,p=0,q=0,sum=0,o=0;
            if(k%2!=0)
             System.out.println("NO");
            else if(k%2==0)
            {
                System.out.println("YES");
                for(int i=2;i<=n;i=i+2)
                {
                 a[j++]=i;
                 sum+=i;
                }
                for(int i=0;i<k-1;i++)
                {
                    b[i]=a[i]-1;
                    o+=b[i];
                }
                b[k-1]=sum-o;
                 for(int i=0;i<k;i++)
                {
                    arr[i]=a[p++];
                }
                 for(int i=k;i<n;i++)
                {
                    arr[i]=b[q++];
                }
                for(int i=0;i<n;i++)
                {
                 System.out.print(arr[i]+" ");
                }
                System.out.println();
            }
        }
    }
}