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.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.Scanner;

public class b {
	
	
	public static void main(String[] args) 
	{
		Scanner s=new Scanner(System.in);

		int t=s.nextInt();
	
		for(int ie=0;ie<t;ie++) {
			
			int n=s.nextInt();
			int k=s.nextInt();
			int[] arr=new int[n];
			for(int i=0;i<n;i++) {
				arr[i]=s.nextInt();
			}

			int[] zero=new int[2*k+1];
			int one[] =new int[2*k+1];
			
			for(int i=0;i<n/2;i++) {
			int sum=arr[i]+arr[n-1-i];
				zero[sum]++;
				int l=sum-Math.max(arr[i], arr[n-1-i])+1;
				int r=sum-Math.min(arr[i], arr[n-1-i])+k;
				one[l]++;
				if(r<2*k) {
					one[r+1]--;
				}
			}
			
			int ans=Integer.MAX_VALUE;
			for(int i=2;i<=2*k;i++) {
				one[i]=one[i]+one[i-1];
				int now=one[i]-zero[i]+2*((n/2)-one[i]);
				if(now<ans) {
					ans=now;
				}
			}
			System.out.println(ans);
			
			
			
			
		}
	}
	
}