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.Scanner;
public class G {
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
			int k=sc.nextInt();
			int a[]=new int [n+2];
			int d[]=new int [2*k+10];
			//注意差分数组求的是改变的次数，而不是求定值x
			for (int i = 1; i <=n; i++) {
				a[i]=sc.nextInt();
			}
			for (int i = 1; i <=n/2; i++) {
				int sum=a[i]+a[n-i+1];
				int max=Math.max(a[i], a[n-i+1]);
				int min=Math.min(a[i], a[n-i+1]);
				//2-min
				d[2]+=2;
				d[min+1]-=2;
				//min+1-sum
				d[min+1]+=1;
				d[sum]--;
				//sum~,max+k;
				d[sum+1]++;
				d[max+k+1]--;
				//max+k+1~2*k
				d[max+k+1]+=2;
				d[2*k+1]-=2;
			}
			int min=Integer.MAX_VALUE;
			for (int i = 3; i <=2*k+1; i++) {
				d[i]+=d[i-1];
				min=Math.min(min, d[i-1]);
			}
			System.out.println(min);
		}
	}
}