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.*;

/**
 * Spend 5 to 10 minutes analyzing different approaches to solve Don't rush and
 * start coding first solution that comes to mind Right now best strategy is
 * analyze for 20 minutes and code 20 minutes 3 question 2 hour. Remaining 30
 * minutes is miscellaneous Simply coding and coding won't solve problem,
 * analyzing then coding will solve it Right now forget speed focus on
 * accuracy****
 */
public class Codeforces {

	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		while (t-- > 0) {
			int n = in.nextInt(), k = in.nextInt();
			int m = 2*k+2;
			int a[] = new int[n];
			for (int i=0;i<n;i++) a[i] = in.nextInt();
			
			int zero[] = new int[m];
			Arrays.fill(zero, 0);
			for (int i=0;i<n/2;i++) {
				int s = a[i] + a[n-i-1];
				zero[s]++;
			}
			
			int one[] = new int[m];
			Arrays.fill(one, 0);
			for (int i=0;i<n/2;i++) {
				int min = Math.min(a[i], a[n-i-1]);
				int max = Math.max(a[i], a[n-i-1]);
				one[min+1]++;
				one[max+k+1]--;
			}
			
			for (int i=1;i<m;i++) one[i] = one[i] + one[i-1];
			
			int min = Integer.MAX_VALUE;
			for (int sum=2;sum<=2*k;sum++) {
				int cnt = (one[sum] - zero[sum]) + (n/2-one[sum])*2;
				min = Math.min(cnt, min);
			}
			System.out.println(min);
		}
		in.close();
	}

}