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 Main {
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        
        int T = in.nextInt();
        int n, k;
        int a[] = new int[200010];
        int min[] = new int[400010];
        int max[] = new int[400010];
        int sum[] = new int[400010];
        
        while(T > 0) {
            T--;
            
            n = in.nextInt();
            k = in.nextInt();

            for(int i = 1; i <= n; ++i) {
                a[i] = in.nextInt();
            }
            for(int i = 1; i <= (k << 1); ++i) 
                min[i] = sum[i] = max[i] = 0;

            for(int i = 1; i <= (n >> 1); ++i) {
                sum[a[i] + a[n - i + 1]]++;
                min[Math.min(a[i], a[n - i + 1]) + 1]++;
                max[Math.max(a[i], a[n - i + 1]) + k]++;
            }

            int ans = n, pre_sum = 0;

            for(int i = 2; i <= (k << 1); ++i) {
                pre_sum += min[i] - max[i - 1];
                ans = Math.min((n / 2 - pre_sum) * 2 + pre_sum - sum[i], ans);
            }
            
            System.out.println(ans);
        }

        in.close();
    }
}