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 Classmain {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while(t-->0){
            int n = in.nextInt(),sum1 = 0,sum2 = 0;
            if (n/2%2 == 1){
                System.out.println("NO");
            }else {
                System.out.println("YES");
                int []A = new int[n];
                for (int i = 0,k=2; i <n/2 ; i++) {
                    A[i] = k;
                    k+=2;
                    sum1 += A[i];
                }
                for (int i = n/2,k=1; i < n -1 ; i++) {
                    A[i] = k;
                    k+=2;
                    sum2 += A[i];
                }
                A[n-1] = sum1 - sum2;
                for (int i = 0; i < n; i++) {
                    System.out.print(A[i] + " ");
                }
                System.out.println();
            }
        }
    }
}