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 Balances_Array {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int test = sc.nextInt();
        for (int i = 0; i < test; i++) {
            int size = sc.nextInt();
            if (size % 4 != 0) {
                System.out.println("NO");
            } else {
                System.out.println("YES");
                int[] elements = new int[size];
                int ecount = 0;
                for (int j = 0, k = 2; j < size / 2; j++, k = k + 2) {
                    elements[j] = k;
                    ecount = ecount + k;
                    System.out.print(elements[j]+"\t");
                }
                int ocount = 0;
                for (int j = size / 2, k = 1; j < size - 1; j++, k = k + 2) {
                    elements[j] = k;
                    ocount = ocount + k;
                    System.out.print(elements[j]+"\t");
                }
                elements[size - 1] = ecount - ocount;
                System.out.println(elements[size-1]);
            }
        }
    }
}