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 sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- != 0) {
            int n, k;
            n = sc.nextInt();
            k = sc.nextInt();
            int[] co = new int[2 * k + 2];
            int[] sum = new int[2 * k + 2];
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = sc.nextInt();
            }
            for (int i = 0; i < n / 2; i++) {
                co[Math.min(arr[i], arr[n - i - 1]) + 1]++;
                co[Math.max(arr[i], arr[n - i - 1]) + k + 1]--;
                sum[arr[i] + arr[n - i - 1]]++;
            }
            int ans = n;
            for (int i = 0; i < k * 2; i++) {
                co[i + 1] += co[i];
                ans = Math.min(ans, 2 * (n / 2 - co[i + 1]) + co[i + 1] - sum[i + 1]);
            }
            System.out.println(ans);
        }
    }
}