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.*;
public class Main {

  public static void main(String[] args) { 
      Scanner sc = new Scanner(System.in);
      int t = sc.nextInt();
      while(t-- > 0) {
        int n = sc.nextInt();
        int k = sc.nextInt();
        int[] segSum = new int[k * 2 + 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++) {
          int num = arr[i] + arr[n - 1 - i];
          int min = Math.min(arr[i], arr[n - 1 - i]);
          int max = Math.max(arr[i], arr[n - 1 - i]);
          segSum[2] += 2; segSum[min + 1] -= 2;
          segSum[min + 1] += 1; segSum[max + k + 1] -= 1;
          segSum[max + k + 1] += 2; segSum[k * 2 + 1] -= 2;
          segSum[num] -= 1; segSum[num + 1] += 1;
        }

        int ans = segSum[2];
        int sum = ans;
        for(int i = 3; i < k * 2 + 1; i++) {
          sum += segSum[i];
          ans = Math.min(ans, sum);
        }
        System.out.println(ans);
      }
      
  }
}