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.io.*;
import java.util.*;



public class Main
{

   public static void main(String[] args) throws IOException
   {
      BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
      int T = Integer.parseInt(input.readLine());
      for(int t = 0; t < T; t++)
      {
         StringTokenizer st = new StringTokenizer(input.readLine());
         int N = Integer.parseInt(st.nextToken());
         int K = Integer.parseInt(st.nextToken());
         int[] arr = new int[N];
         st = new StringTokenizer(input.readLine());
         for(int i = 0; i < N; i++)
         {
            arr[i] = Integer.parseInt(st.nextToken());
         }
         int[] freq = new int[2 * K + 1];
         int[] possible = new int[2 * K + 2];
         for(int i = 0; i < N / 2; i++)
         {
            freq[arr[i] + arr[N - 1 - i]]++;
            possible[1 + Math.min(arr[i], arr[N - 1 - i])]++;
            possible[K + 1 + Math.max(arr[i], arr[N - 1 - i])]--;
         }
         for(int i = 1; i < 2 * K + 2; i++)
         {
            possible[i] += possible[i - 1];
         }
         int result = Integer.MAX_VALUE;
         for(int i = 0; i < 2 * K + 1; i++)
         {
            int alt = (possible[i] - freq[i]) + 2 * (N / 2 - possible[i]);
            if(alt < result) result = alt;
         }
         System.out.println(result);
      }
   }

}