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 MyClass {
    public static void main(String args[]) {
      Scanner sc=new Scanner(System.in);
      int q=sc.nextInt();
      while(q-->0)
      {
          int n=sc.nextInt();
          int k=sc.nextInt();
          int a[]=new int[n];
          for(int i=0;i<n;i++)
          {
              a[i]=sc.nextInt();
          }
          
          //Logic part
          int pres_sum[]=new int[2*k+2];
          int freq[]=new int[2*k+2];
          int min_v=0,max_v=0,l=0,r=0;
          for(int i=0;i<n/2;i++)
          {
              min_v=Math.min(a[i],a[n-i-1]);
              max_v=Math.max(a[i],a[n-i-1]);
              pres_sum[min_v+max_v]++;
              freq[min_v+1]++;
              freq[max_v+k+1]--;
              //System.out.print(freq[i]+" ");
          }
          
          for(int i=2;i<(2*k+1);i++)
          {
              freq[i]=freq[i]+freq[i-1];
              //System.out.print(freq[i]+" ");
          }
          int min_change=Integer.MAX_VALUE;
          int chan_0=0,chan_1=0,chan_2=0,tot_change=0;
          for(int i=2;i<(2*k+1);i++)
          {
              tot_change=0;
              chan_0=pres_sum[i];
              chan_1=freq[i]-chan_0;
              chan_2=(n/2)-chan_1-chan_0;
              if(chan_2>0)
              {
                  tot_change=(chan_2*2);
              }
              tot_change+=chan_1;
              //System.out.println(tot_change+" for "+i+" "+pres_sum[i]+" "+chan_1+" "+chan_2);
              if(min_change>tot_change)
              {
                  min_change=tot_change;
              }
          }
          System.out.println(min_change);
      }
    }
}