Problem: You are given an array a consisting of n integers (it is guaranteed that n is even, i.e. divisible by 2). All a_i does not exceed some integer k.

Your task is to replace the minimum number of elements (replacement is the following operation: choose some index i from 1 to n and replace a_i with some integer in range [1; k]) to satisfy the following conditions:

  * after all replacements, all a_i are positive integers not greater than k; 
  * for all i from 1 to n/2 the following equation is true: a_i + a_{n - i + 1} = x, where x should be the same for all n/2 pairs of elements. 



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 first line of the test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k ≤ 2 ⋅ 10^5) — the length of a and the maximum possible value of some a_i correspondingly. It is guratanteed that n is even (i.e. divisible by 2). The second line of the test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the i-th element of a.

It is guaranteed that the sum of n (as well as the sum of k) over all test cases does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5, ∑ k ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — the minimum number of elements you have to replace in a to satisfy the conditions from the problem statement.

Example

Input


4
4 2
1 2 1 2
4 3
1 2 2 1
8 7
6 1 1 7 6 3 4 6
6 6
5 2 6 1 3 4


Output


0
1
4
2

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);
      }
    }
}