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.Scanner;

/**
 * 21/04/20
 * Created by Himanshu
 **/
public class ConstantPalindromeSum {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int t = s.nextInt();
        while (t-- > 0) {
            int n = s.nextInt();
            int k = s.nextInt();
            int [] arr = new int [n];
            for (int i=0;i<n;i++) arr[i] = s.nextInt();
            long [] sum = new long[2*k+1];
            long [] a = new long[2*k+2];
            for (int i=0;i<n/2;i++) {
                sum[arr[i] + arr[n-1-i]]++;
                int max = Math.max(arr[i],arr[n-1-i]) + k;
                int min = Math.min(arr[i],arr[n-1-i]) + 1;
                a[min]++;
                a[max+1]--;
            }
            for (int i=1;i<a.length;i++) a[i] += a[i-1];
            long max = Integer.MAX_VALUE;
            for (int i=0;i<sum.length;i++) {
                long x = a[i] - sum[i] + (n/2 - a[i])*2;
                max = Math.min(max,x);
            }
            System.out.println(max);
        }
    }
}