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;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- != 0) {
            int n, k;
            n = sc.nextInt();
            k = sc.nextInt();
            int[] co = new int[2 * k + 2];
            int[] sum = new int[2 * k + 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++) {
                co[Math.min(arr[i], arr[n - i - 1]) + 1]++;
                co[Math.max(arr[i], arr[n - i - 1]) + k + 1]--;
                sum[arr[i] + arr[n - i - 1]]++;
            }
            int ans = n;
            for (int i = 0; i < k * 2; i++) {
                co[i + 1] += co[i];
                ans = Math.min(ans, 2 * (n / 2 - co[i + 1]) + co[i + 1] - sum[i + 1]);
            }
            System.out.println(ans);
        }
    }
}