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
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.OutputStreamWriter;
import java.util.Arrays;
import java.util.Scanner;

public class d {
    public static void main(String[] args) throws IOException {
        Scanner std = new Scanner(System.in);
        BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
	    StringBuilder strBu = new StringBuilder();

        int t = std.nextInt();
        while (t-- > 0) {
            int n = std.nextInt();
            int k = std.nextInt();
            int[] vals = new int[n];
            int[] noChange = new int[k*2 + 4];
            int[] oneChange = new int[k*2 + 4];
            for (int i = 0; i < n; i++) {
                vals[i] = std.nextInt();
            }
            for (int i = 0; i < n/2; i++) {
                int v1 = vals[i];
                int v2 = vals[n - 1 - i];

                int total = v1 + v2;
                noChange[total]++;

                int minVal = Math.min(v1, v2) + 1;
                int maxVal = Math.max(v1, v2) + k + 1;
                oneChange[minVal]++;
                oneChange[maxVal]--;
            }
            long prefixSum = 0;
            long ans = Integer.MAX_VALUE;
//            System.out.println(Arrays.toString(noChange));
//            System.out.println(Arrays.toString(oneChange));
            for (int i = 2; i <= 2*k; i++) {
                //i is the x val we will try
                long pairsNoChange = noChange[i]; //# of pairs that are perfect for i
                prefixSum += oneChange[i]; //from 2...i sum
                long pairsOneChange = prefixSum - pairsNoChange; //# of pairs that need 1 value changed for i
                long pairsTwoChange = n/2 - (pairsNoChange + pairsOneChange);
                long iAns = pairsOneChange + pairsTwoChange*2;
                ans = Math.min(ans, iAns);
            }
            log.write(ans + "\n");
        }
        log.flush();
	

    }
}

/**
Remember to check for overflow
Does number need to be greater than a lower bound?
**/

