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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
public class CF_636_D {

	public static void main(String[] args) {

        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int t = in.nextInt(); // Scanner has functions to read ints, longs, strings, chars, etc.
        for (int i = 1; i <= t; ++i) {
         int n = in.nextInt();
         int k = in.nextInt();
         int arr[] = new int[n];
         HashMap<Integer,Integer> hash = new HashMap<Integer, Integer>();
//         int range[][] = new int[n/2][2];
         int[] pre = new int[2*k +2];
         for(int j =0; j <n; j++) {
        	 arr[j] = in.nextInt();
         }
         for(int j =0; j < n/2; j++) {
        	 if(hash.containsKey(arr[j]+ arr[n-j-1])) {
        		 int temp = hash.get(arr[j]+ arr[n-j-1]);
        		 hash.put(arr[j] + arr[n-j-1], ++temp);
        	 }else {
        		 hash.put(arr[j] + arr[n-j-1], 1);  }
        	 
        	 int min = Math.min(arr[j], arr[n-j-1]);
//        	 range[j][0] = min+1;
        	 int max= arr[j]+ arr[n-j-1]- min;
        	 pre[min+1]++;
        	 pre[max+1+k]--;
         }
         
         for(int j = 3; j < 2*k +1; j++) {
        	 pre[j] = pre[j-1] + pre[j];
         }
         int min = Integer.MAX_VALUE;
         for(int k1 = 2; k1 <= 2*k; k1++) {
        	 int a= 0; //number of pairs that require zero change to get k1
        	 if(hash.containsKey(k1)) {
        		 a = hash.get(k1);}
        	 int b = pre[k1] - a; //number of pairs that require 1 change to get k1 

        	 
//        	 #############causing TLE
//        		 for(int j =0; j < n/2; j++) {
//            		 if(k1 >= range[j][0] && k1 <= range[j][1]) {
//            			 if(k1 != range[j][0] -1 + range[j][1] - k) {
//            				b++; 
//            			 }
//            			 
//            		 }
//            	 }
        	 
        	 
        	 int c = n/2 -a - b; //number of pairs needing 2 changes 
        	 min = Math.min(min, b+ 2*c);	 
         }
         System.out.println(min);
        
        }
      }

	
}