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.*;
import java.util.*;

public class Main{
	static int N = 550;
	static int[] a = new int[N];
	static List<int[]> all = new ArrayList<int[]>();
	
	static void swap(int[] a, int i, int j) {
		int tmp = a[i];
		a[i] = a[j];
		a[j] = tmp;
	}
	
	static void work(int[] a, int idx) {
		swap(a, idx+1, idx+2);
		swap(a, idx, idx+1);
	}
	
	public static void main(String[] args) throws IOException{
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		
		int T = Integer.parseInt(in.readLine());
		
		while(T -- > 0) {
			all.clear();
			
			int n = Integer.parseInt(in.readLine());
			String[] arr = in.readLine().split(" ");
			for(int i=1; i<=n; i++) {
				a[i] = Integer.parseInt(arr[i-1]);
				all.add(new int[]{a[i], i});
			}
			
			Collections.sort(all, new Comparator<int[]>() {

				@Override
				public int compare(int[] o1, int[] o2) {
					// TODO Auto-generated method stub
					return o1[0] - o2[0];
				}
				
			});
			
			for(int i=1; i<=n; i++) {
				a[all.get(i-1)[1]] = i;
			}
			
			int cnt = 0; // 统计逆序对的数量
			for(int i=1; i<=n; i++) {
				for(int j=i+1; j<=n; j++) {
					if(a[j] < a[i]) cnt ++;
				}
			}
			
			if(cnt % 2 != 0) {
				for(int i=1; i<n; i++) {
					if(all.get(i-1)[0] == all.get(i)[0]) {
						swap(a, all.get(i-1)[1], all.get(i)[1]);
						cnt ++;
						break;
					}
				}
			}

			List<Integer> ans = new ArrayList<Integer>();
			
			if(cnt % 2 != 0) System.out.println(-1);
			else {
				for(int i=1; i<=n-2; i++) {
					int min = 550; int pos = -1;
					// 找到最小值及其下标
					for(int j=i; j<=n; j++) {
						if(a[j] < min) {
							min = a[j];
							pos = j;
						}
					}
					
					while(pos >= i + 2) {
						work(a, pos-2);
						ans.add(pos - 2);
						pos -= 2;
					}
					
					if(pos != i) {
						work(a, pos-1);
						work(a, pos-1);
						ans.add(pos-1);
						ans.add(pos-1);
					}
				}
				
				System.out.println(ans.size());
				for(int i=0; i<ans.size(); i++) System.out.print(ans.get(i) + " ");
				System.out.println();
			}
		}
	}
}