Problem: You are given an array a consisting of n integers.

In one move, you can choose some index i (1 ≤ i ≤ n - 2) and shift the segment [a_i, a_{i + 1}, a_{i + 2}] cyclically to the right (i.e. replace the segment [a_i, a_{i + 1}, a_{i + 2}] with [a_{i + 2}, a_i, a_{i + 1}]). 

Your task is to sort the initial array by no more than n^2 such operations or say that it is impossible to do that.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (3 ≤ n ≤ 500) — the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 500), where a_i is the i-th element a.

It is guaranteed that the sum of n does not exceed 500.

Output

For each test case, print the answer: -1 on the only line if it is impossible to sort the given array using operations described in the problem statement, or the number of operations ans on the first line and ans integers idx_1, idx_2, ..., idx_{ans} (1 ≤ idx_i ≤ n - 2), where idx_i is the index of left border of the segment for the i-th operation. You should print indices in order of performing operations.

Example

Input


5
5
1 2 3 4 5
5
5 4 3 2 1
8
8 4 5 2 3 6 7 3
7
5 2 1 6 4 7 3
6
1 2 3 3 6 4


Output


0

6
3 1 3 2 2 3 
13
2 1 1 6 4 2 4 3 3 4 4 6 6 
-1
4
3 3 4 4 

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();
			}
		}
	}
}