This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.

There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.

For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.

Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.

Input

The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.

The next two lines contain two binary strings a and b of length n.

It is guaranteed that the sum of n across all test cases does not exceed 1000.

Output

For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.

Example

Input


5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1


Output


3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1

Note

In the first test case, we have 01→ 11→ 00→ 10.

In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.

In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
import java.io.*;
import java.util.*;  
import java.util.Map.Entry;
public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader br  =new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		int t = Integer.parseInt(br.readLine());
		while(t-->0)
		{
			int n = Integer.parseInt(br.readLine());
			String s = br.readLine();
			int[] a = new int[n];
			for (int i = 0; i < a.length; i++) {
				a[i] = Integer.parseInt(s.charAt(i)+"");
			}
			s = br.readLine();
			int[] b = new int[n];
			for (int i = 0; i < b.length; i++) {
				b[i] = Integer.parseInt(s.charAt(i)+"");
			}
			ArrayList<Integer> arr = new ArrayList<Integer>();
			int even=0;
			int odd=a.length-1;
			for (int i = 0; i < b.length; i++) {
				if(i%2==0)
				{
					if(a[even]==b[b.length-1-i])
					arr.add(1);
					even++;
				}
				else if(i%2==1)
				{
					if(a[odd]!=b[b.length-1-i])
					arr.add(1);
					odd--;
				}
				arr.add(b.length-i);
			}
			pw.print(arr.size() + " ");
			for (int i = 0; i < arr.size(); i++) {
				pw.print(arr.get(i) + " ");
			}
			pw.println();
		}
		pw.close();
	}
}