Problem: 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.

Solution (in JAVA):

Solution:
import java.util.*;
public class c1 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		
		while(t-->0)
		{
			int n = sc.nextInt();
			char[] a = sc.next().toCharArray();
			char[] b = sc.next().toCharArray();
			int c = 0;
			ArrayList<Integer> ans = new ArrayList<>();
			for(int i = 0; i < n; i++)
			{
				if(a[0] == b[n-1-i])
				{
					//System.out.print("1 ");
					ans.add(1);
					if(a[0] == '1')
						a[0] = '0';
					else
						a[0] = '1';
					c++;
				}
				
				char[] temp = new char[n-i];
				
				for(int j = 0; j < n-i; j++)
				{
					if(a[n-1-i-j] == '0')
						temp[j] = '1';
					else
						temp[j] = '0';
				}
				
				for(int j = 0; j < n-i; j++)
				{
					a[j] = temp[j];
				}
				
			//	System.out.print((n-i)+" ");
				ans.add((n-i));
				c++;
				
				
				
				
			}
			
			System.out.print(ans.size());
			for(int i = 0; i < ans.size(); i++)
				System.out.print(" "+ans.get(i));
			System.out.println();
		}
	}

}