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.util.*;

public class PrefixFlip {
	static ArrayList<Integer> output = new ArrayList<Integer>();
	public static void main(String[] args) {
		Scanner sc= new Scanner(System.in);
		int t = sc.nextInt();
		while(t-->0)
		{
			int size = sc.nextInt();
			String start = sc.next();
			String end = sc.next();
			recur(start, end, size);
			System.out.print(output.size() + " ");
			for(int i = 0; i < output.size(); i++)
			{
				System.out.print(output.get(i) + " ");
			}
			System.out.println();
			output.clear();
		}
	}
	static void recur(String start, String end, int n)
	{
		if(start.equals(end) )
		{
			return;
		}
		if(start.charAt(0) != end.charAt(n - 1))
		{
			String newString = change(start, n);
			//System.out.println(newString);
			output.add(n);
			recur(newString, end, n - 1);
		}
		else
		{
			String newString = change(start, 1);
			//System.out.println(newString);
			output.add(1);
			recur(newString, end, n);
		}
	}
	static String change(String str,int prefix)
	{
		StringBuilder output = new StringBuilder();
		for(int i = 0; i < prefix; i++)
		{
			output.append(convert(str.charAt(prefix - i - 1)));
		}
		output.append(str.substring(prefix, str.length()));
		return output.toString();
	}
	static char convert(char x)
	{
		if(x == '0')
		{
			return '1';
		}
		return '0';
	}
}
