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.*;
public class Ishu
{
	static String r(String str)
	{
	int n = str.length();
	for(int i=0;i<n;++i)
		{
		if(str.charAt(i) == '0')
			str = str.substring(0,i) + "1" + str.substring(i+1);
		else
			str = str.substring(0,i) + "0" + str.substring(i+1);
		}
	StringBuffer buf = new StringBuffer(str);
	buf.reverse();
	str = String.valueOf(buf);
	return str;
	}
	public static void main(String[] args)
	{
	Scanner scan=new Scanner(System.in);
	int t;
	t = scan.nextInt();
	while(t-- > 0)
		{
		int n = scan.nextInt();
		String a,b;
		a = scan.next();
		b = scan.next();
		
		int i = 0, j = n-1;
		int  k = 0;
		int[] ans = new int[2*n];
		while(j>=0)
			{
			if(a.charAt(j) == b.charAt(j))
				{
				--j;
				continue;
				}
			if(a.charAt(i) != a.charAt(j))
				{
				if(a.charAt(i) == '0')
					a = "1" + a.substring(1);
				else
					a = "0" + a.substring(1);
				ans[k++] = 1;
				}
			ans[k++] = j+1;
			a = r(a.substring(0,j+1)) + a.substring(j+1);
			--j;
			}
		System.out.print(k+" ");
		for(i=0;i<k;++i)
			System.out.print(ans[i]+" ");
		System.out.println();
		}	
	}
}
