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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class Scc {
  
	public static void main(String[] args) {
		
    myscanner in  = new myscanner();
     int t = in.nextInt();
   for(int z = 0;z<t;z++)
    {
    	int n  = in.nextInt();
    	String a = in.next();
    	String b = in.next();
    	int count = 0;
    	ArrayList<Integer> x = new ArrayList<>();
    	int backIndex = a.length()-1;
        while(a.length()>0)
        {
        	//System.out.println("im here"+a.length()+" "+b.length()+" "+backIndex);
        	while(backIndex>=0 && a.charAt(backIndex)==b.charAt(backIndex))
        		  backIndex--;
        	a = a.substring(0,backIndex+1);
        	b = b.substring(0,backIndex+1);
        	int i = 0;
        	while(backIndex>=0 && a.charAt(backIndex) != b.charAt(backIndex) && i<a.length() &&  a.charAt(backIndex)==a.charAt(i))
        	{
        		backIndex--;
        		i++;
        	}
        	if(i>0)
        	{
        		count++;
            	x.add(a.length());
        		b = b.substring(0,backIndex+1);
            	a = revinv(a.substring(i));
        	}
        	i= 0;
        	if(backIndex>=0 && a.charAt(backIndex) != b.charAt(backIndex) && i<a.length() &&  b.charAt(backIndex)==a.charAt(i))
        	{
                 b= b.substring(0,backIndex);
                 a = revinv(a.substring(i+1));
                 count+=2;
                 x.add(i+1);
                 x.add(backIndex+1);
                 backIndex--;
        	}
        }
        System.out.print(count+" ");
        for(int j : x)
        	System.out.print(j+" ");
    }
 }
	 public static String revinv(String a)
	   {
		   String x = "";
		  for(int i = 0;i<a.length();i++)
		  {
			 x = String.valueOf((1-Integer.parseInt(String.valueOf(a.charAt(i)))))+x;  
		  }
		  return x;
	   }
}
class myscanner
{
	BufferedReader bf;
	StringTokenizer st;
	
	public myscanner()
	{
		bf = new BufferedReader(new InputStreamReader(System.in));
		st = null;
	}
	
	public  String next()
	{
		while(st == null || !st.hasMoreElements())
		{
			try
			{
				st = new StringTokenizer(bf.readLine());
			}
			catch(IOException e)
			{
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}
	public int nextInt()
	{
		return Integer.parseInt(next());
	}
	public long nextLong()
	{
		return Integer.parseInt(next());
	}
}
