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 C40
{

        

	
	public static void main(String[] args)
	{
	  Scanner s=new Scanner(System.in);
	  int t=s.nextInt();
	  while(t>0)
	  {	
	    int n=s.nextInt();
	    String a=s.next();
	    String b=s.next();
	     char[] ar=a.toCharArray();
	     char[] br=b.toCharArray();
	     ArrayList<Integer> ans=new ArrayList<>();
	     for(int i=n-1;i>-1;i--)
	     {
		if(ar[i]!=br[i])
		{
	         if(i!=0)
		{
		  if(ar[0]!=ar[i])
		  {		
			ans.add(1);
			operation(ar);
			ans.add((i+1));
			Operation(i,ar);
		  }
		  else
		  {
		    ans.add(i+1);
		    Operation(i,ar);
		  }
		}
		else
		{
		  ans.add(1);
		  operation(ar);
		}
	       }
	     
		//String k=new String(ar);
		//System.out.println(k);
    	     }
	
	    System.out.print(ans.size());
	   
	    for(int i=0;i<ans.size();i++)
	   {
		System.out.print(" "+ans.get(i));
	   }
	    System.out.println();
            t--;
	  }
	}
	static void Operation(int n,char ar[])
	{
		int i=n;
		
	       for(int j=0;j<=i;j++)
		  {
		   if(ar[j]=='1')
		    ar[j]='0';
		   else
		    ar[j]='1';
		  }
		  //String k=new String(ar);
		//System.out.println(k);
		  int end=i;
		  int start=0;
		  while(start<end)
		  {
		    char temp=ar[start];
		    ar[start]=ar[end];
		    ar[end]=temp;
		    start++;
		    end--;
		  }
	      //k=new String(ar);
		//System.out.println(k);
		  
	}
	  static void operation(char ar[])
	  {
		
	       if(ar[0]=='1')
		ar[0]='0';
		else
		ar[0]='1'; 
		 //String k=new String(ar);
		//System.out.println(k);
	       
	  }
}
