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.util.ArrayList;
import java.util.StringTokenizer;
import java.io.InputStreamReader;
import java.io.IOException;
 
public class codeforces {
	
	public static void main(String[] args) {	
         scanner in  = new scanner();
         int t = in.nextInt();
         for(int i = 0;i<t;i++)
         {
        	int n = in.nextInt();
        	String a = in.next();
        	String b = in.next();
        	ArrayList<Integer> xb = new ArrayList<>();
        	ArrayList<Integer> xa = new ArrayList<>();
        	int j = 1;
        	char A = a.charAt(n-1);
        	char B = b.charAt(n-1);
        	while(j<n)
        	{
        		if(a.charAt(j)!=a.charAt(j-1))
        			xa.add(j);
        		if(b.charAt(j)!=b.charAt(j-1))
        			xb.add(j);
        		j++;
        	}
        	if(A!=B)
        		xb.add(n);
        	System.out.print(xa.size()+xb.size()+" ");
        	for(int k : xa)
        		System.out.print(k+" ");
        	for(int k = xb.size()-1;k>=0;k--)
        		System.out.print(xb.get(k)+" ");
        	System.out.println();
         }
	}
}
class scanner
{
	BufferedReader br ;
	StringTokenizer st;
	
	public scanner()
	{
		br = new BufferedReader(new InputStreamReader(System.in));
		st = null;
	}
	 public String next() {
	        while (st == null || !st.hasMoreElements()) {
	            try {
	                st = new StringTokenizer(br.readLine());
	            } catch (IOException e) {
	                e.printStackTrace();
	            }
	        }
	        return st.nextToken();
	    }
	public int nextInt()
	{
		return Integer.parseInt(next());
	}
	public long nextLong()
	{
	  return	Long.parseLong(next());
	}
}