Problem: 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.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.lang.*;
public class Main{
     static class FastReader{
	BufferedReader br;
	StringTokenizer st;
      public FastReader(){
         br=new BufferedReader(new InputStreamReader(System.in));
         }
      public FastReader(String s)throws FileNotFoundException{
        br=new BufferedReader(new FileReader(new File(s)));
        }
       String next(){
          while(st==null || !st.hasMoreElements()){
            try{
                  st=new StringTokenizer(br.readLine());
                  }
            catch(IOException e){
                   e.printStackTrace();
		}

               }
             return st.nextToken();
         }
        String nextLine(){
             String str="";
             try{
                 str=br.readLine();
                 }
            catch(IOException e){
              e.printStackTrace();
                     }
                return str;
           }
	int nextInt(){
            return Integer.parseInt(next());
                }
         long nextLong(){
            return Long.parseLong(next());
             }
         double nextDouble(){
            return Double.parseDouble(next());

            }
  } 
     public static char reverse(char c){
       if(c=='1') return '0';
         return '1';
     }
	public static void main(String[] args)throws IOException {
	     FastReader in=new FastReader();
            int T=in.nextInt();
            while(T-->0){
            int n=in.nextInt();
            String a=in.next(),b=in.next();
            Queue<Integer> q=new LinkedList<>();
            for(int i=1;i<n;i++){
              if(a.charAt(i)!=a.charAt(i-1)){
                q.add(i);
              }
            }
            char c=a.charAt(n-1);
            for(int i=n-1;i>=0;i--){
                if(b.charAt(i)!=c){
                 q.add(i+1);
                   c=reverse(c);
                }
            }
            int size=q.size();
            System.out.print(size+" ");
            for(int i=0;i<size;i++){
               System.out.print(q.poll()+" ");
            
            }
            System.out.println(); 
              
         } 
      }
}