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 MyClass {
    
    
    static String rev(String a,int ei)
    {
        String res="";
        
        for(int i = ei+1;i>=0;i--)res += a.charAt(i)=='0'?"1":"0";
        return res;
    }
    
    public static void main(String args[]) 
    {
        Scanner s = new Scanner(System.in);
        int t = s.nextInt();
        StringBuilder str = new StringBuilder();
        
        while(t-->0)
        {
            int n = s.nextInt();
            String a = s.next();
            String b = s.next();
            
            if(a.equals(b))
            {
                str.append("0"+"\n");
                continue;
            }
            
            int ei = n-1;
            String sa = "";
            int count = 0;
            while(true)
            {
                //System.out.println(ei+ a+" **** "+b);
                while(ei > -1 && a.charAt(ei) == b.charAt(ei)) ei--;
                
                if(ei==-1) break;
                
                if(b.charAt(ei) == a.charAt(0))
                {
                    sa += " 1 "+(ei+1);
                    count+=2;
                }
                else
                {
                    sa += " "+(ei+1);
                    count++;
                }
                
                ei--;
                a = rev(a,ei); 
            
            }
            
            sa = count+sa;
            str.append(sa+"\n");
            
            
        }
        
        System.out.println(str);
    }
}
 
