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.*;
import java.util.*;

public class Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int t=Integer.parseInt(bu.readLine()),x=t;
        while(t-->0)
        {
            int n=Integer.parseInt(bu.readLine());
            char a[]=bu.readLine().toCharArray(),b[]=bu.readLine().toCharArray();
            int i,j,z=0,o=0;
            ArrayList<Integer> al=new ArrayList<>();
            for(i=0;i<n;i++)
            {
                if(a[i]=='0') z++;
                else o++;
            }
            if(z>o)
            {
                if(a[0]!='0') al.add(0);
                for(i=1;i<n;i++)
                if(a[i]!='0') {al.add(i-1); al.add(i);}
                char cur='0';
                for(i=n-1;i>=0;i--)
                if(b[i]!=cur)
                {
                    al.add(i);
                    if(cur=='0') cur='1';
                    else cur='0';
                }
            }
            else
            {
                if(a[0]!='1') al.add(0);
                for(i=1;i<n;i++)
                if(a[i]!='1') {al.add(i-1); al.add(i);}
                char cur='1';
                for(i=n-1;i>=0;i--)
                if(b[i]!=cur)
                {
                    al.add(i);
                    if(cur=='1') cur='0';
                    else cur='1';
                }
            }
            sb.append(al.size()+" ");
            for(i=0;i<al.size();i++)
            sb.append(al.get(i)+1+" ");
            sb.append("\n");
        }
        System.out.print(sb);
    }
}