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.lang.*;
import java.io.*;

public class Main
{
	PrintWriter out = new PrintWriter(System.out);
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tok = new StringTokenizer("");
    String next() throws IOException {
        if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
        return tok.nextToken();
    }
    int ni() throws IOException { return Integer.parseInt(next()); }
    long nl() throws IOException { return Long.parseLong(next()); }
  
    void solve() throws IOException {
        for (int tc=ni();tc>0;tc--) {
            int n=ni();
            char[]A=next().toCharArray();
            char[]B=next().toCharArray();
            ArrayList<Integer>C=new ArrayList();
            int p=1;
            int q=n-1;
            int c=1;
            char z=A[0];
            
            for (int i=n-1;i>0;i--) {
                if (z==B[i]) {
                    C.add(1);
                }
                C.add(i+1);
                if (c==1) {
                    if (A[q]=='1') z='0';
                    else z='1';
                    c=2;
                    q--;
                }
                else {
                    z=A[p];
                    c=1;
                    p++;
                }
            }
            if (z!=B[0]) C.add(1);
            
            out.print(C.size()+" ");
            for (int i=0;i<C.size();i++) out.print(C.get(i)+" ");
            out.println();
        }
        out.flush();
    }
    
    public static void main(String[] args) throws IOException {
        new Main().solve();
    }
}