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.io.*;
import java.util.*;

public class Solution {
    static String perform(String str1,int j,int n){
        StringBuilder sb=new StringBuilder();
        for(int p=0;p<=j;p++){
            sb.append(1-Integer.parseInt(Character.toString(str1.charAt(p))));
        }
        sb=sb.reverse();
        for(int p=j+1;p<n;p++)
            sb.append(Integer.parseInt(Character.toString(str1.charAt(p))));
        return sb.toString();
    }
    public static void main(String[] args) throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw=new BufferedWriter(new OutputStreamWriter(System.out));
        int t=Integer.parseInt(br.readLine());
        for(int i=1;i<=t;i++){
            int n=Integer.parseInt(br.readLine());
            String str1=br.readLine();
            String str2=br.readLine();
            int cnt=0;
            ArrayList<Integer> al=new ArrayList<Integer>();
            for(int j=n-1;j>=0;j--){
                if(str1.charAt(j)!=str2.charAt(j)){
                    if(str1.charAt(0)==str2.charAt(j)){
                        cnt+=2;
                        al.add(1);
                        str1=perform(str1,0,n);
                        al.add(j+1);
                        str1=perform(str1,j,n);
                    }
                    else{
                        cnt++;
                        al.add(j+1);
                        str1=perform(str1,j,n);
                    }
                }
            }
            bw.write(cnt+" ");
            if(al.size()!=0)
                for(int j:al)
                    bw.write(j+" ");
            bw.write("\n");
        }
        bw.flush();
    }
}