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.*;
public class Main{
    public static void main(String args[])throws IOException{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int t=Integer.parseInt(br.readLine().trim());
        StringBuffer result=new StringBuffer();
        while(t-->0){
            int n=Integer.parseInt(br.readLine().trim());
            char arr[]=br.readLine().trim().toCharArray();
            char brr[]=br.readLine().trim().toCharArray();
            ArrayList<Integer> ans=new ArrayList<>();
            
            for(int i=n-1;i>=0;i--){
                if(arr[i] == brr[i])
                continue;
                
                if(arr[0] != arr[i]){
                    ans.add(0);
                    invertArray(arr,0);
                }
                ans.add(i);
                invertArray(arr,i);
            }
            
            result.append(ans.size()+" ");
            for(int i:ans)
            result.append((i+1)+" ");
            result.append("\n");
        }
        System.out.print(result);
    }
    static void invertArray(char arr[],int in){
        for(int i=0;i<=in/2;i++){
            char temp = arr[i] == '0'?'1':'0';
            char temp2 = arr[in-i] == '0'?'1':'0';
            arr[i]=temp2;
            arr[in-i]=temp;
        }
    }
}