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.
//  package Div2_658;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class ProblemC {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        StringBuilder print=new StringBuilder();
        int test=Integer.parseInt(br.readLine());
        while(test--!=0){
            int n=Integer.parseInt(br.readLine());
            char a[]=br.readLine().toCharArray();
            char b[]=br.readLine().toCharArray();
            StringBuilder temp=new StringBuilder();
            int moves=0;
            for(int i=n-1;i>=0;i--){
//                if(a[i]==b[i]){
//                    continue;
//                }
                if(a[0]==b[i]){
                    temp.append(1+" ");
                    moves++;
                    a[0]=a[0]=='0'?'1':'0';
                }
                temp.append((i+1)+" ");
                moves++;
                int d=n-i;
                if(d%2!=0){
                    int j=n-1-d/2;
                    a[0]=a[j]=='0'?'1':'0';
                }
                else{
                    int j=d/2;
                    a[0]=a[j];
                }
//                int s=0,e=i;
//                for(int j=0;j<=i;j++){
//                    a[j]=a[j]=='0'?'1':'0';
//                }
//                while(s<e){
//                    char t=a[s];
//                    a[s]=a[e];
//                    a[e]=t;
//                    s++;
//                    e--;
//                }
//                for(int j=0;j<n;j++){
//                    System.out.print(a[j]);
//                }
//                System.out.println(" "+i);
            }
            print.append(moves).append(" ").append(temp.toString()).append("\n");
        }
        System.out.print(print.toString());
    }
}
