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.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class ProblemC {
    public static void main(String[] args) throws Exception {
        Scanner scanner = new Scanner(System.in);
        int tot = scanner.nextInt();
        while (tot-- > 0){
            int n = scanner.nextInt();
            String startss = scanner.next();
            String endss = scanner.next();
            List<Integer> start = new ArrayList<>();
            List<Integer> end = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                start.add(startss.charAt(i) - '0');
                end.add(endss.charAt(i) - '0');
            }
            List<Integer> howmanys = new ArrayList<>();
            dfs(start, end, start.size()-1, howmanys);
            System.out.print(howmanys.size() + " ");
            for (int i = 0; i < howmanys.size(); i++) {
                System.out.print(howmanys.get(i) + " ");
            }
            System.out.println();
        }
    }
    public static void dfs(List<Integer> start, List<Integer> end, int k, List<Integer> howmanys){
        if(k==0){
            if(start.get(0) == end.get(0)){
            }else{
                start = execute(start, 0, howmanys);
            }
            return;
        }

        if(start.get(k) != end.get(k)){
            if(start.get(0) == start.get(k)){
                start = execute(start, k, howmanys);
                dfs(start, end, k-1, howmanys);
            }else{
                start = execute(start, 0, howmanys);
                dfs(start, end, k, howmanys);
            }
        }else{
            dfs(start, end, k-1, howmanys);
        }
    }
    public static List<Integer> execute(List<Integer> start, int k, List<Integer> howmanys){
        List<Integer> newString = new ArrayList<>();
        for (int i = k; i >= 0 ; i--) {
            newString.add(start.get(i) == 0 ? 1: 0);
        }
        for (int i = k+1; i < start.size() ; i++) {
            newString.add(start.get(i));
        }
        print(k+1, howmanys);
        return newString;
    }
    public static void print(int k, List<Integer> howmanys){
        howmanys.add(k);
    }
}
