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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        solution(in, out);
        in.close();
        out.close();
    }

    public static void solution(Scanner in, PrintWriter out) {
        int T = Integer.parseInt(in.nextLine());
        for(int i=0;i<T;i++) {
            algo(in, out);
        }
    }

    public static void algo(Scanner in, PrintWriter out) {
        int N = Integer.parseInt(in.nextLine());
        String a = in.nextLine(), b = in.nextLine();
        int[] A = new int[N], B = new int[N];
        for(int i=0;i<N;i++) {
            A[i] = Integer.parseInt("" + a.charAt(i));
            B[i] = Integer.parseInt("" + b.charAt(i));
        }
        ans = new ArrayList<>();
        for(int i=N-1;i>=0;i--) {
            if(B[i] != A[i]) {
                if(B[i] == A[0]) {
                    operation(A, 0);
                }
                operation(A, i);
            }
        }
        out.print(ans.size() + " ");
        for(int i=0;i<ans.size();i++) {
            out.print(ans.get(i) + " ");
        }
        out.println();
    }

    static ArrayList<Integer> ans;

    public static void operation(int[] A, int index) {
        ans.add(index+1);
        for(int i=0;i<=index/2;i++) {
            int t = A[i]^1;
            A[i] = A[index-i]^1;
            A[index-i] = t;
        }
    }
}
