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.Scanner;

public class A1381 {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int T = in.nextInt();
        for (int t=0; t<T; t++) {
            int N = in.nextInt();
            char[] A = in.next().toCharArray();
            char[] B = in.next().toCharArray();
            StringBuilder output = new StringBuilder();
            int count = 0;
            int start = 0;
            boolean reversed = false;
            for (int n=N-1; n>=0; n--) {
                if (flip(A[idx(n,n,start,reversed)],reversed) != B[n]) {
                    if (flip(A[idx(0,n,start,reversed)],reversed) == B[n]) {
                        count++;
                        output.append(" 1");
                    }
                    count++;
                    output.append(' ').append(n+1);
                    reversed = !reversed;
                }
                if (reversed) {
                    start++;
                }
            }
//            for (int n=N-1; n>=0; n--) {
//                if (A[n] != B[n]) {
//                    if (A[0] == B[n]) {
//                        count++;
//                        output.append(" 1");
//                        A[0] = (A[0] == '0') ? '1' : '0';
//                    }
//                    count++;
//                    output.append(' ').append(n+1);
//                    char[] C = new char[N];
//                    for (int i=0; i<=n; i++) {
//                        C[i] = (A[n-i] == '0') ? '1' : '0';
//                    }
//                    A = C;
//                }
//            }
            System.out.println(count + "" + output);
        }
    }

    private static char flip(char c, boolean reversed) {
        return reversed ? (c == '0' ? '1' : '0') : c;
    }

    private static int idx(int index, int n, int start, boolean reversed) {
        if (reversed) {
            index = n - index;
        }
        index += start;
        return index;
    }

}