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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author sofiane
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskA solver = new TaskA();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskA {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int t = in.nextInt();
            while (t-- > 0) {
                int n = in.nextInt();
                String a = in.next();
                String b = in.next();
                int l = 0, r = 0, c = 0;
                ArrayList<Integer> ans = new ArrayList();
                for (int i = n - 1; i >= 0; i--) {
                    if (c % 2 == 0) {
                        if (b.charAt(i) == a.charAt(n - r - 1)) {
                            r++;
                            continue;
                        } else {
                            if (b.charAt(i) == a.charAt(l)) {
                                ans.add(1);
                            }
                            ans.add(i + 1);
                            c++;
                            l++;
                        }
                    } else {
                        if (b.charAt(i) != a.charAt(l)) {
                            l++;
                            continue;
                        } else {
                            if (b.charAt(i) != a.charAt(n - r - 1))
                                ans.add(1);
                            ans.add(i + 1);
                            r++;
                            c++;
                        }
                    }
                }
                out.printf("%d ", ans.size());
                for (int i = 0; i < ans.size(); i++) {
                    if (i != 0) out.printf(" ");
                    out.printf("%d", ans.get(i));
                }
                out.printf("\n");
            }
        }

    }
}

