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.*;
import java.io.*;

public class Prefix_Flip {

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}

	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastReader t = new FastReader();
		PrintWriter o = new PrintWriter(System.out);
		int test = t.nextInt();
		StringBuilder sb = new StringBuilder();

		while (test-- > 0) {
			int n = t.nextInt();
			char a[] = t.next().toCharArray();
			char b[] = t.next().toCharArray();
			List<Integer> list = new ArrayList<>();
			int x = 0, y = n - 1;
			boolean flag = false;

			for (int i = n - 1; i >= 0; --i) {
				if (!flag) {
					if (a[x] == b[i]) {
						list.add(1);
					}

					++x;

					list.add(i + 1);
				} else {
					if (a[y] != b[i]) {
						list.add(1);
					}

					--y;

					list.add(i + 1);
				}

				flag = !flag;
			}

			sb.append(list.size() + " ");

			for (int val : list) {
				sb.append(val + " ");
			}

			sb.append("\n");
		}

		o.println(sb);
		o.flush();
		o.close();
	}
}