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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class C {

	PrintWriter out;

	public static void main(String[] args) throws Exception {
		new C().run();
	}

	public void run() throws Exception {
		FastScanner file = new FastScanner();
		out = new PrintWriter(System.out);

		//C1 and C2
		int times = file.nextInt();
		for (int asdf = 0; asdf < times; asdf++) {
			int ln = file.nextInt();
			String a = file.next(), b = file.next();
			if (a.equals(b)) {
				out.println(0);
				continue;
			}
			ArrayList<String> sol = new ArrayList<String>();
			for (int i = 0; i < ln - 1; i++) {
				if (a.charAt(i) != a.charAt(i + 1))
					sol.add(String.valueOf(i + 1));
			}
			if (a.charAt(ln - 1) != '0')
				sol.add(String.valueOf(ln));
			int curr = 48;
			for (int i = ln - 1; i >= 0; i--) {
				if (curr != b.charAt(i)) {
					sol.add(String.valueOf(i + 1));
					if (curr == 48)
						curr = 49;
					else
						curr = 48;
				}
			}
			out.println(sol.size() + " " + String.join(" ", sol));
		}

		file.close();
		out.flush();
	}

	static class FastScanner {

		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public FastScanner() {
			reader = new BufferedReader(new InputStreamReader(System.in), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

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

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

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

		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}

		public void close() throws IOException {
			reader.close();
		}
	}
}