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.*;
public class code {
	public static void main(String[] args)throws IOException {
		FastReader sc = new FastReader();
		PrintWriter pw = new PrintWriter(System.out);

		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			String a = sc.next();
			String b = sc.next();
			Vector<Integer> v1 = new Vector<>();
			Vector<Integer> v2 = new Vector<>();
			a = a + "0";
			b = b + "0";
			for (int i = 1; i <= n; i++) {
				if (a.charAt(i) != a.charAt(i - 1)) {
					v1.add(i);
				}
				if (b.charAt(i) != b.charAt(i - 1)) {
					v2.add(i);
				}
			}
			for (int i = v2.size() - 1; i >= 0; i--) {
				v1.add(v2.get(i));
			}
			System.out.print(v1.size() + " ");
			for (Integer i : v1)
				System.out.print(i + " ");
			System.out.println();
		}


	}
}





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;
	}
}