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.io.*;
import java.util.*;

public class Main {
	static Scanner sc = new Scanner(System.in);
	static PrintWriter out = new PrintWriter(System.out);

	static void solve() throws Exception {
		int n = sc.nextInt();
		char a[] = sc.nextLine().toCharArray();
		char b[] = sc.nextLine().toCharArray();
		boolean reversed = false;
		ArrayList<Integer> ans = new ArrayList<>();

		int l = 0, r = n - 1;
		for (int i = n - 1; i > 0; i--) {
			if (!reversed) {
				if (a[r] == b[i]) {
					r--;
					continue;
				}

				if (a[l] == b[i]) {
					ans.add(1);
				}
				
				l++;
				ans.add(i + 1);
				reversed = true;
				
			}else {
				if(a[l] != b[i]) {
					l++;
					continue;
				}
				
				if(a[r] != b[i]) {
					ans.add(1);
				}
				
				r--;
				ans.add(i + 1);
				reversed = false;
			}
		}
		
		if(!reversed && a[l] != b[0]) {
			ans.add(1);
		}else if(reversed && a[r] == b[0]) {
			ans.add(1);
		}
		
		out.print(ans.size() + " ");
		for (int x : ans) out.print(x + " ");
		out.println();

	}

	public static void main(String[] args) throws Exception {
		int t = sc.nextInt();
		while (t-- > 0) {
			solve();
		}
		out.close();
	}
}

class Scanner {
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream system) {
		br = new BufferedReader(new InputStreamReader(system));
	}

	public Scanner(String file) throws Exception {
		br = new BufferedReader(new FileReader(file));
	}

	public String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}

	public String nextLine() throws IOException {
		return br.readLine();
	}

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

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

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