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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.ArrayList;
import java.util.List;
import java.util.Locale;
import java.util.StringTokenizer;

public class Solution implements Runnable {
	
	private PrintStream out;
	private BufferedReader in;
	private StringTokenizer st;
	
	public void solve() throws IOException {
		long time0 = System.currentTimeMillis();
		
		int t = nextInt();
		for (int test = 1; test <= t; test++) {
			int n = nextInt();
			String a = next();
			String b = next();
			int[] ans = solve(n, a, b);
			out.print(ans.length);
			for (int i = 0; i < ans.length; i++) {
				out.print(' ');
				out.print(ans[i]);
			}
			out.println();
		}
		
		System.err.println("time: " + (System.currentTimeMillis() - time0));
	}
	
	private int[] solve(int n, String astr, String bstr) {
		boolean[] a = decode(astr);
		boolean[] b = decode(bstr);
		List<Integer> answer = new ArrayList<Integer>();
		for (int i = n - 1; i >= 0; i--) {
			if (b[i] == a[i]) {
				continue;
			}
			if (a[0] == b[i]) {
				doopp(1, answer, a);
			}
			doopp(i + 1, answer, a);
		}
		int[] result = new int[answer.size()];
		for (int i = 0; i < answer.size(); i++) {
			result[i] = answer.get(i);
		}
		return result;
	}

	private void doopp(int cnt, List<Integer> answer, boolean[] a) {
		answer.add(cnt);
		for (int i = 0; i < cnt; i++) {
			a[i] = !a[i];
		}
		for (int i = 0; i < cnt - 1 - i; i++) {
			int j = cnt - 1 - i;
			boolean tmp = a[i];
			a[i] = a[j];
			a[j] = tmp;
		}
	}

	private boolean[] decode(String s) {
		boolean[] b = new boolean[s.length()];
		for (int i = 0; i < s.length(); i++) {
			b[i] = s.charAt(i) == '1';
		}
		return b;
	}

	public double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
	
	public long nextLong() throws IOException {
		return Long.parseLong(next());
	}
	
	public int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
	
	public String next() throws IOException {
		while (!st.hasMoreTokens()) {
			String line = in.readLine();
			if (line == null) {
				return null;
			}
			st = new StringTokenizer(line);
		}
		return st.nextToken();
	}
	
	@Override
	public void run() {
		try {
			solve();
			out.close();
		} catch (Throwable e) {
			throw new RuntimeException(e);
		}
	}
	
	public Solution(String name) throws IOException {
		Locale.setDefault(Locale.US);
		if (name == null) {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintStream(new BufferedOutputStream(System.out));
		} else {
			in = new BufferedReader(new InputStreamReader(new FileInputStream(name + ".in")));
			out = new PrintStream(new BufferedOutputStream(new FileOutputStream(name + ".out")));
		}
		st = new StringTokenizer("");
	}
	
	public static void main(String[] args) throws IOException {
		new Thread(new Solution(null)).start();
	}
}