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.BufferedReader;
import java.io.BufferedWriter;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class Main {
	private BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
	private BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
	private Long MOD = 1000000007L;

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

	private void solve() throws Exception {

		int testCases = 1;
		testCases = Integer.parseInt(in.readLine().trim());
		for (int i = 1; i <= testCases; i++) {
			solve(i);
		}

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

	private void solve(int testCaseNumber) throws Exception {

		int n = Integer.parseInt(in.readLine().trim());
//		StringTokenizer st = new StringTokenizer(in.readLine());
		String s1 = in.readLine();
		String s2 = in.readLine();
//		int n
		int diff = 0;
		for(int i = 0; i < n; i += 1) {
			if(s1.charAt(i) != s2.charAt(i)) diff++;
		}
		
		ArrayList<Integer> list  = new ArrayList<>();
		
		int half = n / 2 ;
		int idx = n - 1;
		for(int i = 0 ; i < half; i ++) {
			
			int first = i;
			if(s1.charAt(first) == s2.charAt(idx)) {
				list.add(1);
				list.add(idx + 1);
			} else {
				list.add(idx + 1);
			}
			idx -= 1;
			
			int second = n - 1 - i;
			// but complement
			if(s1.charAt(second) != s2.charAt(idx)) {
				list.add(1);
				list.add(idx + 1);
			} else {
				list.add(idx + 1);
			}
			
			idx -= 1;
		}
		
		if(n % 2 == 1) {
			// do one more step
			int first = half;
			if(s1.charAt(first) == s2.charAt(idx)) {
				list.add(1);
				list.add(idx + 1);
			} else {
				list.add(idx + 1);
			}
		}
		
		out.write(list.size() + " ");
		for(int x : list) {
			out.write(x + " ");
		}
		out.write("\n");
		
		
	}

}