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.util.*;
import java.io.*;
public class c1 {
	static ArrayList<Integer> ans;
	public static void main(String[] args) throws IOException {
		FastScanner sc = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int T = sc.nextInt();
		
		while(T-- > 0) {
			int n = sc.nextInt();
			String a = sc.next(), b = sc.next();
			ans = new ArrayList<>();
			for (int i = n-1 ; i>= 0 ;i--) {
				// not correct, have to change to other bit
				if(b.charAt(i) != a.charAt(i)) { 
					
					
					// first bit will turn into same thing as this so flip bit at index 0 first
					if(a.charAt(i) != a.charAt(0)) {
						//out.println("flip first");
						//out.println(a);
						a = flipandinvert(a, 1);
						//out.println(a);
					}
					
					//out.println("flip index " + i);
					//out.println(a);
					
					a = flipandinvert(a,i+1);
					//out.println(a);
				}
			}
			
			
			out.print(ans.size() + " ");
			for(Integer e : ans) {
				out.print(e + " ");
			}
			out.println();
			
			//out.println(a);
			//out.println(b);
			
		}
		
		out.close();
	}
	
	static String flipandinvert(String a, int n) {
		StringBuilder sb = new StringBuilder();
		
		// flip and reverse
		for(int i = n-1 ; i>= 0 ; i--) {
			if(a.charAt(i) == '1') {
				sb.append('0');
			} else {
				sb.append('1');
			}
		}
		
		// get unchanged substring
		for(int i = n; i< a.length(); i++) {
			sb.append(a.charAt(i));
		}
		
		ans.add(n);
		
		return sb.toString();
	}

	static class FastScanner {
	    BufferedReader br;
	    StringTokenizer st;
		
	    public FastScanner(InputStream i) {
	        br = new BufferedReader(new InputStreamReader(i));
	        st = new StringTokenizer("");
	    }
				
	    public String next() throws IOException {
	        if(st.hasMoreTokens())
	            return st.nextToken();
	        else
	            st = new StringTokenizer(br.readLine());
	        return next();
	    }

	    public int nextInt() throws IOException {
	        return Integer.parseInt(next());
	    }
	    public long nextLong() throws IOException {
	        return Long.parseLong(next());
	    }
	    public double nextDouble() throws IOException {
	        return Double.parseDouble(next());
	    }
	}
}