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.*;
import java.util.*;

public class main{
	static Read input;

	static void solve(){

		int n = input.nextInt();
		ArrayList<Integer> ans = new ArrayList<>();
		StringBuffer a = new StringBuffer(input.next()),
		b = new StringBuffer(input.next());

		for (int i=n-1;i>=0;i--){
			char ac = a.charAt(i), bc = b.charAt(i), beg = a.charAt(0);

			if (ac!=bc){

				if (beg==bc) {
					ans.add(1); 
					if (a.charAt(0)=='0') a.setCharAt(0,'1');
					else a.setCharAt(0,'0');
				}


				ans.add(i+1);

				for (int j=0;j<=i;j++){
					char m = a.charAt(j);
					if (m=='0') a.setCharAt(j,'1');
					else a.setCharAt(j,'0');
				}

				StringBuffer aux = new StringBuffer(a.substring(0,i+1));
				aux.reverse();
				a = new StringBuffer(a.substring(i+1,n));
				a = aux.append(a);
			}

		}

		System.out.print(ans.size());
		for (Integer x : ans) System.out.print(" "+x);
		System.out.println();

	}

	public static void main(String args[]){

		input = new Read();
		int t = input.nextInt();

		while (t-->0)
			solve();

	}

	private static class Read{
		BufferedReader br;
		StringTokenizer st;
		public Read(){
			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();
		}
		String line(){
			String ans = "";
			try { ans = br.readLine(); }
			catch (IOException e) { e.printStackTrace(); }
			return ans;
		}
		int nextInt(){return Integer.parseInt(next());}
		double nextDouble(){return Double.parseDouble(next());}
		long nextLong(){return Long.parseLong(next());}
	}

}