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 yo {
	public static PrintWriter out;
	public static InputReader in;
	public static void main(String[] args)throws IOException {
        in = new InputReader(System.in);
        out = new PrintWriter(System.out);
        
		int cases = in.nextInt();
		
		for(int t = 0; t < cases; t++){	
			int n = in.nextInt();
			int a[] = new int[n];
			int b[] = new int[n];
			LinkedList<Integer> ll = new LinkedList<Integer>();
			String s = in.next();
			for(int i=0;i<n;i++){
				a[i]=(s.charAt(i)-'0');
				ll.add(a[i]);
			}
			s = in.next();
			for(int i=0;i<n;i++){
				b[i]=(s.charAt(i)-'0');
			}
			int pos = n-1; int ct=0;
			ArrayList<Integer> ans = new ArrayList<Integer>();
			for(int i=0;i<n;i++){
				int bval = b[n-1-i];
				if(i==n-1){
					int val = ll.pollLast();
					if(ct%2!=0)
						val = 1-val;
					if(val!=bval)
						ans.add(1);
					break;
				}
				if(pos==n-1){
					int val = ll.peekLast();
					int v0 = ll.peekFirst();
					if(ct%2!=0){
						v0 = 1-v0;
						val = 1-val;
					}
					if(val!=bval){
						if(val!=v0)
							ans.add(1);
						ll.pollFirst();
						ct++;
						ans.add(n-i);
						pos = 0;
					}
					else{
						ll.pollLast();
					}
				}
				else{
					int val = ll.peekFirst();
					int v0 = ll.peekLast();
					if(ct%2!=0){
						v0 = 1-v0;
						val = 1-val;
					}
					if(val!=bval){
						if(val!=v0)
							ans.add(1);
						ll.pollLast();
						ct++;
						ans.add(n-i);
						pos = n-1;
					}
					else{
						ll.pollFirst();
					}
				}
			}
			out.println(ans.size());
			// out.println(ans);
			for(int i : ans)
				out.print(i+" ");
			out.println();
		}

        out.close();
	}
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

    }

}