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 A {

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int T = sc.nextInt();
		StringBuilder res = new StringBuilder();
		while(T-- > 0) {
			int n = sc.nextInt();
			char[] s = sc.next().toCharArray();
			char[] t = sc.next().toCharArray();
			StringBuilder sb = new StringBuilder();
			int c = 0;
			for(int i = n-1; i >= 0; i--) {
				if(s[i] == t[i]) continue;
				else {
					if(i >= 1) {
						if(s[i-1] == t[i-1]) { //01 -> 00, 11 -> 10, 10 -> 11 or 00 -> 01
							sb.append(" "+(i+1)+" 1 "+(i+1)); c += 3;
						}
						else {
							if(s[i-1] == t[i]) { //01 -> 10 or 10 -> 01
								if(i >= 2) {
									sb.append(" "+i+" "+(i+1)+" "+i+" "+(i-1)); c += 4;
								}
								else {
									sb.append(" "+i+" "+(i+1)+" "+i); c += 3;
								}
							}
							else { //00 -> 11 or 11 -> 00
								sb.append(" "+(i+1)+" 2 "+(i+1)); c += 3;
							}
						}
						i--;
					}
					else {
						sb.append(" 1"); c++;
					}
				}
			}
			res.append(c+""+sb.toString()+"\n");
		}
		PrintWriter pw = new PrintWriter(System.out);
		pw.print(res.toString());
		pw.flush();
	}
	
	static class FastScanner {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public FastScanner() {
            reader = new BufferedReader(new InputStreamReader(System.in), 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());
        }
        public double nextDouble() {
        	return Double.parseDouble(next());
        }
        public String nextLine() {
        	try {
        		return reader.readLine();
        	} catch(IOException e) {
        		throw new RuntimeException(e);
        	}
        }
    }

}
