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 File {
    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner() {
            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();
        }

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
	
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
		
		
		
		
		
		int T = sc.nextInt();
		
		
		for (int t = 1; t <= T; t++) {
			int n = sc.nextInt();
			
			char[] a = sc.next().toCharArray();
			char[] b = sc.next().toCharArray();
			
			List<Integer> aToZero = new ArrayList<Integer>();
			List<Integer> bToZero = new ArrayList<Integer>();
			
			// Transform a into all 0's.
			for (int i = 0; i + 1 < n; i++) {
				if (a[i] != a[i+1]) {
					if (a[i] == '1') {
						a[i] = '0';
					}
					else {
						a[i] = '1';
					}
					aToZero.add(i+1);
				}
			}
			if (a[n-1] == '1') {
				aToZero.add(n);
			}
			
			// Transform b into all 0's'.
			for (int i = 0; i + 1 < n; i++) {
				if (b[i] != b[i+1]) {
					if (b[i] == '1') {
						b[i] = '0';
					}
					else {
						b[i] = '1';
					}
					bToZero.add(i+1);
				}
			}
			if (b[n-1] == '1') {
				bToZero.add(n);
			}
			
			// Answer is steps a to 0's, b's to 0's reversed.
			out.print((aToZero.size() + bToZero.size()) + " ");
			for (int index : aToZero) {
				out.print(index + " ");
			}
			for (int i = bToZero.size() - 1; i >= 0; i--) {
				out.print(bToZero.get(i) + " ");
			}
			
			out.println();
		}
		
		
		
		
		
		
		
		out.close();
	}
}