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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class PrefixFlipHardVersion {
    public static void main(String[] args) throws NumberFormatException, IOException {
        BufferedReader file = new BufferedReader(new InputStreamReader(System.in));
        int inputs = Integer.parseInt(file.readLine());
        while (inputs-- > 0) {
            int n = Integer.parseInt(file.readLine());
            char[] a = file.readLine().toCharArray();
            char[] b = file.readLine().toCharArray();
            int moves = 0;
            int left = 0;
            int right = n - 1;
            StringBuilder ans = new StringBuilder();
            for (int i = 0; i < n; i++) {
                int c = -1;
                if (i % 2 == 0) {
                    c = a[left++];
                } else {
                    c = a[right--];
                }
                c = (c + i - '0') % 2;
//				System.out.println(i + " " + c);
//				System.out.println(c + " " + (b[n-i-1]-'0'));
                if (c == b[n - i - 1] - '0') {
                    moves++;
                    ans.append("1 ");
                }
                moves++;
                ans.append((n - i) + " ");
            }
            System.out.println(moves + " " + ans);
        }
    }
    static class Reader {
    		static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    		static StringTokenizer tokenizer = new StringTokenizer("");
    		static String nextLine() throws IOException
    		{// 读取下一行字符串
    			return reader.readLine();
    		}
    		static String next() throws IOException {// 读取下一个字符串
    			while (!tokenizer.hasMoreTokens()) {
    				tokenizer = new StringTokenizer(reader.readLine());
    			}
    			return tokenizer.nextToken();
    		}

    		static int nextInt() throws IOException {// 读取下一个int型数值
    			return Integer.parseInt(next());
    		}
    		static long nextLong() throws IOException {// 读取下一个int型数值
    			return Long.parseLong(next());
    		}
    		static double nextDouble() throws IOException {// 读取下一个double型数值
    			return Double.parseDouble(next());
    		}
    	}
}