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 mok{
	static class FastReader {
 
        BufferedReader br;
        StringTokenizer st;
 
        private FastReader() {
            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());
        }
 
	 int[] nextIntArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
	}
 
	int[] nextIntArrayOne(int n) {
				int[] a = new int[n+1];
				for (int i = 1; i < n+1; i++)
					a[i] = nextInt();
				return a;
	}
 
        long nextLong() {
            return Long.parseLong(next());
        }
 
        double nextDouble() {
            return Double.parseDouble(next());
        }
 
        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
 
	public static void main(String[] args) {
		FastReader s =new FastReader();
		int t = s.nextInt();
		StringBuilder str = new StringBuilder();
		while(t-- >0) {
			int n = s.nextInt();
			StringBuilder a = new StringBuilder(s.nextLine());
			StringBuilder b = new StringBuilder(s.nextLine());
			StringBuilder ans = new StringBuilder();
			int count = 0 ;
			int invert = 0;
			int direction = 1;
			
			int z = 0;
			int c = n - 1;
			
			for(int i = n - 1; i >=0 ;i--) {
				
				if((a.charAt(c) -'0' ^ invert) == b.charAt(i) -'0') {
					c -= direction;
					continue;
				}
				
				
				if(((a.charAt(z) - '0') ^ invert) == b.charAt(i) -'0') {
					a.setCharAt(z, a.charAt(z) == '0' ? '1' :'0');
					ans.append("1 ");
					count++;
					if(i == 0) break;
				}
				
				ans.append((i + 1) + " ") ;
				count++;
				direction *= -1;
				invert ^= 1;
				int temp = z;
				z = c;
				c = temp;
				
				c -= direction;
			}
			
			str.append(count +" "+ ans +"\n");
		}
		System.out.println(str);
	}
 
	
 
}