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.lang.*;
import java.util.*;
 
public class Main {
	/*public static char[] change(char[] s, int poz) {
		for (int i = 0; i < poz / 2; ++i) {
			char aux = s[i];
			s[i] = s[poz - 1 - i];
			s[poz - 1 - i] = aux;
		}
 
		for (int i = 0; i < poz; ++i)
			s[i] = (s[i] == '0' ? '1' : '0');
		return s;
	}*/
 
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
 
        int t = in.nextInt();
        while (t-- > 0) {
        	int n = in.nextInt();
        	String sa = in.next(), sb = in.next();
        	char[] a = sa.toCharArray(), b = sb.toCharArray();
        	Vector<Integer> ans = new Vector<>();
 
        	for (int i = 0; i < n - 1; ++i) {
        		if (a[i] != a[i + 1])
        			ans.add(i + 1);
        	}
        	if (a[n - 1] == '1')
        		ans.add(n);
 			
        	char prev = '0';
        	for (int i = n - 1; i >= 0; --i) {
        		if (b[i] != prev) {
        			ans.add(i + 1);
        			prev = (prev == '1' ? '0' : '1');
        		}
        	}
        	System.out.print(ans.size() + " ");
        	for (int it : ans)
        		System.out.print(it + " ");
        	System.out.println();
        }
    }
}