human
I found an interesting problem on https://leetcode.com/problems/binary-trees-with-factors/:
Given an array of unique integers, each integer is strictly greater than 1.
We make a binary tree using these integers and each number may be used for any number of times.
Each non-leaf node's value should be equal to the product of the values of it's children.
How many binary trees can we make?  Return the answer modulo 10 ** 9 + 7.
Example 1:
Input: A = [2, 4]
Output: 3
Explanation: We can make these trees: [2], [4], [4, 2, 2]
Example 2:
Input: A = [2, 4, 5, 10]
Output: 7
Explanation: We can make these trees: [2], [4], [5], [10], [4, 2, 2], [10, 2, 5], [10, 5, 2].
 
Note:

1 <= A.length <= 1000.
2 <= A[i] <= 10 ^ 9.

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python code you requested.

class Solution:
    def numFactoredBinaryTrees(self, A: List[int]) -> int:
        A = sorted(A)
        mems = {}
        for ix in range(len(A)):
            mems[A[ix]] = ix
            
        dp = [1] * len(A)
        
        big_mod = 10**9 + 7
        
        for i in range(len(A)):
            extras = 0
            for j in range(i):
                d, mo = divmod(A[i], A[j])
                if mo == 0:
                    if d in mems:
                        extras += dp[j] * dp[mems[d]]
                        extras = extras % big_mod

            dp[i] += extras
        
        result = sum(dp) % big_mod
        
        return result