<p>Given an array of unique integers, <code>arr</code>, where each integer <code>arr[i]</code> is strictly greater than <code>1</code>.</p>

<p>We make a binary tree using these integers, and each number may be used for any number of times. Each non-leaf node&#39;s value should be equal to the product of the values of its children.</p>

<p>Return <em>the number of binary trees we can make</em>. The answer may be too large so return the answer <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [2,4]
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can make these trees: <code>[2], [4], [4, 2, 2]</code></pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [2,4,5,10]
<strong>Output:</strong> 7
<strong>Explanation:</strong> We can make these trees: <code>[2], [4], [5], [10], [4, 2, 2], [10, 2, 5], [10, 5, 2]</code>.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length &lt;= 1000</code></li>
	<li><code>2 &lt;= arr[i] &lt;= 10<sup>9</sup></code></li>
	<li>All the values of <code>arr</code> are <strong>unique</strong>.</li>
</ul>

class Solution:
    def getNumberOfFactoredBinTrees(self, root: int, prods_to_factors):
        count = 1  # Just me
        if root not in prods_to_factors:
            return count
        
        for factor1, factor2 in prods_to_factors[root]:
            if factor1 != factor2:
                count += 2 * self.getNumberOfFactoredBinTrees(factor1, prods_to_factors) * self.getNumberOfFactoredBinTrees(factor2, prods_to_factors)
            else:
                count += self.getNumberOfFactoredBinTrees(factor1, prods_to_factors) ** 2
        
        return count % (10**9 + 7)

    def numFactoredBinaryTrees(self, arr: List[int]) -> int:
        prods_to_factors = {}
        arr_set = set(arr)
        for a in arr:
            for b in arr:
                if a*b in arr_set:
                    if a*b not in prods_to_factors:
                        prods_to_factors[a*b] = set()
                    if a <= b:
                        prods_to_factors[a*b].add((a,b))
                    else:
                        prods_to_factors[a*b].add((b,a))
        
        count = 0
        for el in arr:
            print(el, self.getNumberOfFactoredBinTrees(el, prods_to_factors))
            count = (count + self.getNumberOfFactoredBinTrees(el, prods_to_factors)) % (10**9 + 7)
        return count
                    


        
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