<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 {
public:
    unordered_map<int, vector<pair<int, int>>> factors;
    unordered_map<int, bool> freq;
    long long int mod = 1e9 + 7;
    void findDivisors(int num) {
        vector<pair<int, int>> ans;
        for (int i = 2; i <= sqrt(num); i++) {
            if (num % i == 0) {
                ans.push_back({i, num / i});
            }
        }
        factors[num] = ans;
    }
    long long fun(int num) {
        if (factors[num].empty()) {
            return 1;
        } else {
            long long ans = 1;
            for (auto i : factors[num]) {
                if (freq[i.first] && freq[i.second]) {
                    if (i.first == i.second)
                        ans += (fun(i.first) * fun(i.second)) % mod;
                    else
                        ans += (2 * fun(i.first) * fun(i.second)) % mod;
                }
            }
            return ans;
        }
    }
    int numFactoredBinaryTrees(vector<int>& arr) {
        for (auto i : arr) {
            freq[i] = true;
            findDivisors(i);
        }
        long long int ans = 0;
        for (auto i : arr) {
            ans = (ans + fun(i)) % mod;
        }
        return ans;
    }
};
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