<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:
    int numFactoredBinaryTrees(vector<int>& arr) {
        
        sort(arr.begin(), arr.end());
        
        vector<long long> dp(arr.size(), 1);
        long long mod = 1e9 + 7;
        
        for(int i = 1; i < arr.size(); i++) {

            int s = 0, e = i - 1;
            while(s <= e) {
                
                long long mul = 1ll*arr[s]*arr[e];
                if(mul < arr[i]) s++;
                else if(mul > arr[i]) e--;
                else {
                    if (s == e) {
                        // Case where both factors are the same
                        dp[i] = (dp[i] + (dp[s] * dp[e]) % mod) % mod;
                    } else {
                        // Case where factors are different
                        dp[i] = (dp[i] + (2 * dp[s] * dp[e]) % mod) % mod;
                    }
                    s++; e--;
                }
            }
        }
        long long ans = 0;
        for(long long n: dp) {
            
            ans += n;
            ans = ans % mod;
        }
        return (int) ans;
    }
};
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