<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) {

 const int MOD = 1e9 + 7;
    int n = arr.size();
    sort(arr.begin(), arr.end());  // Sort the array to ensure correct product factorization.

    unordered_map<int, long> dp;  // Stores the number of trees with arr[i] as the root.

    // Each element in the array is initially a tree by itself.
    for (int i = 0; i < n; i++) {
        dp[arr[i]] = 1;  // Each element can at least form a tree with just itself.
    }

    // For each element, try to form it using two other elements.
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < i; j++) {
            // Check if arr[i] can be factored as arr[j] * arr[k].
            if (arr[i] % arr[j] == 0) {  // arr[i] is divisible by arr[j]
                int other = arr[i] / arr[j];
                if (dp.find(other) != dp.end()) {
                    // If both arr[j] and other exist, add the contribution
                    dp[arr[i]] = (dp[arr[i]] + dp[arr[j]] * dp[other]) % MOD;
                }
            }
        }
    }

    long result = 0;
    // Sum up all the ways to form trees with each element as the root.
    for (const auto& entry : dp) {
        result = (result + entry.second) % MOD;
    }

    return result;
}

};
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