<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) {
    constexpr int kMod = 1'000'000'007;
    const int n = arr.size();
    // dp[i] := the number of binary trees with arr[i] as the root
    vector<long> dp(n, 1);
    unordered_map<int, int> numToIndex;

    ranges::sort(arr);

    for (int i = 0; i < n; ++i)
      numToIndex[arr[i]] = i;

    for (int i = 0; i < n; ++i)  // arr[i] is the root
      for (int j = 0; j < i; ++j)
        if (arr[i] % arr[j] == 0) {  // arr[j] is the left subtree
          const int right = arr[i] / arr[j];
          if (const auto it = numToIndex.find(right); it != numToIndex.cend()) {
            dp[i] += dp[j] * dp[it->second];
            dp[i] %= kMod;
          }
        }

    return accumulate(dp.begin(), dp.end(), 0L) % kMod;
  }
};
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