<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,long long>mp;
unordered_set<int>nums;
int mod=1e9+7;
long long f2(int x){
  if(mp.count(x))
      return mp[x];
  
  long long res=1;
  for (auto i : nums)
    if(x % i == 0 && nums.count(x/i)){
    auto ans =  f2(i) * f2(x/i) ;
    ans%=mod;
    res+=ans;
    res%=mod; 
    }
  
  return  mp[x]=res;
}

int numFactoredBinaryTrees(vector<int>& arr) {  
  long long res=0;
  nums.clear();
  mp.clear();
  nums.insert(arr.begin(),arr.end());
  for (auto  i: nums){
      res+=f2(i);
      res%=mod;
  }
  return res;
}

};
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