<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());
        map<int, long long int> mp;
        long long int mod = 1e9 + 7;
        mp[arr[0]] = 1;
        if(arr.size() == 1)return(1);

        for(int i=1;i<arr.size();i++){
            int num = arr[i];
            int k = sqrt(num) + 1e-12;
            mp[num] = 1;
            if(num == k*k){
                mp[num] = (mp[num] + (mp[k] * mp[k])%mod)%mod;
                k--;
            }
            for(int j=2;j<=k;j++){
                if(num%j==0)mp[num] = (mp[num] + 2*(mp[j] * mp[num/j])%mod)%mod;
            }
        }

        long long ans = 0;
        for(int i=0;i<arr.size();i++){
            ans += mp[arr[i]];
        }
        ans = ans % mod;
        return(ans);
    }
};
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