<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>

map<int,long long>mp,memo;
int mx=-1,mod=1000000007;
vector<int>vec;

long long dp(int val)
{
    
    if(memo.find(val) == memo.end())
    {
        memo[val]=1;
        for (int num : vec)
            if(val%num == 0 && mp[val/num])
            {
                memo[val]=(memo[val] + (dp(val/num) * dp(num))%mod)%mod;
            }

    }

    return memo[val];
}

class Solution {
public:
    int numFactoredBinaryTrees(vector<int>& arr) {

        mp.clear();
        for(int val :arr)
        {
            mp[val]=1;
            mx=max(mx,val);
        }
        memo.clear();
        vec=arr;

        int ans=0;

        for(int val : vec)
        {
            ans=(ans+dp(val))%mod;
        }

       

        return ans;

    }
};
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