<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 f(int n,vector<int>&arr){
        int ans=0,mod=1000000007;
        vector<int>dp(n,0);
        for(int i=0;i<n;i++){
            for(int j=0;j<i;j++){
                if(arr[i]%arr[j]==0){
                    auto it=arr.begin()+j;
                    int idx=lower_bound(arr.begin()+j,arr.end(),arr[i]/arr[j])-arr.begin();
                    if(idx!=n && arr[idx]==arr[i]/arr[j]){
                        int a=arr[j],b=arr[i]/arr[j];
                        int l=dp[j],r=dp[idx],cnt=0;
                        if(l==0 && r==0) cnt=1;
                        else if((l>0 && r==0) || (l==0 && r>0)){
                            if(r>0) cnt=(cnt+r)%mod;
                            else cnt=(cnt+l)%mod;
                            cnt=(cnt+1)%mod;
                        }
                        else{
                            cnt=(cnt+r)%mod;
                            cnt=(cnt+l)%mod;
                            cnt=(cnt+(long long)l*r)%mod;
                            cnt=(cnt+1)%mod;
                        }
                        if(a!=b) cnt=(cnt*2)%mod;
                        dp[i]=(dp[i]+cnt)%mod;
                       // cout<<arr[i]<<" "<<a<<" "<<b<<" "<<j<<" "<<idx<<" "<<l<<" "<<r<<" "<<cnt<<endl;
                    }
                }
            }
        }
        for(int i=0;i<n;i++){
            ans=(ans+dp[i])%mod;
            //cout<<dp[i]<<" ";
        }
        return (ans+n)%mod;
    }
    int numFactoredBinaryTrees(vector<int>& arr) {
        int n=arr.size();
        sort(arr.begin(),arr.end());
        return f(n,arr);
    }
};
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