<p>Given an array of integers arr, find the sum of <code>min(b)</code>, where <code>b</code> ranges over every (contiguous) subarray of <code>arr</code>. Since the answer may be large, 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 = [3,1,2,4]
<strong>Output:</strong> 17
<strong>Explanation:</strong> 
Subarrays are [3], [1], [2], [4], [3,1], [1,2], [2,4], [3,1,2], [1,2,4], [3,1,2,4]. 
Minimums are 3, 1, 2, 4, 1, 1, 2, 1, 1, 1.
Sum is 17.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [11,81,94,43,3]
<strong>Output:</strong> 444
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= arr[i] &lt;= 3 * 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int sumSubarrayMins(vector<int>& arr) 
    {
        int n=arr.size();
        int nse[n];
        int mod=1e9+7;
        for(int i=(n-1);i>=0;i--)
        {
            nse[i]=i+1;
            while(nse[i]!=n && arr[i]<=arr[nse[i]])
            {
                nse[i]=nse[nse[i]];
            }
        } 

        int pse[n];
        for(int i=0;i<n;i++)
        {
            pse[i]=i-1;
            while(pse[i]!=-1 && arr[i]<arr[pse[i]])
            {
                pse[i]=pse[pse[i]];
            }
        } 

        long long ans=0;
        for(int i=0;i<n;i++)
        {
            long long freq=1LL*(i-pse[i])*(nse[i]-i);
            long long num=1LL*freq*arr[i];
            ans=(ans+num)%mod;
        }   
        return ans;

    }
};
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