<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) {
        stack<pair<int,int>>left;
        stack<pair<int,int>>right;
        long long sum=0;
        int n=arr.size();
        int countl,countr;
        int lef[n],righ[n];
        for(int i=0;i<n;i++)
        {
            countl=1;
            countr=1;
            pair<int,int>p;
          while(!left.empty() && arr[i]<left.top().first)
          {
            countl+=left.top().second;
            left.pop();
          }
          left.push({arr[i],countl});
          lef[i]=countl;
          while(!right.empty() && arr[n-i-1]<=right.top().first)
          {
            countr+=right.top().second;
            right.pop();
          }
          right.push({arr[n-1-i],countr});
          righ[n-1-i]=countr;
        }
        for(int i=0;i<n;i++)
        {
            sum=(sum+(long long)arr[i]*righ[i]*lef[i])%(1000000000+7);
        }
        return sum;
    }
};
44000