<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 mod = 1e9+7;
        stack<pair<int,int>>s;
        
        s.push({arr[0],1});
        int sum = arr[0]%mod;
        int last = arr[0];
        for(int i = 1; i < arr.size(); i++){
            sum = (sum + arr[i])%mod;
           
            if(arr[i] > s.top().first){
                sum = (sum + last)%mod;
                last = (last+ arr[i])%mod;
                s.push({arr[i],1});
            }
            else{
                int factor = 0;
                while(!s.empty() && arr[i] <= s.top().first){
                    last = last - s.top().first*s.top().second;
                    factor += s.top().second;
                    s.pop();
                }
                sum = (sum + arr[i]*factor)%mod;
                sum = (sum + last)%mod;
                last = (last + arr[i]*(factor+1))%mod;
                s.push({arr[i],factor+1});
            }
        }
        return sum; 
    }
};
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