<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 size = arr.size();
        stack<int> s;
        vector<pair<int, int>> b(size);

        for(int i=0; i!=size; i++) {
            while(!s.empty() && arr[i] < arr[s.top()]) {
                s.pop();
            }

            if(s.empty()) b[i].first = i+1;
            else b[i].first = i-s.top();

            s.push(i);
        }

        while(!s.empty()) s.pop();

        for(int i=size-1; i>=0; i--) {
            while(!s.empty() && arr[i] <= arr[s.top()]) {
                s.pop();
            }

            if(s.empty()) b[i].second = size-i;
            else b[i].second = s.top()-i;
            
            s.push(i);
        }

        int MOD = 1e9 + 7;
        int sum = 0;
        for(int i=0; i!=size; i++) {
            unsigned long int prod = arr[i]*b[i].first;
            prod *= b[i].second;
            sum = sum%MOD + (prod)%MOD;
        }
        return sum;
    }
};
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