<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();
        vector<int> left(n,0) , right(n,0);
        stack<pair<int , int>> sLeft , sRight;  //initialize pair of stacks 

        for(int i=0; i<n; i++){
            int cnt = 1;
            while(!sLeft.empty() && sLeft.top().first > arr[i]){
                cnt += sLeft.top().second;
                sLeft.pop();
            }

            sLeft.push({arr[i] , cnt});
            left[i] = cnt;   //update the left
        }

        
        for(int i=n-1; i>=0; i--){
            int cnt = 1;
            while(!sRight.empty() && sRight.top().first >= arr[i]){
                cnt += sRight.top().second;
                sRight.pop();
            }

            sRight.push({arr[i] , cnt});
            right[i] = cnt;
        }

        int ans = 0; 
        const int mod = 1e9 + 7; 
        for(int i=0; i<n; i++){
            ans = (ans + (arr[i] * (long long) (left[i] * right[i])%mod)%mod)%mod;
        }

        return ans;
    }
};
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