<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>& nums) {
        stack<int>  s;
        int mod = 1e9+7;
        vector<int> left(nums.size(),-1);
        vector<int> right(nums.size(),nums.size());
        for(int i=0;i<nums.size();i++) {
            while(!s.empty() && nums[s.top()]>nums[i]) {
                right[s.top()] = i;
                s.pop();
            }
            if(s.size()) 
                left[i] = s.top();
            s.push(i);
        }
        while(!s.empty()) {
            right[s.top()] = nums.size(); 
            int top = s.top() ; s.pop();
            left[top] = s.size() ? s.top() : -1;
        }

        long long minsum = 0;
        for(int i=0;i<nums.size();i++) {
            minsum = (minsum + (1LL)*nums[i]*(i-left[i])*(right[i]-i)) % mod;
        }
        return minsum;
    }
};
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