<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();
        stack<int>right;
        stack<pair<int,int>>left;
        vector<int>aleft(n,0),aright(n,0);
        for(int i=0;i<arr.size();i++)
        {
            int cnt=1;
            while(!left.empty()&& left.top().first>arr[i])
            {
                cnt=cnt+left.top().second;
                left.pop();
            }
            left.push({arr[i],cnt});
            aleft[i]=cnt;
        }
        for(int i=arr.size()-1;i>=0;i--)
        {
            while(!right.empty() && arr[right.top()]>=arr[i])
            {
                right.pop();
            }
            
            aright[i]=right.empty()? n:right.top();
            right.push(i);
        }

        int ans=0;
        
        int mod=1000000007;
        for(int i=0;i<arr.size();i++)
        {
            ans=(ans+(arr[i]*(long long)(aleft[i]*(aright[i]-i))%mod)%mod)%mod;
        }
        //ans=ans%1000000007;
        return ans;
    }
};
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