<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();
        int mod = 1e9+7;
        stack<int> st1;
        vector<int> prevSmall(n,-1);
        for(int i = 0;i<n;i++){
            while(!st1.empty() && arr[st1.top()]>arr[i]) st1.pop();
            if(!st1.empty()) prevSmall[i] = st1.top();
            st1.push(i);
        }
        stack<int> st2;
        vector<int> nextSmall(n,n);
        for(int i = 0;i<n;i++){
            if(!st2.empty() && arr[st2.top()]<arr[i]) st2.push(i);
            while(!st2.empty() && arr[st2.top()]>arr[i]){
                nextSmall[st2.top()] = i;
                st2.pop();
            }
            st2.push(i);
        }
        int res = 0;
        for(int i = 0;i<n;i++){
            int a = prevSmall[i];
            int b = nextSmall[i];
            res = (res + ((arr[i])*((static_cast<long long>(i-a)*(b-i)))))%mod;
        }
        return res;
    }
};

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