<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:
    const int mod = 1e9+7;
    int sumSubarrayMins(vector<int>& arr) {
        vector<int> inc; // increasing stack of elements, vector of their indexes
        vector<int> p; // prefix sum p[x] = sum j=0 to inc[x] min(arr[j..i])
        int sum = arr[0];
        inc.push_back(0);
        p.push_back(arr[0]);
        for (int i = 1; i < arr.size(); i++) {
            while (!inc.empty() && arr[inc.back()] >= arr[i]) {
                inc.pop_back();
                p.pop_back();
            }
            // number of new subarrays that have min = arr[i]
            int nelem = i - (inc.empty() ? -1 : inc.back());
            sum += nelem * arr[i];
            sum %= mod;
            if (!inc.empty()) {
                sum += p.back();
                sum %= mod;
            }
            p.push_back(((p.empty() ? 0 : p.back()) + nelem * arr[i]) % mod);
            inc.push_back(i);
        }
        return sum;
    }
};
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