<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>

#include <bits/stdc++.h>

using namespace std;

class Solution {
public:
    int sumSubarrayMins(vector<int> &arr) {
        int res = 0, n = arr.size(), M = 1e9 + 7;
        int l[n], stk[n], tt = -1;
        for (int i = 0; i < n; i++) {
            while (~tt && arr[stk[tt]] >= arr[i]) tt--;
            l[i] = ~tt ? stk[tt] + 1 : 0;
            stk[++tt] = i;
        }
        tt = -1;
        for (int i = n - 1; i >= 0; i--) {
            while (~tt && arr[stk[tt]] > arr[i]) tt--;
            int ri = ~tt ? stk[tt] - 1 : n - 1;
            stk[++tt] = i;
            res = (res + (long long) (i - l[i] + 1) * (ri - i + 1) % M * arr[i]) % M;
        }
        return res;
    }
};
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