<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 {
#define MOD 1000000007
public:
    int sumSubarrayMins(vector<int>& arr) {
        const int n = arr.size();
        int result = 0;

        for (int i = 0; i < n; i++) {
            int l = i - 1,
                r = i + 1;
            for (; l >= 0 && arr[l] > arr[i]; l--);
            for (; r < n && arr[r] >= arr[i]; r++);
            result = (result + ((static_cast<long long>(r - i) * static_cast<long long>(i - l))% MOD * static_cast<long long>(arr[i]))% MOD) % MOD;
        }
        return result;
    }
};
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