<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();
        vector<int> left(n), right(n);
        stack<pair<int, int>> s1, s2;

        // Calculate the number of subarrays where arr[i] is the minimum on the left
        for (int i = 0; i < n; ++i) {
            int count = 1;
            while (!s1.empty() && s1.top().first > arr[i]) {
                count += s1.top().second;
                s1.pop();
            }
            s1.push({arr[i], count});
            left[i] = count;
        }

        // Calculate the number of subarrays where arr[i] is the minimum on the right
        for (int i = n - 1; i >= 0; --i) {
            int count = 1;
            while (!s2.empty() && s2.top().first >= arr[i]) {
                count += s2.top().second;
                s2.pop();
            }
            s2.push({arr[i], count});
            right[i] = count;
        }

        // Calculate the result with modulo 10^9 + 7
        const int MOD = 1e9 + 7;
        long long result = 0;
        for (int i = 0; i < n; ++i) {
            result = (result + (long long) arr[i] * left[i] * right[i]) % MOD;
        }

        return result;
    }
};
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