<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 = 1'000'000'007;
    int sumSubarrayMins(vector<int>& arr) {
        int n = arr.size();
    
    // Arrays to store the index of the previous and next less elements
    std::vector<int> ple(n, -1); // Previous Less Element
    std::vector<int> nle(n, n);  // Next Less Element
    
    // Monotonic stack for Previous Less Element
    std::stack<int> stack;
    for (int i = 0; i < n; ++i) {
        while (!stack.empty() && arr[stack.top()] >= arr[i]) {
            stack.pop();
        }
        if (!stack.empty()) {
            ple[i] = stack.top();
        }
        stack.push(i);
    }
    
    // Clear stack for the next use
    while (!stack.empty()) {
        stack.pop();
    }

    // Monotonic stack for Next Less Element
    for (int i = n - 1; i >= 0; --i) {
        while (!stack.empty() && arr[stack.top()] > arr[i]) {
            stack.pop();
        }
        if (!stack.empty()) {
            nle[i] = stack.top();
        }
        stack.push(i);
    }
    
    // Calculate the result
    long long result = 0;
    for (int i = 0; i < n; ++i) {
        int left = i - ple[i];
        int right = nle[i] - i;
        result = (result + static_cast<long long>(arr[i]) * left * right) % MOD;
    }
    
    return static_cast<int>(result);

    }
};
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