<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>

const auto __ = []() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  return true;
}();
class Solution {
public:
    int sumSubarrayMins(vector<int>& arr) {
    constexpr int32_t kMod = 1e9 + 7;
    uint64_t result = 0;
    stack<uint64_t> stack;
    const auto count = static_cast<uint64_t>(arr.size());
    for (uint64_t i = 0; i < count; ++i) {
      while (!stack.empty() && arr[i] < arr[stack.top()]) {
        const auto idx = stack.top();
        stack.pop();
        auto left = stack.empty() ? idx + 1 : idx - stack.top();
        auto right = i - idx;
        result = (result + arr[idx] * left * right) % kMod;
      }
      stack.push(i);
    }
    while (!stack.empty()) {
      const auto idx = stack.top();
      stack.pop();
      auto left = stack.empty() ? idx + 1 : idx - stack.top();
      auto right = count - idx;
      result = (result + arr[idx] * left * right) % kMod;
    }
    return static_cast<int32_t>(result);
    }
};
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