<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) {
        stack<pair<int, pair<int, int>>> st;  // Stack to store (element, (sum of subarray mins, count))
        int mod = 1e9 + 7;  // Use the correct modulus
        long long res = 0;

        for (int i = 0; i < arr.size(); ++i) {
            if (st.empty()) {
                st.push({arr[i], {arr[i], 1}});
                res = (res + arr[i]) % mod;
            } else {
                if (arr[i] > st.top().first) {
                    res = (res + arr[i]) % mod;
                    auto p = st.top();
                    res = (res + p.second.first) % mod;
                    int k = (p.second.first + arr[i]) % mod;
                    st.push({arr[i], {k, 1}});
                } else {
                    int curr = 1;
                    while (!st.empty() && arr[i] < st.top().first) {
                        curr += st.top().second.second;
                        st.pop();
                    }
                    if (st.empty()) {
                        res = (res + (i + 1) * (long long)arr[i]) % mod;
                        int k = ((i + 1) * (long long)arr[i]) % mod;
                        st.push({arr[i], {k, curr}});
                    } else {
                        int k = (curr * (long long)arr[i]) % mod;
                        k = (k + st.top().second.first) % mod;
                        res = (res + k) % mod;
                        st.push({arr[i], {k, curr}});
                    }
                }
            }
        }

        return (int)res;
    }
};

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