<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 long long mod = 1e9 + 7;
class Solution {
public:
    int sumSubarrayMins(vector<int>& arr) {
        int n = arr.size();
        deque<int> st;
        vector<pair<long long, long long>> a(n);
        for (int i = 0; i < n; i++) {
            while (!st.empty() && arr[st.back()] > arr[i]) {
            a[st.back()].second = i-1;
            st.pop_back();
            }

            if (!st.empty()) a[i].first = st.back() + 1;
            else a[i].first = 0;

            st.push_back(i);
        }

        long long ans = 0;
        for (int i = 0; i < n; i++) {
            if (a[i].second == 0 && !(n >= 2 && i == 0 && arr[i+1] < arr[i])) 
                a[i].second = n-1;
            long long l = i - a[i].first + 1, r = a[i].second - i + 1;
            ans += (l * r * (long long)arr[i]) % mod;
            ans %= mod;
        }
        return ans;
    }
};
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