<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>& nums) {
        ios_base::sync_with_stdio(false);
        int i, n = nums.size();
        long long int out = 0, l, r, temp;
        int mod = (int)1e9+7;
        vector<int> nge(n, -1);
        stack<pair<int, int>> st;

        for (i = 0; i < n; i++) {
            while (st.size() && st.top().first > nums[i]) {
                st.pop();
            }
            nge[i] = st.size() ? st.top().second : -1;
            st.push({nums[i], i});
        }
        st = stack<pair<int, int>>();

        for(i = n-1; i >= 0; i--) {
            while (st.size() && st.top().first >= nums[i]) {
                st.pop();
            }
            r = st.size() ? st.top().second-i : n-i;
            l = i-nge[i];
            out = (out + (((l*r)%mod) * nums[i]) % mod) % mod;
            // cout << nums[i] << " " << out << " " << l << " " << r <<  endl;
            st.push({nums[i], i});
        }
        return out % mod;
    }
};
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