<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) {
        int n = arr.size();
        vector<int> nse(n);
        vector<int> pse(n);
        int ans = 0, mod = int(1e9 + 7);

        stack<int> stk;
        pse[0] = -1;
        stk.push(0);
        for(int i = 1;i < n;i++) {
            if(arr[stk.top()] <= arr[i]) {
                pse[i] = stk.top();
                stk.push(i);
            }
            else {
                while(!stk.empty() && arr[i] < arr[stk.top()]) stk.pop();
                if(stk.empty()) pse[i] = -1;
                else pse[i] = stk.top();
                stk.push(i);
            }
        }
        while(!stk.empty()) stk.pop();
        stk.push(n - 1);
        nse[n - 1] = n;
        for(int i = n - 2;i >= 0;i--) {
            if(arr[stk.top()] < arr[i]) {
                nse[i] = stk.top();
                stk.push(i);
            }
            else{
                while(!stk.empty() && arr[i] <= arr[stk.top()]) stk.pop();
                if(stk.empty()) nse[i] = n;
                else nse[i] = stk.top();
                stk.push(i);
            }
        }
        // for(auto it : pse) cout << it << " ";
        // cout <<endl;
        // for(auto it : nse) cout << it << " ";
        // cout << endl;
        for(int i = 0;i < n;i++) {
            // int temp = nse[i] - pse[i] - 1;
            ans = ((long)ans + ((long)arr[i] * (long) ((long)(nse[i] - i) * (long) (i - pse[i])) % mod) % mod) % mod;
        }

        return ans;
    }
};
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