<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();
        int nse1[n];
        fill(nse1, nse1+n, n);

        int nse2[n];
        fill(nse2, nse2+n, -1);

        stack<int> stk;
        for(int i = n-1; i>=0; i--){
            while(stk.size() and arr[i] <= arr[stk.top()]) stk.pop();
            if(stk.size()) nse1[i] = stk.top();
            stk.push(i);
        }
        stack<int> sk;
        for(int i = 0; i<n; i++){
            while(sk.size() and arr[i] < arr[sk.top()]) sk.pop();
            if(sk.size()) nse2[i] = sk.top();
            sk.push(i);
        }
        for(int i = 0; i<n; i++){
            cout << nse1[i] << " " << nse2[i] <<" \n";
        }
        long long ans = 0;
        int M = 1e9 + 7;
        for(int i = 0; i<n; i++){
            int i1 = nse2[i];
            int i2 = nse1[i];

            i1 = i1 + 1;
            i2 = i2 - 1;
            // [i1:i] and [i:i2]
            long long val = (i-i1+1)*(i2-i+1);
            ans = (ans + val*arr[i])%M;
        }

        return ans;
    }
};
43100