<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:
    vector<int> findnse(vector<int>arr){
        int n = arr.size();
        vector<int>nse(n);
        stack<int>st;

        for(int i=n-1; i>=0; i--){
            while(!st.empty() && arr[st.top()] >= arr[i])
             st.pop();

            nse[i] = st.empty() ? n : st.top();

            st.push(i);
        }
        return nse;
    }

     vector<int> findPsee(vector<int>arr){
      int n = arr.size();
              vector<int>pse(n);
              stack<int>st;
      
              for(int i=0; i<n; i++){
                  while(!st.empty() && arr[st.top()] > arr[i])
                   st.pop();
      
                  pse[i] = st.empty() ? -1 : st.top();
      
                  st.push(i);
              }
              return pse;
     }
    int sumSubarrayMins(vector<int>& arr) {
        int n = arr.size();
        vector<int> nse = findnse(arr);
        vector<int> pse = findPsee(arr);
        int total = 0;
        int mod = (int)1e9 + 7;
        for(int i=0; i<n; i++){
            int left = i-pse[i];
            int right = nse[i]-i;
            total = (total + (right*left*1LL*arr[i])%mod)%mod;
        }
        return total;
    }
};
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