<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:
    void findnse(int nse[],vector<int>arr){
        stack<int> st;
        int n = arr.size();
        for(int i = arr.size()-1;i>=0;i--){
            while(!st.empty() && arr[i]<= arr[st.top()]){
                st.pop();
            }
            nse[i] = st.empty()?n:st.top();
            st.push(i);

        }
    }
    void findpsee(int psee[],vector<int>& arr){
        stack<int>st;
        int n = arr.size();
        for(int i=0;i<n;i++){
            while(!st.empty() && arr[i]<arr[st.top()]){
                st.pop();
            }

            psee[i] = st.empty()?-1:st.top();
            st.push(i);
        }
    }
    int sumSubarrayMins(vector<int>& arr) {
        int n=arr.size();
        int nse[n];
        int psee[n];
        findnse(nse,arr);
        findpsee(psee,arr);
        int MOD = 1e9 + 7;

        long long total = 0;
        for(int i=0;i<n;i++){
            int l = i -psee[i];
            int r = nse[i] - i;
            total = (total + (l * r * 1LL * arr[i])%MOD)%MOD;
        }
        return total;

    }
};
44700