<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 MOD=1e9+7;
    int sumSubarrayMins(vector<int>& arr) {
        int n=arr.size(),ans=0;
        stack<pair<int,int>> st;
        stack<pair<int,int>> st2;
        vector<int> pse(n,-1),nse(n,n);

        for(int i=0;i<n;i++){
            if(st.empty()){
                st.push({arr[i],i});
                continue;
            }
            while(st.size() && st.top().first>arr[i]){
                st.pop();
            }
            if(st.size()){
                pse[i]=st.top().second;
                st.push({arr[i],i});
            }
            else{
                pse[i]=-1;
                st.push({arr[i],i});
            }
           
        }

        for(int i=n-1;i>=0;i--){
            if(st2.empty()){
                st2.push({arr[i],i});
            }
            else{while(st2.size() && st2.top().first>=arr[i]){
                st2.pop();
            }
            if(st2.size()){
                nse[i]=st2.top().second;
                st2.push({arr[i],i});
            }
            else{
                nse[i]=n;
            }
            st2.push({arr[i],i});
            }
        }

        for(int i=0;i<n;i++){
            int left=i-pse[i];
            int right=nse[i]-i;
            ans=(ans+((long)left*right*arr[i])%MOD)%MOD;
        }
        return ans;
    }
};
48000