<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();
        long long pse[n];
        long long nse[n];
        stack<long long>st1,st2;
        for(long long i=0;i<n;i++){
            long long j=n-1-i;

            pse[i]=-1;
            while(!st1.empty() && arr[st1.top()]>arr[i]){
                st1.pop();
            }
            if(!st1.empty()){
                pse[i]=st1.top();
            }
            st1.push(i);

            nse[j]=n;
            while(!st2.empty() && arr[st2.top()]>=arr[j]){
                st2.pop();
            }
            if(!st2.empty()){
                nse[j]=st2.top();
            }
            st2.push(j);
        }
        long long ans=0,mod=1e9+7;
        for(long long i=0;i<n;i++){
            // cout<<pse[i]<<" "<<nse[i]<<endl;
            long long left=i-pse[i];
            long long right=nse[i]-i;
            // cout<<left<<" "<<right<<endl<<endl;
            ans=(ans+(((left*right)%mod)*arr[i])%mod)%mod;
        }

        
        return ans;
    }
};
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