<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:
    const int mod=1e9+7;
    int sumSubarrayMins(vector<int>& arr) {
        int n=arr.size();
        stack<int>st1;
        int l[n];
        for(int i=0;i<n;i++){
            while(!st1.empty() && arr[st1.top()]>arr[i]){
                st1.pop();
            }
            if(st1.empty())l[i]=0;
            else l[i]=st1.top()+1;
            st1.push(i);
        }
        stack<int>st2;
        int r[n];
        for(int i=n-1;i>=0;i--){
            while(!st2.empty() && arr[st2.top()]>=arr[i]){
                st2.pop();
            }
            if(st2.empty())r[i]=n-1;
            else r[i]=st2.top()-1;
            st2.push(i);
        }
        /*for(int i=0;i<n;i++){
            cout<<l[i]<<" "<<r[i]<<endl;
        }*/
        long long ans=0;
        for(int i=0;i<n;i++){
            (ans+=(1ll*(i-l[i]+1)*(r[i]-i+1)*arr[i]))%=mod;
        }
        return ans;
    }
};
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