<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;
	// find next smaller element
        vector<long long> nextSmaller(vector<int>&arr){
        int n=(int)arr.size();
        vector<long long> nse(n);
        for(int i=n-1;i>=0;i--){
            nse[i]=i+1;
            while(nse[i]!=n && arr[i]<=arr[nse[i]]){
                nse[i]=nse[nse[i]];
            }
        }
        return nse;
    }
	
	// find previous smaller element 
    vector<long long> prevSmaller(vector<int> &arr){
        int n=arr.size();
         vector<long long> pse(n);
        for(int i=0;i<n;i++){
            pse[i]=i-1;
            while(pse[i]!=-1&&arr[i]<arr[pse[i]]){
                pse[i]=pse[pse[i]];
            }
        }
         return pse;
    }
	
    int sumSubarrayMins(vector<int>& arr) {
        long long ans=0;
         vector<long long>next_sm=nextSmaller(arr);
        auto prev_sm=prevSmaller(arr);
        for(int i=0;i<(int)arr.size();i++){
            ans+=(arr[i]*((i-prev_sm[i])*(next_sm[i]-i)));
        }
         return ans%mod;
    }
};
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