<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 modu=1e9+7;
    int sumSubarrayMins(vector<int>& arr) {
      int n=arr.size();
vector<int>sum(n,0);
sum[n-1]=arr[n-1];
int mini=arr[n-1];
for(int i=n-2;i>=0;i--){
   int mnm=arr[i];
   if(mnm<=mini){
       sum[i]+=(n-i)*arr[i];
   }
   else{
       sum[i]+=arr[i];
     //  sum[i]+=sum[i+1];
     int j=i+1;
     while(j<n && arr[i]<arr[j]){
         sum[i]+=arr[i];
         j++;
     }
     if(j<n)
     sum[i]+=sum[j];
   }
   mini=min(mini,arr[i]);
}
int ans=0;
for(int i=0;i<n;i++){
    ans=((ans%modu)+(sum[i]%modu))%modu;;
}
return ans;
    }
};
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