<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 MOD = 1e9+7;
        int n = arr.size();
      

        stack<int> st;
       
        int ans =0 ;
        for(int i = 0 ; i < n ; i++)
        {   
            while( !st.empty() && arr[st.top()] > arr[i]   )
            {
                int x = arr[st.top()] ; 
                int y = st.top();
                st.pop() ; 
                int idx ; 
                if( st.empty()) idx =-1; 
                else{ 
                    
                    idx = st.top();
                    
                }
                int k = (i-idx-1);
                int com =( x*( ((i-y)*(y-idx))%MOD ) )%MOD ;
                ans  = (ans +  com) %MOD;
                // cout<<k <<' '<<com<<endl;

           
            }
            st.push(i);
        }
        while( !st.empty())
        {
                int x = arr[st.top()] ; 
                int y = st.top() ; 
                st.pop() ; 
                int idx ; 
                if( st.empty()) idx =-1; 
                else{ 
                    
                    idx = st.top();
                    
                }
                int k = (n-idx-1);
               int com =( x*1LL*( ((n-y)*(y-idx))%MOD ) )%MOD ;
                 ans  = (ans +  com) %MOD;
                // cout<<k <<' '<<com<<endl;



        }


        return ans;













    }
};
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