<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;
    vector<int> prev_smaller_equal_element_index(vector<int>& arr, int n)
    {
        vector<int> ans(n,0);
        stack<int> st;
        for(int i=0 ; i<n ; i++)
        {
            while(!st.empty() && arr[st.top()]> arr[i]) 
            {
                st.pop();
            }
            ans[i] = st.empty()? -1: st.top();
            st.push(i);
        }
        return ans;
    }
    vector<int> next_smaller_element_index(vector<int>& arr, int n)
    {
        vector<int> ans(n,0);
        stack<int> st;
        for(int i=n-1 ; i>=0 ; i--)
        {
            while(!st.empty() && arr[st.top()]>= arr[i])
            {
                st.pop();
            }
            ans[i] = st.empty()? n: st.top();
            st.push(i);
        }
        return ans;
    }
    int sumSubarrayMins(vector<int>& arr) {
        int n = arr.size();
        vector<int> prev(n);
        vector<int> next(n);
        prev= prev_smaller_equal_element_index(arr,n); 
        next= next_smaller_element_index(arr,n); 
        long long sum =0 ; 
        for(int i=0 ; i< arr.size() ; i++)
        {
            long long  left = i - prev[i];
            long long  right= next[i]- i;
            sum = (sum +(left*right*arr[i])%mod)%mod;
        }
        return sum;
        
    }
};
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