<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 {
    int mod = 1000000007;
public:
    int sumSubarrayMins(vector<int>& ar) {
        int n = ar.size();
        vector<int>ans(n, 0);
        int j = n - 1;
        stack<int>st;
        int ant = 0 ;
        while(j >= 0){
            while(!st.empty() && ar[j] < ar[st.top()]){
                st.pop();
            }
            if(st.empty()){
                ans[j] = ((n - j) * ar[j])%mod;
            } else{
                ans[j] = (((st.top() - j) * ar[j])%mod + ans[st.top()])%mod;
            }
            ant = (ant + ans[j])%mod;
            st.push(j);
            j--;
        }
        return ant;
    }
};
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