<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 n=arr.size(), mod=1e9+7;
        vector<long long> dp(n, -1); // dp[i]=sum of subarrays ending at i
        vector<int> stack;
        long long ans=0;
        for(int i=0; i<n; i++){
            while(!stack.empty() && arr[i]<=arr[stack.back()])
                stack.pop_back();
            if (!stack.empty()){
                int j=stack.back();
                dp[i]=dp[j]+arr[i]*(i-j);
            }
            else dp[i]=arr[i]*(i+1);
            stack.push_back(i);
            ans=(ans+dp[i])%mod;
        }
        return ans;      
    }
};
44800