<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) {
        vector <int> stack;
        int mod = 1e9+7;
        stack.push_back(-1);
        int ans = 0;
        int n = arr.size();
        for (int i = 0;i < n; ++i) {
            while(stack.back() != -1 && arr[i] <= arr[stack.back()]) {
                int curr = stack.back();
                stack.pop_back();
                ans += (arr[curr]*(i - stack.back() - 1ll) % mod);
                ans = ans % mod;
                ans += (arr[curr] * ((curr - stack.back() -1ll) * (i - curr -1ll) )%mod ) % mod;
                ans = ans % mod;
            }
            stack.push_back(i);
        }
        int i = n;
        while(stack.back() != -1) {
            int curr = stack.back();
            stack.pop_back();
            //ans += (arr[curr]*(i - stack.back() - 1ll) % mod);
            ans = ans % mod;
            ans += (arr[curr] * ((curr - stack.back()*1ll) * (i - curr*1ll) )%mod ) % mod;
            ans = ans % mod;
        }
        return ans;
    }
};
42000