<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>

#define ll long long
class Solution {
public:
    int sumSubarrayMins(vector<int>& arr) {
        ll n=arr.size();
        stack<pair<ll,ll> > st;
        st.push({-1,-1});
        vector<pair<int,int> > ranges(n);
        for(int i=0;i<n;i++){
            while(st.top().first>arr[i]){
                st.pop();
            }
            ranges[i].first=st.top().second;
            st.push({arr[i],i});
        }
        stack<pair<ll,ll> > st2;
        st2.push({-1,n});
        for(int i=n-1;i>=0;i--){
            while(st2.top().first>=arr[i]){
                st2.pop();
            }
            ranges[i].second=st2.top().second;
            st2.push({arr[i],i});
        }
        ll sum=0;
        ll d=1e9+7;
        for(int i=0;i<n;i++){
            ll el=arr[i];
            sum+=el*(i-ranges[i].first)*(ranges[i].second-i);
            sum=sum%d;
        }
        return (int)sum;
    }
};
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