<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();
        long long int mod=1e9+7;
        vector<long long int> next(n,n),prev(n,-1);
        stack<int> s1,s2;
        s1.push(n);
        s1.push(n-1);
        s2.push(-1);
        s2.push(0);
        next[n-1]=1;
        prev[0]=1;
        for(int i=n-2;i>=0;i--){
            if(arr[i]<=arr[s1.top()]){
               while(s1.top()!=n&&arr[s1.top()]>=arr[i]){
                s1.pop();
               } 
            }
            
                next[i]=s1.top()-i;
                s1.push(i);
        }
        for(int i=1;i<n;i++){
            if(arr[i]<arr[s2.top()]){
               while(s2.top()!=-1&&arr[s2.top()]>arr[i]){
                s2.pop();
               } 
            }
            
                prev[i]=i-s2.top();
                s2.push(i);
        }
        long long int ans=0;
        for(auto i:next){cout<<i<<" ";}cout<<endl;
        for(auto i:prev){cout<<i<<" ";}cout<<endl;
        for(int i=0;i<n;i++){
            int temp=(((next[i]*prev[i])%mod)*(arr[i])%mod);
            ans=(ans+temp)%mod;
        }
        return ans;

    }
};
48900