<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>

using ll = long long ;
using vll = vector<pair<ll,ll>> ; 

class Solution {
public:
    int sumSubarrayMins(vector<int>& a) {
       
        stack<ll> st; 
        ll n = a.size();vll v(n,{-1,n});


        for(ll i=0;i<n;i++){

          while(!st.empty() and a[st.top()]>=a[i]){
             v[st.top()].second = i;st.pop();
          }
          if(!st.empty()){
             v[i].first = st.top();
          }

          st.push(i);

        }
        
        ll flag = 0;ll mod = 1e9+7; 
        for(ll i=0;i<n;i++)flag=(flag+(i-v[i].first)*(v[i].second-i)*a[i]%mod)%mod;


        return flag;

        
    }
};
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