<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 TLESolve(vector<int>&arr){
        int sum = 0;
        int n = arr.size();
        int M = 1e9 + 7;
        for(int i=0;i<n;i++){
            priority_queue<int,vector<int>,greater<int>> pq;
            pq.push(arr[i]);
            sum = (sum + pq.top())%M;
            for(int j=i+1;j<n;j++){
                pq.push(arr[j]);
                sum = (sum + pq.top())%M;
            }
        }
        return sum%M;
    }

    int sumSubarrayMins(vector<int>& arr) {
        int mod= (1e9+7);
        long sum=0;
        long sumc= 0;
        deque<pair<int,int>>dq;
        for(int i=0;i<arr.size();i++){
            int popped= 1;
            while(dq.size()&& dq.back().first >= arr[i]){
                popped+= dq.back().second;
                sumc-= (dq.back().first * dq.back().second);
                dq.pop_back();
            }
            sum+= sumc;
            dq.push_back({arr[i],popped});
            sum= (sum + arr[i]*popped)% mod;
            sumc+= arr[i]*popped;
        }  
        return sum;
    }
};
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