<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 mod 1000000007
#define ll long long
class Solution {
public:
    vector<int> ps, ns; // vector storing indexes of prev smaller and next smaller element w.r.t current element. 
    void prevSmaller(vector<int>& arr){
        stack<int> st;
        st.push(-1); // Prev Smaller of 1st element will be at index -1.
        for(int i = 0; i<arr.size(); i++){
            while(st.top() != -1 && arr[st.top()] > arr[i]) st.pop();
            ps.push_back(st.top());
            st.push(i);
        }
    }

    void nextSmaller(vector<int>& arr){
        int n = arr.size();
        stack<int> st;
        st.push(n); // Next smaller of last element will be at index n.
        for(int i = arr.size()-1; i>=0; i--){
            while(st.top() != n && arr[st.top()] >= arr[i]) st.pop();
            ns.push_back(st.top());
            st.push(i);
        }
        reverse(ns.begin(), ns.end());
    }

    int sumSubarrayMins(vector<int>& arr) {
        int n = arr.size();
        prevSmaller(arr); // Finding vector containing indexes of prev smaller element.
        nextSmaller(arr); // Finding vector containing indexes of prev smaller element.

        for(auto i : ps) cout << i <<" ";
        cout << endl;
        for(auto i : ns) cout << i <<" ";

        int ans = 0;
        for(int i = 0; i<n; i++){
            ll leftCnt = i - ps[i]; // Count before prev smaller element is found.
            ll rightCnt = ns[i] - i; // Count before next smaller element is found.

            ll t = (leftCnt%mod * rightCnt%mod)%mod; // Total number of subarray containing element arr[i] as the smallest element.

            ans = (ans + (arr[i]%mod * t%mod)%mod)%mod;
        }

        return ans;
    }
};
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