<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:
    vector<int> findNse(vector<int>& arr) {
        int n = arr.size();
        vector<int> nse(n);
        stack<int> st;
        for(int i = arr.size()-1 ; i >= 0 ; i--) {
            while(!st.empty() && arr[st.top()] >= arr[i])
                st.pop();
            if(st.empty())
                nse[i] = arr.size();
            else
                nse[i] = st.top();
            st.push(i);
        }
        return nse;
    }

    vector<int> findPsee(vector<int>& arr) {
        vector<int> psee;
        stack<int> st;
        int n = arr.size();
        for(int i = 0 ; i < n ; i++) {
            while(!st.empty() && arr[st.top()] > arr[i])
                st.pop();
            if(st.empty())
                psee.push_back(-1);
            else
                psee.push_back(st.top());
            st.push(i);
        }
        return psee;
    }

    int sumSubarrayMins(vector<int>& arr) {
        vector<int> nse = findNse(arr);
        vector<int> psee = findPsee(arr);
        int sum = 0;
        int mod = 1e9+7;
        for(int i = 0 ; i < arr.size() ; i++) {
            int left = i - psee[i];
            int right = nse[i] - i;
            sum = (sum + ((1LL * left * right * arr[i]) % mod)) % mod;
        }
        return sum;
    }
};
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