<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(int ind, int n, vector<int> v) {
        stack<int> s;
        vector<int> ans(n);
        for (int j = n - 1; j >= 0; j--) {
            while (!s.empty() && v[s.top()] >= v[j]) {
                s.pop(); // we find next smaller element so we pop all element
                         // which are greater than current
            }
            if (s.empty()) {
                ans[j] = n;
            } else {
                ans[j] = s.top();
            }
            s.push(j);
        }
        return ans;
    }

    vector<int> findlsee(int i, int n, vector<int>& v) {
        stack<int> s;
        vector<int> ans(n);
        for (int i = 0; i < n; i++) {
            while (!s.empty() && v[s.top()] > v[i]) {
                s.pop(); // because we find previous smaller or equal element
            }
            if (s.empty()) {
                ans[i] = -1;
            } else {
                ans[i] = s.top();
            }
            s.push(i);
        }
        return ans;
    }

    int sumSubarrayMins(vector<int>& v) {
        long long ans = 0, n = v.size();
        vector<int> nse, lsee;
        nse = findnse(0, n, v);
        lsee = findlsee(0, n, v);
        // for (auto c : nse)
        //     cout << c << ' ';
        // cout << endl;
        // for (auto c : lsee)
        //     cout << c << ' ';
        // cout<<endl;
        long long mod = (long long)1e9 + 7;
        for (int i = 0; i < v.size(); i++) {
            int left = i - lsee[i];
            int right = nse[i] - i;
            // cout << left << ' ' << right << endl;
            long long multi = (left * right)%mod;
            multi = (multi * v[i]) % mod;
            ans = (ans + multi) % mod;
        }
        return ans;
    }
};
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