<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 sumSubarrayMins(vector<int>& arr) {
        stack<int> stk;
        int n = arr.size();
        vector<int> dp(n, 0);
        dp[0] = arr[0];
        stk.push(0);
        int ans = dp[0];
        int MOD = 1e9 + 7;
        for (int i = 1; i < n; i++) {
            while (!stk.empty() && arr[i] < arr[stk.top()]) {
                stk.pop();
            }
            if (stk.empty())
                dp[i] = arr[i] * (i + 1);
            else 
                dp[i] = dp[stk.top()] + arr[i] * (i - stk.top());
            cout << dp[i] << " ";
            // if (arr[i] > arr[stk.top()]) {
            //     dp[i] = dp[i - 1] + arr[i];
            // } else {
            //     while (!stk.empty() && arr[i] < arr[stk.top()]) {
            //         stk.pop();
            //     }
            //     if (stk.empty()) {
            //         dp[i] = arr[i] * (i + 1);
            //     } else {
            //         dp[i] = arr[stk.top()] * (stk.top() + 1) + dp[i] * (i - stk.top());
            //     }
                
            // }
            ans += dp[i];
            ans %= MOD;
            stk.push(i);
        }
        return ans;
    }
};
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