<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 {
    int MOD = 1e9+7;
    long long solve(vector<int>& arr, int low, int high){
        if(low>high) return 0;
        if(low==high) return arr[low];
        int minv = INT_MAX;
        int mini = INT_MAX;
        long long ans = 0;
        for(int i=low;i<=high;i++){
            if(minv>arr[i]){
                minv = arr[i];
                mini = i;
            }
        }
        long long temp = minv;
        temp = ((high-mini+1)*temp) %MOD;
        temp = ((mini-low+1)*temp) %MOD;
        ans = (ans+temp) %MOD;
        ans = (ans+solve(arr,low,mini-1)) %MOD;
        ans = (ans+solve(arr,mini+1,high)) %MOD;
        return ans %MOD;
    }
public:
    int sumSubarrayMins(vector<int>& arr) {
        int n = arr.size();
        return solve(arr,0,n-1);
    }
};
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