<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 f[2][30001];
    int g[30001];
    int mod=1e9+7;
    int sumSubarrayMins(vector<int>& a) {
        int n=a.size();
        int l=0;
        for (int i=0;i<n;i++) {
            while (l>0&&a[g[l]]>a[i]) l--;
            if (l==0) f[0][i]=-1;
            else f[0][i]=g[l];
            l++;
            g[l]=i;
        }
        l=0;
        for (int i=n-1;i>=0;i--) {
            while (l>0&&a[g[l]]>=a[i]) l--;
            if (l==0) f[1][i]=n;
            else f[1][i]=g[l];
            l++;
            g[l]=i;
        }
        int res=0;
        for (int i=0;i<n;i++) {
            res=(res+a[i]*(f[1][i]-i)*(long long)(i-f[0][i]))%mod;
        }
        return res;
    }
};
41700