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