<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 mod = 1e9+7;
    int sumSubarrayMins(vector<int>& nums) {
        int n = nums.size();
        vector<int>left(n,-1),right(n,n);
        stack<pair<int,int>>st;
        for(int i = 0;i<n;i++){
            while(!st.empty() && st.top().first >= nums[i]){
                st.pop();
            }
            if(!st.empty()){
                left[i] = st.top().second;
            }
            st.push({nums[i],i});
        }
        while(!st.empty())st.pop();
        for(int i = n-1;i>=0;i--){
            while(!st.empty() && st.top().first > nums[i]){
                st.pop();
            }
            if(!st.empty()){
                right[i] = st.top().second;
            }
            st.push({nums[i],i});
        }
        int ans = 0;
        for(int i = 0;i<n;i++){
            long long leftel = i-left[i];
            long long  rightel = right[i] - i;
            ans = (ans + (leftel * rightel % mod) * nums[i] % mod) % mod;
        }
        
        return ans%mod;
    }
};
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