<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 ans=0;
        const int MOD=1e9+7;
        int n=arr.size();

        vector<int> nsl(arr.size());
        vector<int> nsr(arr.size());

        stack<int> st;
        for(int i=0;i<n;++i){
            if(st.empty()){
                nsl[i]=-1;
                st.push(i);
                continue;
            }
            while(!st.empty() && arr[st.top()]>=arr[i]){
                st.pop();
            }
            nsl[i]=st.empty()?-1:st.top();
            st.push(i);
        }

        st=stack<int>();
        
        for(int i=n-1;i>=0;--i){
            if(st.empty()){
                nsr[i]=n;
            }
            else{
                while(!st.empty() && arr[st.top()]>arr[i]){
                    st.pop();
                }
                nsr[i]=st.empty()?n:st.top();
            }
            st.push(i);
        }
        
        for(int i=0;i<n;++i){
            long long l=i-nsl[i];
            long long r=nsr[i]-i;
            long long m=(l%MOD * r%MOD) % MOD;
            ans=(ans + (arr[i] * m) % MOD) % MOD;
        }

        return ans;
    }
};
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