<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();
        vector<int> vec(n);
        stack<int> st;
        st.push(0);
        vec[0] = arr[0];
        int mod = 1e9 + 7;

        for(int i = 1; i < n; i++) {
            if(arr[st.top()] < arr[i]) {
                vec[i] = ((long long)arr[i] * (i - st.top()) + vec[st.top()]) % mod;
                st.push(i);
            }
            else {
                while(!st.empty() && arr[st.top()] >= arr[i]) {
                    st.pop();
                }

                if(st.empty()) {
                    vec[i] = ((long long)(i + 1) * arr[i]) % mod;
                }
                else {
                    vec[i] = ((long long)arr[i] * (i - st.top()) + vec[st.top()]) % mod;
                }

                st.push(i);
            }
        }

        int ans = 0;
        for(auto z: vec) {
            ans = (ans + z) % mod;
        }

        return ans;
    }
};

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