<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>

def getRightSmallestElement(arr):
    n = len(arr)
    nse = [-1] * n
    st = []

    for i in range(n - 1, -1, -1):
        while st and arr[st[-1]] >= arr[i]:
            print(arr[i])
            st.pop()
        
        if st:
            nse[i] = st[-1]
        else:
            nse[i] = n
        
        st.append(i)
    
    return nse

def getLeftSmallestOrEqualElement(arr):
    n = len(arr)
    lsee = [-1] * n
    st = []

    for i in range(n):
        while st and arr[st[-1]] > arr[i]:
            st.pop()

        if st:
            lsee[i] = st[-1]
        else:
            lsee[i] = -1
        st.append(i)
    
    return lsee

class Solution:
    def sumSubarrayMins(self, arr: List[int]) -> int:
        n = len(arr)
        nse = getRightSmallestElement(arr)
        lsee = getLeftSmallestOrEqualElement(arr)

        print(nse)
        print(lsee)
        mod = (10 ** 9) + 7
        ans = 0
        for i in range(n):
            s = ((i - lsee[i]) * (nse[i] - i) * arr[i]) % mod
            ans += s
            ans = ans % mod
        
        return ans
            

         
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