<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:
    def find_nse(self,arr):
        n=len(arr)
        i=n-1
        ans=[n]*n
        st=[]

        while i>=0:
            while st and arr[i]<=arr[st[-1]]:
                st.pop()
            
            if st:
                ans[i]=st[-1]
            
            st.append(i)
            i-=1
        
        return ans


    def find_pse(self,arr):
        n=len(arr)
        i=0
        ans=[-1]*n
        st=[]

        while i<n:
            while st and arr[i]<arr[st[-1]]:
                st.pop()
            
            if st:
                ans[i]=st[-1]
            
            st.append(i)
            i+=1
        
        return ans




    def sumSubarrayMins(self, arr: List[int]) -> int:
        nse=self.find_nse(arr)
        pse=self.find_pse(arr)
        print(nse,pse)
        mod=10**9 + 7
        sum=0
        for i in range(len(arr)):
            sum+=((nse[i]-i)*(i-pse[i])*arr[i])%mod
            sum %= mod
        return sum
22200