<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 sumSubarrayMins(self, arr: List[int]) -> int:
        #appending [value,ocuurance,replaces n items]
        total=0
        stack=[]
        for i in arr:
            if not stack:
                stack.append([i,1,0])
                #print(stack[-1])
            else:
                count=0
                while stack and stack[-1][0]>i:
                    a,b,c=stack.pop()
                    total+=(a*(count+b))+(count*c*a)
                    #print(total)
                    count+=b
                stack.append([i,1+count,count])
                #print(stack[-1])
        count=0
        while stack:
            a,b,c=stack.pop()
            #print(a,b,c)
            total+=(a*(count+b))+(count*c*a)
            #print(total)
            count+=b
        m=(10**9)+7
        return total%m

                


        
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