Calculating the total sum of inversion count for all subarrays

Question

My approach was to visit all inversion count pair and count how many subarrays these pair contribute. Visiting every pair requires \$\mathcal{O}(n^2)\$ time, but I want an optimized version of this, something like \$\mathcal{O}(n \log n)\$.

Can I do something with the Fenwick tree? Inversion in an array Fenwick tree

e.g. if arr = [1,3,4,2] , then total inversion count for all subarrays = 5.

number = int(input("Input number := "))
main_list = list(map(int,input().split()))
answer = 0
for i in range(number):
    for j in range(number):
        if i<j and main_list[i]>main_list[j]:
            k = (i+1) * (number-j)
            answer += k
print(answer)