You are given two lists A and $B$ both containing the numbers $1$dotsn in some

order. Your task is to count how many of the numbers $1$dotsn occur earlier on the

list A than on the list $B.$

In this task, $n$ can be large and an efficient algorithm is required. The time

complexity should be $O(n).$

In a file

twolists.py$,$ implement a function count that returns the desired count.

def count$($a$,$ b$)$:

# TODO

if $=="""$:

print$($count$([2,3,4,1],[1,2,3,4]))$ # $3$

print $($count $([1,2,3,4],[1,2,3,4])$ # $0$

print $($count $([4,7,3,1,6,2,5],[5,6,1,2,4,3,7])$ # $3$

print $($count $([5,4,9,1,8,3,2,6,7],[6,2,8,4,9,1,5,7,3])$

Explanation: In the first test, the numbers $2,3$ and $4$ occur earlier on the list $A$

than on the list $B.$