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 $\_\_$name$\_\_=="\_\_$main$\_\_"$:

a $=$ list$($range$(1,10**5+1))$

b $=$ list$($range$(1,10**5+1))$

print$($count$($a$,$b$))$

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

than on the list $B.$