What is the base case for the following recursion:

def f$($b$,$ q$,$ n$)$:

if n $==0$:

return $0$

else:

return b$*$q$**($n$-1)+$ f$($b$,$ q$,$ n$-1)$

print$($f$($b$=2,$ q$=3,$ n$=4))$

a$.$

n is $0$

b$.$

n is $1$

c$.$

n is greater than $1$

d$.$

no base case