Problem:
Score : 400 points

Problem Statement

Given are positive integers A and B.

Let us choose some number of positive common divisors of A and B.

Here, any two of the chosen divisors must be coprime.

At most, how many divisors can we choose?

Definition of common divisor

An integer d is said to be a common divisor of integers x and y when d divides both x and y.

Definition of being coprime

Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.

Definition of dividing

An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.

Constraints

All values in input are integers.

1 \leq A, B \leq 10^{12}

Input

Input is given from Standard Input in the following format:

A B

Output

Print the maximum number of divisors that can be chosen to satisfy the condition.

Sample Input 1

12 18

Sample Output 1

3

12 and 18 have the following positive common divisors: 1, 2, 3, and 6.

1 and 2 are coprime, 2 and 3 are coprime, and 3 and 1 are coprime, so we can choose 1, 2, and 3, which achieve the maximum result.

Sample Input 2

420 660

Sample Output 2

4

Sample Input 3

1 2019

Sample Output 3

1

1 and 2019 have no positive common divisors other than 1.

Solution in Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

integer(16) A,B
integer(16) g,gf,i
integer(16) cnt

read*,A,B
g=gcd(A,B)

gf=g
cnt=1
do i=2,gf
  if(mod(g,i)==0)then
    do while(mod(g,i)==0)
      g=g/i
    end do
    cnt=cnt+1
  endif
end do
print"(i0)",g

contains
pure function gcd(a,b)
integer(16),intent(in)::a,b
integer(16) gcd
integer(16) s,t
s=a
gcd=b

t=mod(s,gcd)
do while(t/=0)
  s=gcd
  gcd=t
  t=mod(s,gcd)
end do
end function gcd
end