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)

program main
    integer::p,ans
    integer(8):: a,b,i, sosu
    read*, a,b
    sosu = 0
    i = 1
    do while(real(i) <= min(sqrt(real(a)),sqrt(real(b))))
        i=i+1
        if (.not.((mod(a,i) == 0) .and. (mod(b,i) == 0))) cycle
        sosu=sosu+1

        do while((mod(a,i) == 0) .and. (mod(b,i) == 0))
            a = a/(i)
            b = b/(i)
        end do
        
    end do
    print*, sosu+1
 


end program