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
    implicit none
    integer,pointer:: an(:),bn(:)
    integer(8):: a,b,i,j,ans
    read*, a,b
    allocate(an(1),bn(1))
    call calc_pn(a,an)
    call calc_pn(b,bn)
    ! print*, an
    ! print*, bn
    j=1
    ans = 1
    do i =1,size(an)
        if (an(i) == 0)cycle
        do while(bn(j) < an(i))
            j = j + 1
        end do
        if (bn(j) == an(i)) ans=ans+1
    end do

    print*, ans

contains
subroutine calc_pn(num,pn_list)
    integer, pointer:: pn_list(:)
    integer(8):: num
    integer:: p,i
    p = 0
    do i = 2, int(sqrt(real(num)))
        if (mod(num,i)/=0) cycle
        p = p + 1
        if (p > size(pn_list)) call append(pn_list)
        pn_list(p) = i
        do while(mod(num,i) == 0)
            num= num/i
        end do
    end do
    if (num == 1)return
    p = p + 1
    if (p > size(pn_list)) call append(pn_list)
    pn_list(p) = int(num)
    
    
end subroutine 



subroutine append(arr) 
    integer, pointer:: arr(:)
    integer, pointer:: tmp(:)
    integer:: size_
    size_ = size(arr)
    allocate(tmp(size_))
    tmp(1:size_) = arr(1:size_)
    deallocate(arr)
    allocate(arr(2*size_))
    arr(1:size_) = tmp(1:size_)
    arr(size_+1:2*size_) = 0
    deallocate(tmp)
    return
end subroutine


end program