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: Accepted)

program abc142
    use,intrinsic :: iso_fortran_env
    implicit none
    integer(int64):: a,b,g,div,divmax,cnt,ans

    read*, a,b
    g = gcd(a,b)
    divmax = ceiling(sqrt(dble(g)))
    ans=1
    do div=2,divmax
        cnt=1
        if (mod(g,div) == 0) ans=ans+1
        do while(mod(g,div)==0)
            g=g/div
            cnt=cnt+1
        end do
    end do
    if (g>1)ans=ans+1
    print'(i0)', ans
contains
    function lcm(x,y) result(ret)
        integer(int64):: x,y,ret
        ret=x*y/gcd(x,y)
    end function
    recursive function gcd(x,y) result(ret)
        integer(int64):: x,y,ret
    
        if (mod(x,y) == 0) then
            ret = y
            return
        end if
    
        ret = gcd(y,mod(x,y))
    end function
end program abc142