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 Julia (0.5.0) (judge verdict: Wrong Answer)

parseInt(x) = parse(Int, x)
parseBigInt(x) = parse(BigInt, x)
parseFloat(x) = parse(Float64, x)
parseBigFloat(x) = parse(BigFloat, x)
parseMap(x::Array{SubString{String},1}) = map(parseInt, x)

function main()
    a,b=parseMap(split(readline()))
    soinsu1=Set{Int}([1,a])
    soinsu2=Set{Int}([1,b])
    for i in 2:trunc(Int,sqrt(a))+1
        while a%i==0
            a÷=i
            push!(soinsu1,i)
        end
    end
    for i in 2:trunc(Int,sqrt(b))+1
        while b%i==0
            b÷=i
            push!(soinsu2,i)
        end
    end
    ans=0
    for i in soinsu1
        if i in soinsu2
            ans+=1
        end
    end
    println(ans)
end
main()