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)
parseMap(x::Array{SubString{String},1})=map(parseInt,x)

function main()
	a,b = readline() |> split |> parseMap
	q = gcd(a,b)
	a = Int[]
	i = 1
	while i<=sqrt(q)
		if q%i == 0
			push!(a,i)
			push!(a,div(q,i))
		end
		i+=1
	end
	a = sort(a)
	f = ones(Int,length(a))
	for j in 1:length(a)
		for k in j+1:length(a)
			if gcd(a[j],a[k])>1
				f[k] = 0
			end
		end
	end
	f[length(a)] = 0
	println(sum(f))
end

main()