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 Python (3.4.3) (judge verdict: Wrong Answer)

def mkou(A,B):
    x = A%B
    if  x== 0:
        return B
    else:
        return mkou(B,x)
        
def prime_F(k):
    a =[1,]
    end = int(pow(k,0.5))
    if k % 2 == 0:
        k /=2
        a.append(2)
    k =int(k)

    for x in range(3,k+1,1):
        if(k % x == 0):
            while k %x == 0:
                k/=x
                a.append(x)
    return a
            
A,B = map(int,input().split())
divA_array = []
divB_array = []
if A<B:
    A,B =B,A

kouyaku = mkou(A,B)
print(kouyaku)
soinsu =prime_F(kouyaku)
print(soinsu)
print(len(soinsu))