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 C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

# include <iostream>
# include <cstdio>

using namespace std;

bool prime (int x)
{
    if (x==1) return true;
    for (int i = 2;i <=x / i;i++ )
    {
        if (x % i == 0)  return false;
    }
    return true;
}


int main ()
{
    int a,b,cnt=0;
    cin>>a>>b;
    
    while (b!=0)
    {
        int t=a%b;
        a=b;
        b=t;
    }
    
    for (int i = 1;i <= a;i++)
    {
        if (a % i == 0)
        {
            if (prime(i))
            cnt++;
        }
    }
    
    cout<<cnt<<endl;
    return 0;
}