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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
long gdc(long a, long b)
{
    long temp;
    for(;b>0;)
    {
        temp=b;
        b=a%b;
        a=temp;
    }
    return a;

}
int main()
{
    long a,b,gcom,i,j,k,c=1,l=0,m=0,n=0;
    scanf("%ld%ld",&a,&b);
    gcom=gdc(a,b);
    for(i=2;gcom>= i*i;)
    {
        for(j=0;gcom%i==0;j++)
            gcom/=i;
            if(j>0)
                c++;
            if(i==2)
                i++;
            else i=i+2;
    }
    if(gcom>=i)
        c++;
    printf("%ld\n",c);
    return 0;
}