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: Wrong Answer)

#include<stdio.h>
long gcd(long m,long n)
{
    if(n==0)
        return m;
        printf("%ld %ld\n",n,m%n);
    return gcd(n,m%n);

}

int main()
{
    long  a,b,gCommon,i,j,k,cnt=1;
    scanf("%ld%ld",&a,&b);

    gCommon = gcd(a,b);
    

        for(i=2; gCommon>1 && gCommon>=i*i ; )
        {
            for(j=0; gCommon%i==0; j++)
                gCommon/=i;
                    if(j>0)
                        cnt++;
                    if(i==2)
                        i++;
                    else
                        i+=2;
        }

        if(gCommon>=i)
            cnt++;
        printf("%ld\n",cnt);
        return 0;
}