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: Time Limit Exceeded)

#include <stdio.h>
int main(void){
    long long a,b,t,x[100000],p=1,ans=0,i,j,k;
    scanf("%lld %lld", &a, &b);
    if(a>b)
    {
        t=b;
        b=a;
        a=t;
    }
    if(a==1)
    ans=1;
    x[0]=1;
    for(i=2; i<=a; i++)
    {
        if(a%i==0&&b%i==0)
        {
            x[p]=i;
            p++;
        }
    }
    for(i=1; i<p; i++)
    {
        a=1;
        for(j=i+1; j<p; j++)
        {
            for(k=2; k<x[i]; k++)
            {
                if(x[i]%k==0&&x[j]%k==0)
                {
                    a=0;
                    break;
                }
            }
            if(a<1)
            break;
        }
        ans+=a;
    }
    printf("%lld\n", ans);
    
    return 0;
}