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: Runtime Error)

// CPP program to print prime factors and their
// powers.
#include <stdio.h>
int sum = 0;

// function to calculate all the prime factors and
// count of every prime factor
void primecounter(long long n)
{
    int count = 0;

    // count the number of times 2 divides
    while (!(n % 2)) {
        n >>= 1; // equivalent to n=n/2;
        count++;
    }

    // if 2 divides it
    if (count){
       // printf("%d %d\n",2,count);
        sum++;
    }
    // check for all the possible numbers that can
    // divide it
    for (long long i = 3; i <= sqrt(n); i += 2) {
        count = 0;
        while (n % i == 0) {
            count++;
            n = n / i;
        }
        if (count){
         //   printf("%d %d\n",i,count);
            sum++;
        }
    }

    // if n at the end is a prime number.
    if (n > 2){
       // printf("%d %d\n",n,1);
        sum++;
    }
}
long long GCD(long long a,long long b)
{
    if(a%b==0)
        return b;
    else
        return GCD(b,a%b);
}
// driver program to test the above function
int main(void)
{
    long long int a,b;
    scanf("%d%d",&a,&b);
    long long n = GCD(a,b) ;
    primecounter(n);
    printf("%d\n",sum+1);
    return 0;
}