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>
#include <math.h>

void change(long* x, long* y){
    long z = *x;
    *x = *y;
    *y = z;
}

void gcd(long* x, long* y){
    long z;

    if (*x < *y){
        change(x, y);
    }

    while(*x % *y){
        z = *x % *y;
        *x = *y;
        *y = z;
    }
}

int main(){
    long a,b;
    int t = 1;

    scanf("%ld%ld", &a, &b);  //a > b

    gcd(&a, &b);
   
    for(long i = 2; i <= sqrt(b); i ++){
        if(b % i == 0){
            while(b % i == 0){
                b /= i;
            }

            t++;
        }
    }
    printf("%d",t);

    return 0;
}