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>
#include<math.h>

int main(void){
    long A,B;
    scanf("%ld%ld",&A,&B);
    long base;

    int i;

    if(A<B){
        base=A;
        A=B;
        B=base;
    }

    long r;

    r=A%B;
    while(r!=0){
        A=B;
        B=r;
        r=A%B;
    }

    long gcm=B;

    int count=1;

    if(gcm%2==0){
        count++;
        while(gcm%2==0 && gcm!=0){
            gcm/=2;
        }
    }

    for(i=3;i<=gcm;i+=2){
        if(gcm<i){
            break;
        }
        if(gcm%i==0){
            count++;
            while(gcm%i==0 && gcm!=0){
                gcm/=i;
            }
        }
    }

    printf("%d",count);

    return 0;

}