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)

#include<stdio.h>

int main(){
    int i;
    int A,B,GCM;
    scanf("%d %d",&A,&B);
    int R;
    if(A >= B){
        R = A % B;
        while(R != 0){
            A = B;
            B = R;
            R = A % B;
        }
        GCM = B;
    }else if(A < B){
        R = B % A;
        while(R != 0){
            B = A;
            A = R ;
            R = B % A;
        }
        GCM = A;
    }
    int ch=1;
    for(i = 2 ; i <= GCM ; i++){
        if(GCM % i == 0){
            ch++;
            while(GCM % i == 0){
                GCM /= i;
            }
        }
    }
    printf("%d",ch);
    return 0;
}