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>

unsigned long func(unsigned long h, unsigned long l){
    unsigned long a = h%l;
    if(a != 0){
        l = func(l, a);
    }
    return l;
}

int subfunc(unsigned long h, unsigned long* l, unsigned long len){
    for(unsigned long i=0; i<len; i++){
        if(func(h, *(l+i)) != 1){
            return 0;
        }
    }
    return 1;
}

int main(){
    unsigned long a, b;
    scanf("%lu%lu", &a, &b);
    unsigned long h, l, g;
    if(a>b){
        h = a;
        l = b;
    }else{
        h = b;
        l = a;
    }
    g = func(h, l);
    unsigned long k[g];
    unsigned long next = 0;
    for(unsigned long i=1; i<=g; i++){
        if(h%i == 0 && l%i == 0){
            if(subfunc(i, k, next) == 1){
                k[next++] = i;
                //printf("%lu ", i);
            }
        }
    }
    printf("%lu", next);

    return 0;
}