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>

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

unsigned long sub_func(unsigned long x){
    unsigned long c = 0;
    int flag = 1;
    for(unsigned long i=2; i<=x; i++){
        if(x%i == 0){
            if(flag == 1){
                c++;
                flag = 0;
            }
            x = x/i;
            i--;
        }else{
            flag = 1;
        }
    }
    return c;
}

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);
    printf("%lu", sub_func(g) + 1);

    return 0;
}