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: Accepted)

#include<stdio.h>

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

unsigned long sub_func(unsigned long x){
    unsigned long c = 0;
    if(x % 2 == 0){
        c++;
        while(x%2 == 0){
            x = x/2;
        }
    }
    if(x % 3 == 0){
        c++;
        while(x%3 == 0){
            x = x/3;
        }
    }
    if(x % 5 == 0){
        c++;
        while(x%5 == 0){
            x = x/5;
        }
    }
    if(x % 7 == 0){
        c++;
        while(x%7 == 0){
            x = x/7;
        }
    }
    for(unsigned long i=5; i*i<=x; i += 2){
        if(i%3 == 0 || i%5 == 0 || i%7 == 0){
            continue;
        }
        if(x%i == 0){
            c++;
            while(x%i == 0){
                x = x/i;
            }
        }
    }
    if(x>1){
        c++;
    }

    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;
}