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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
long long detmax(long long x,long long y)
{
    if (x > y) {
      return x;
    } else {
      return y;
    }   
}
long long detmin(long long x,long long y)
{
    if (x > y) {
      return y;
    } else {
      return x;
    }   
}
long long detc(long long x,long long y)
{
    long long max = detmax(x,y);
    long long min = detmin(x,y);
    long long tmp1,tmp2;
    while(max != min) {
        tmp1 = max - min;
        tmp2 = min;
        max = detmax(tmp1,tmp2);
        min = detmin(tmp1,tmp2);
    }
    return max;
}

int main()
{
    long long a,b;
    scanf("%lli%lli",&a,&b);
    long long num = detc(a,b);
    int count=1;
    if (num == 1) {
        printf("%i",count);
        return 0;
    }
    for (long long i=2;i*i<num;i++) {
        if (num%i==0) {
            count++;
            while(num%i==0) {
                num/=i;
            }
        }
    }
    printf("%i",count);
    return 0;
}