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: Wrong Answer)

#include <stdio.h>
#include <math.h>
long long int isprime(long long int n)
{
    long long int  i;
    if(n==2 || n==1)
    return 1;
    for (i = 2; i <= sqrt(n); ++i)
    {
        // condition for non-prime
        if (n % i == 0)
        {
            return 0;
        }
    }
    return 1;
}
int main()
{
    long long int a, b, i, min, max, count = 0;
    scanf("%lld %lld",&a,&b);
    if(a==1 || b==1)
    {
        printf("1");
        return 0;
    }
    if(a>b)
    {
        max = a;
        min = b;
    }
    else
    {
        max = b;
        min = a;
    }
    for(i=1;i<sqrt(min);i++)
    {
        if(a%i==0 && b%i==0)
        if(isprime(i))
        count++;
    }
    printf("%lld",count);
    return 0;
}