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 min(long long int a,long long int b)
{
    if(a<b)
        return a;
    else
        return b;
}
int isPrime(int x)
{
    int up=sqrt(x);
    if(x<=2)
        return 1;
    else
        for(int i =2;i<=up;i++)
        {
            if(x%i==0)
                return 0;
        }
    return 1;
}
int main()
{
    long long int a,b;
    long long int ans=1;
    long long int up;
    int all[1000000]={0};
    long long int counter=0;
    scanf("%lld %lld",&a,&b);
    up=min(a,b);
    if(a%2==0&&b%2==0)
    {
        //all[counter++]=i;
        ans++;
    }
    for(long long int i = 3;i<=up/2;i+=2)
    {
        if(a%i==0&&b%i==0&&i%2!=0)
        {
            if(isPrime(i))
               {
                   all[counter++]=i;
                   ans++;
                }
            else
               continue;
        }
    }
    for(int i = 0;i<counter;i++)
        printf("%lld ",all[i]);
    printf("%lld\n",ans);
    return 0;
}