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>
int nextprime(int p);
int main()
{
    long long a,b,tmp;
    int div=3;
    int div_max;
    scanf("%lld %lld",&a,&b);
    int count=1;
    if(a>b)
    {
        tmp=a;
        a=b;
        b=tmp;
    }
    div_max=(int)sqrt(a);
    if(a%2==0&&b%2==0)
    {
        count++;
        a/=2;
    }
    while(a/div>=1)
    {   
        if(div>div_max)break;
        if(a%div==0&&b%div==0)
        {
            count++;
            a/=div;
        }
        div=nextprime(div);
        printf("%d prime %d\n",count,div);
    }
    printf("%d",count);
    return 0;
}
  int nextprime(int p)
  { 
      int count=1;
      while (count!=0)
      {
          p+=2;
          count=0;
          int div=3;
          while (div<sqrt(p))
          {
              if(p%div==0){
                  count=1;
                  break;
              }
              div+=2;
          }
      }
      return p;
  }