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 sk(long,long);
int main(void){
  long a,b;
  scanf("%ld%ld",&a,&b);
  if(a<b){long w=a;
          a=b;
          b=w;
         }
 long max=sk(a,b);
  long count=0;
 // printf("%ld\n",max);
  for(long i=1;i<=sqrt(max);i++){
    if(max%i==0){count++;
                 printf("%ld\n",i);
                 max/=i;
                }
  }
  
printf("%ld ",count+1);
  return 0;
}
long sk(long a,long b){
  long c=a%b;
  if(c!=0)return sk(b,c);
  else return b;
}