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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <vector>
#include <stdio.h>
int gcd(long a,long b){
  long r;
  if(a<b){
  r=a;a=b;b=r;
  }
  r=a%b;
  while(r!=0){
    a=b;b=r;r=a%b;
  }
  return b;
}

int main(){
  long a,b,r,i,j,ans=1;
  std::vector<long> c;
  scanf("%ld%ld",&a,&b);
  b=gcd(a,b);
  for(i=1;i<=b;i++){
    if(0==b%i){
      c.emplace_back(i);
    }
  }
  for(i=1;i<c.size()-1;i++){
    r=0;
    for(j=2;j<c[i];j++){
      if(c[i]%j==0){
        r=1;
        break;
      }
    }
    if(r==0){
      ans++;
    }
  }
  printf("%ld",ans);
}