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>

int gozyo(int A,int B){
  int r,tmp;
  r=A%B;
  while(r!=0){
    tmp=A;
    A=B;
    r=tmp%B;
    if(r!=0){
      B=r;
    }
  }
  return B;
}
int prime(int N){
  int count=0;
  if(N==2){
    return 1;
  }
  for(int i=2;i<N;i++){
    if(N%i==0){
      count++;
    }
  }
  if(count==0){
    return 1;
  }else{
    return 0;
  }
}

int count(int N){
  int count=1;
  if(N==1){
    return 1;
  }
  for(int i=2;i<=N;i++){
    if(prime(i)==0){
      continue;
    }
    if(N%i==0){
      count++;
    }

  }
  return count;
}

int main(void){
  int A,B,ans;
  scanf("%d",&A);
  scanf("%d",&B);
  if(A>B){
    ans=count(gozyo(A,B));
  }else if(A<B){
    ans=count(gozyo(B,A));
  }
  printf("%d\n",ans);

}