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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
long prime(int a){
  long flag,ret;
  if(a==1){
    ret=2;
  }else{
      for(long i=prime(a-1)+1;i>0;i++){
      flag=0;
      for(int j=1;j<=a-1;j++){
        if(i%prime(j)==0){
          flag=1;
          break;
        }
      }
      if(flag==0){
        ret=i;
          break;
      }
    }
  }
    return ret;
}

int main(){
  long a,b,ans=1;
  scanf("%ld %ld",&a,&b);
  
  if(a>b){
    long s=a;
    a=b;
    b=s;
  }
  
  if(a%2==0 && b%2==0){
    ans+=1;
  }
  
  while(a%2==0){
    a/=2;
  }
  while(b%2==0){
    b/=2;
  }
  
  for(int i=2;prime(i)<=a;i++){
    if(a%prime(i)==0 && b%prime(i)==0){
      ans+=1;
      while(a%prime(i)==0){
        a/=prime(i);
      }
      while(b%prime(i)==0){
        b/=prime(i);
      }
    }
  }
  
  printf("%ld",ans);
}