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 GCDisGOD(long n,long m){
  long a = n;
  long b = m;
  long c = 0;
  while(1){
    c = a%b;
    if(c==0){
      return b;
    }else{
      a=b;
      b=c;
    }
  }
}

int main(){
  long a = 0;
  long b = 0;
  long gcd = 0;
  long cgcd = 0;
  long i = 0;
  long j = 0;
  long primes[1000000] = {};
  long divprime=0;
  long primecnt=0;
  
  int isprime = 0;
  
  scanf("%ld %ld",&a,&b);
  gcd = GCDisGOD(a,b);
  cgcd = gcd;
  for(i=2;i<=(long)sqrt(gcd)+1 && cgcd != 1;i++){
    isprime = 1;
    for(j=0;j<primecnt;j++){
      if(i % primes[j] == 0){
        isprime=0;
        break;
      }
    }
    if(isprime){
      primes[primecnt] = i;
      primecnt++;
      //printf("%ld ",i);
      if(cgcd % i == 0){
        divprime++;
      }
      while(1){
        if(cgcd % i==0){
          cgcd = cgcd / i;
          //printf("%ld ",cgcd);
        }else{
          break;
        }
      }
    }
    //printf("%ld ",cgcd);
  }
  
  printf("%ld",divprime+1);

  return 0;
}