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: Runtime Error)

#include <stdio.h>
#include <math.h>

unsigned long long int gcd(unsigned long long int x, unsigned long long int y){
  while(x != 0){
    unsigned long long int z = x;
    x = y%x;
    y = z;
  }
  return y;
} 

unsigned long long int isPrimeNum(unsigned long long int x )
{
    unsigned long long int n;

    if( x == 2 ) return 1;
    if( x < 2 || x % 2 == 0 ) return 0;
    for( n = 3; n <= sqrt((double)x); n += 2 )
        if( x % n == 0 ) return 0;
    return 1;
}

unsigned long long int rekkyo(unsigned long long int d, unsigned long long int B[]){
  unsigned long long int i, c, e, j, k;
  c = 0;
  e = 0;
  k = 0;
  for(i=1; i<=d; i++){
    if(d%i == 0){
      B[c] = i;
      c = c+1;
    }
  }
  for(i=1; i<c; i++){
  if (isPrimeNum(B[i]) == 1){
    k = k+1;
  }
  }
  for(i=1; i<c-1; i++){
    for(j=i+1; j<c; j++){
       if(B[j]%B[i] != 0){
             e = e+1;
       }
    }
    }
    e = e + k;
    return e;
  }


  unsigned long long int main(){
  unsigned long long int a, b, g, t;
  scanf("%llu %llu", &a, &b);
  g = gcd(a, b);
  unsigned long long int B[g];
  if (g == 1) {
    printf("%d", g);
  }
  else{
     t = rekkyo(g, B);
     printf("%llu", t);
  }
  return 0;
}