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>

int main(void)
{
  long a, b;
  scanf("%ld %ld", &a, &b);
  
  long gcd;
  long m = a, d = b;
  while (d > 0){
    long temp = d;
    d = m % d;
    m = temp;
  }
  gcd = m;
  
  long count = 0;
  long dvs[gcd];
  for (long i = 2; i <= gcd; i++){
    if (gcd % i == 0){
      int isPrime = 1;
      for (long j = 0; j < count; j++){
        if (i % dvs[j] == 0){
          isPrime = 0;
          break;
        }
      }
      if (isPrime == 1){
        dvs[count] = i;
        count++;
      }
    }
  }
  
  printf("%ld\n", count + 1);
  return 0;
}