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>

long long int gcd(long long int x, long long int y)
{
  while(x > 0 && y > 0) {
    if(x > y) {
      x -= y;
    }
    else {
      y -= x;
    }
  }

  if(x == 0) {
    return y;
  }
  else {
    return x;
  }
}

int divides(long long int a, long long int b)
{
  while(b > 0) {
    b -= a;
  }

  if(b == 0)
    return 1;
  else
    return 0;
}

int main(void)
{
  long long int a, b;
  scanf("%lld%lld", &a, &b);

  long long int g = gcd(a, b);
  char factor[g+1];

  for(int i = 0; i < g + 1; i++) {
    factor[i] = 1;
  }

  for(int i = 2; i < g + 1; i++) {
    if(factor[i] == 1) {
      if(!divides((long long int)i, g)) {
        factor[i] = 0;
      }
      int j = i + i;
      while(j <= g) {
        factor[j] = 0;
        j += i;
      }
    }
  }

  long long int ans = 0;
  for(int i = 1; i < g + 1; i++) {
    ans += (long long int)factor[i];
  }

  printf("%lld\n", ans);

    

  return 0;
}