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 9.2.1) (judge verdict: Runtime Error)

#include <stdio.h>

int gojo(int a, int b){
  if(a % b)
    gojo(b, a%b);
  else
    return b;
}

int main(){
  int a, b;  scanf("%d %d", &a, &b);
  if(b > a){
    int t = a; a = b; b = t;
  }

  int yaku = gojo(a, b);
  int ans = 1;
  
  for(int i=2; i*i<=yaku; i++){
    if(yaku%i == 0){
      ans++;
      while(yaku % i == 0)  yaku /= i;
    }
  }
  if(yaku != 1)
    ans++;
  
  printf("%d\n", ans);
  return 0;
}