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 <string.h>
#include "math.h"


int main(int argc, char *argv[]){

  long long int i, j, k, l, m;
  long long int a, b;
  long long int min;

  scanf("%lld %lld", &a, &b);

  if(a <= b){
    min = a;
  }else{
    min = b;
  }

  int num[min];
  k = 0;

  for(i=min; i>=1; i--){
    if(a%i==0 && b%i==0){
      num[k] = i;
      k++;
    }
  }
  /*
  for(i=0; i<=k-1; i++){
    printf("%d ", num[i]);
  }
  printf("\n");
  */

  for(i=0; i<=k-2; i++){
    for(j=i+1; j<=k-2; j++){
      if(num[i] % num[j] == 0){
        num[i] = 0;
      }
    }
  }

  m = 0;
  for(i=0; i<=k-1; i++){
    if(num[i] != 0){
      m++;
    }
  }
  printf("%lld\n", m);

  return 0;
}