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<stdlib.h>

int gcdf(int a,int b){
  int r, tmp;
  /* 自然数 a > b を確認・入替 */
  if(a<b){
    tmp = a;
    a = b;
    b = tmp;
  }
 
  /* ユークリッドの互除法 */
  r = a % b;
  while(r!=0){
    a = b;
    b = r;
    r = a % b;
  }
  return b;
}


int main(void){
    int A,B,gcd,c=1,n;
    scanf("%d %d",&A,&B);
    gcd = gcdf(A,B);
    n = gcd;
    for(int i=2;i<=n;i++){
        if(gcd == 1) break;
        if(gcd % i == 0){
            while(gcd % i == 0){
                gcd = gcd / i;
            }
            c++;
        }
    }
    printf("%d",c);
    return 0;
}