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: Accepted)

#include<stdio.h>
long gcd(long a,long b){
  if(a<b){
    long w=a;
    a=b;
    b=w;
  }
  if(a%b!=0)return gcd(b,a%b);
  else return b;
}
int main(void){
  long a,b;
  scanf("%ld%ld",&a,&b);
  long ab=gcd(a,b);
 // printf("%ld\n",ab);
  long w=ab;
  long c=0;
  for(long q=2;q*q<=ab;q++){
		if(w%q==0){//printf("%dを因子に持つ\n",q);
          c++;
                   while(w%q==0){w/=q;           
                  }
  }
}
  if(w>1)c++;
printf("%ld",c+1);
return 0;
}