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>
#include<math.h>

long long gcd(long long x,long long y){
  long long tmp;
  if(x<y){
    tmp=x;
    x=y;
    y=tmp;
  }
  if(y==0) return x;
  return gcd(x%y,y);
}

int main(){
  long long a,b,i,n,cnt=0;
  scanf("%lld%lld",&a,&b);
  n=gcd(a,b);
  for(i=2;i<=sqrt(n);i++) {
    if(n==1) break;
    if(n%i==0){
      cnt++;
      while(n%i==0){
				n/=i;
			}
    }
  }
  if(n==1){
    printf("%lld",cnt+1);
  }
  else{
    printf("%lld",cnt+2);
  }
  return 0;
}