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>

int main(void) {
	long int A,B,tmp;
	long int i,s,j;
	scanf("%ld %ld",&A,&B);
	long int N[tmp];
	if(A>B){
		tmp=B;
	}else{
    	tmp=A;
    }
	s=0;
	for(i=1;i<tmp;i++){
		if(A%i==0&&B%i==0){
			N[s]=i;
			s=s+1;
		}
	}

	int k,count=0,ans=1;
	k=N[s-1];
	count=0;
	for(i=1;i<s;i++){
		count=0;	
      	for(j=1;j<=k;j++){
			if(N[i]%j==0){
				count++;
			}
		}
		if(count==2){
			ans++;
		}
	}
	printf("%d",ans);
    
	
}