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 long int A,B,N,T,i,j;
	int count, ans;
	scanf("%lld %lld",&A,&B);
	if(A>B){
		N=B;
	}else{
		N=A;
	}
	long long int yakusu[N];
  	j=0;
	for(i=2;i<=N;i++){
		if(A%i==0 && B%i==0){
    //      printf("%lld  ",i);
          yakusu[j]=i;
          j++;
		}
	}
	//printf("\n");
	T=j-1;
    for(i=0;i<=T;i++){
    	count=0;
    	for(j=2;j<=yakusu[T];j++){
    		if(yakusu[i]>=j){
    			if(yakusu[i]%j==0){
    				count++;
    			}
    		}
    	}
    	if(count==1){
    		ans++;
    		//printf("%lld  ",yakusu[i]);
    	}
    }
    //printf("\n");
    printf("%d",ans+1);
	return 0;
}