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 gcd(int x,int y);
int main(void) {
	long int A,B,t,i,j;
	long int n[t];
	int k,m, count=0;
	scanf("%ld%ld", &A,&B);
	t=B;
	for(i=2;i<t+1;i++){
		if(A%i==0&&B%i==0){
			n[k]=(i);
			k++;
		}
	}
	for(i=0;i<k;i++){
		for(j=i+1;j<k+1;j++){
			m=gcd(n[i],n[j]);
			if(m==1){
				count++;
			}
		}
	}
	printf("%d",count);
	
	
	return 0;
}

int gcd(int x,int y){
	int r;
	while((r==x%y)){
		x=y;
		y=x%y;
	}
	return y;
}