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>
#include <math.h>
int Gcd(int a,int b,int d[])
{
	int i,j=0,c[(int)sqrt(a)+10],k,e=0;
	for(i=2;i<=(int)sqrt(a);i++)
	{
		if(a%i==0&&b%i==0)
		{
			d[j] = i;j++;
			if(i != b/i&&a%(b/i) == 0){
				c[e] = b/i;
				e++;
			}
		}
	}
	k = e;
	for(i = j; i < j+e;i++)
	{
		k--;
		d[i] = c[k];
	}
	return (e+j);
}
int sosu(int num)
{
	int i,j=0,c[(int)sqrt(num)+10],k,e=0;
	for(i=1;i<=(int)sqrt(num);i++)
	{
		if(num%i==0)
		{
			j++;
			if(i != num/i){
				e++;
			}
		}
	}
	return (e+j);
}
int main(void){
	int A,B;
	scanf("%d %d",&A,&B);
	int tmp;
	if(A < B){
		tmp = A; A = B; B = tmp;
	}
	int d[A];
	int i;
	int ans = Gcd(A,B,d);
	//ans = sosu(d);
	int kate = 0;
	int w;
	for(i=0;i<ans;i++){
		w = sosu(d[i]);
		if(w == 2)  kate++;
	}
	kate++;
	printf("%d\n",kate);
	return 0;
}