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: Wrong Answer)

#include<stdio.h>
#include<math.h>
long gcd2(long a, long b){
	char g = 0;
	while(a){
		if((a&1) && (b&1)){
			if(a<b){
				b = (b-a)>>1;
			}else{
				a = (a-b)>>1;
			}
		}else if(a&1){
			b >>= 1;
		}else if(b&1){
			a >>= 1;
		}else{
			a >>= 1;
			b >>= 1;
			g++;
		}
	}
	return b<<g;
}

long a, b;
int i;
int f;
int main(void){
	scanf("%ld%ld", &a, &b);
	a = gcd2(a, b);
	if(a==1){
		puts("1");
		return 0;
	}
	for(i=2; i<=sqrt(a); i++){
		if(a%i==0){
			f++;
			a/=i;
			while(a%i==0){
				a/=i;
			}
		}
	}
	printf("%d\n", f+2);
}