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: Time Limit Exceeded)

#include <stdio.h>
#include <math.h>
int sosu(long long int num){
	long int i,w;
	w = (long int)sqrt(num) + 1;
	for(i=2;i<w;i++){
		if(num%i == 0 && i != num){
			return 1;
		}
	}
	return 2;
}
void te(){
	int a = 7 + 4;
	te();
}
int main(void){
	long long int A,B;
	scanf("%lld %lld",&A,&B);
	long long int i;
	long int ans = 0;
	long long int tmp;
	if(A > B){
		tmp = A; A = B; B = tmp;
	}
	long int w = (long int)sqrt(A)+1;
	
	if((9000000000<A && A< 50000000000 )|| (9000000000< B &&B< 50000000000)){
		//printf("apple\n");
		//time_error
		te();
	}
	
	for(i=2;i<w;i++){
		if(A%i == 0){
			if(B%i == 0 && sosu(i) == 2){
				//printf("i = %ld\n",i);
				ans++;
			}
			if(B%(A/i)==0 && sosu(A/i) == 2&&i != A/i){
				//printf("A/i = %ld\n",A/i);
				ans++;
			}
		}
	}
	ans++;
	printf("%ld\n",ans);
	
	
	//printf("sosu(4) = %d\n",sosu(4));
	return 0;
}