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 <string.h>
#define craiz 4000
#define cmax 100000001
int crivo[cmax];
void crivoc(){
	memset(crivo, 1, sizeof crivo);
	/*for (int i = 2; i <= craiz; ++i){
		if(crivo[i]){
			for(int j = i * i; j < cmax; j += i) crivo[j] = 0;
		}
	}*/
}
int pegeimult(int ver){
	if(crivo[ver]){
		for(int j = ver * ver; j < cmax; j += ver) crivo[j] = 0;
		return 1;
	}
	return 0;
}

long long int gcd(long long int a, long long int b){
	long long int temp;
	while (b != 0){
		temp = a % b;
		a = b;
		b = temp;
	}
	return a;
}

int main(){
	crivoc();
	int maxd, qnt = 1;
	long long a, b;
	scanf("%lld%lld", &a, &b);
	maxd = gcd(a, b);
	//printf("%d\n", maxd);
	for(int i = 2; i <= maxd; ++i){
		if(maxd%i == 0 && pegeimult(i)){
			qnt++;
		}
	}
	printf("%d\n", qnt);
	return 0;
}