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>

int gcd(int x, int y) {
	if (y == 0) return x;
	return gcd(y, x%y);
}

int main() {
	int A, B, G_Common;
	scanf("%d %d",&A,&B);
	G_Common = gcd(A,B);

	int tmp = G_Common, cnt = 0;
	for (int i = 2; i <= sqrt(A) + sqrt(B); i++) {
		if (tmp == 1) break;
		int flag = 0;
		while (1) {
			if (tmp % i != 0) break;
			flag++;
			tmp /= i;
		}
		if (flag != 0) cnt++;
	}
	printf("%d\n",cnt+1);
	return 0;
}