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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
long long a, b, c;
int d[1000000];
int ans;
long long gcd(long long x, long long y) {
	if (x < y) {
		long long z;
		z = x;
		x = y;
		y = z;
	}
	if (x % y)return gcd(y, x % y);
	else return y;
}
int main() {
	scanf("%lld%lld", &a, &b);
	c = gcd(a, b);
	for (long long i = 2; i * i <= c; i++) {
		if (c % i == 0) {
			d[i]++;
			c /= i;
			i = 1;
		}
	}
	ans = 1;
	for (int i = 0; i < 1000000; i++) {
		if (d[i])ans++;
	}
	if (c != 1)ans++;
	printf("%d\n", ans);
}