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 <stdlib.h>

long long gcd(long long a, long long b)
{
	if (a > b) {
		a += b;
		b = a - b;
		a -= b;
	}
	if (b % a == 0) return a;
	else return gcd(b % a, a);
}

int main()
{
	long long A, B;
	scanf("%lld %lld", &A, &B);
	
	int p, ans = 1;
	long long G = gcd(A, B);
	for (p = 2; p * p <= G; p++) {
		if (G / p * p == G) {
			ans++;
			for (G /= p; G / p * p == G; G /= p);
		}
	}
	if (G > 1) ans++;
	
	printf("%d\n", ans);
	fflush(stdout);
	return 0;
}