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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		try (Scanner sc = new Scanner(System.in)) {
			long a = sc.nextLong();
			long b = sc.nextLong();
			long gcd = gcd(a, b);
			int ans = 1;
			if (gcd % 2 == 0) {
				gcd = divide(gcd, 2);
				ans++;
			}
			for (long i = 3; i <= gcd; i = i + 2) {
				if (gcd % i == 0) {
					gcd = divide(gcd, i);
					ans++;
				}
			}
			System.out.println(ans);
		}
	}

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

	static long divide(long a, long b) {
		while ((a % b) == 0) {
			a = a / b;
		}
		return a;
	}
}