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 long gcd(long m, long n) {
		if(n > m) {
			long tmp = m;
			m = n;
			n = tmp;
		}
		if(m % n == 0)
			return n;
		else {
			m = (m % n);
			return gcd(n, m);
		}
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long a = sc.nextLong();
		long b = sc.nextLong();
		long div = gcd(a, b);
		long ans = 1;
		boolean flag = false;
		while(div % 2 == 0) {
			flag = true;
			div /= 2;
		}
		if(flag)
			ans++;
		for(long i = 3; i < div + 1; i += 2) {
			flag = false;
			while(div % i == 0) {
				flag = true;
				div /= i;
			}
			if(flag)
				ans++;
		}
		System.out.println(ans);
	}

}