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: Accepted)

import java.util.Scanner;

public class Main {
    public static void main(String[] args) throws Exception {

        Scanner sc = new Scanner(System.in);
        long a = sc.nextLong();
        long b = sc.nextLong();

        long gcd = gcd(a, b);

        int count = 1;

		long d = gcd;
		for (int i = 2; i <= Math.sqrt(gcd); i++) {
			if (d % i == 0) {
				count++;
				while (d % i == 0) {
					d = d / i;
				}
			}
		}

		if (d != 1)
			count++;
		System.out.println(count);
    }

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

}