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: Wrong Answer)

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;

public class Main {

    public static void main(String args[]) {

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

        if (a == 1 || b == 1) {
            System.out.println(1);

        } else {
            List<Long> primeFactor = new ArrayList<>();
            Long target = gcd(a, b);
            for (Long i = 2L; i < target; i++) {
                if (target % i == 0) {
                    primeFactor.add(i);
                    target /= i;
                    i = 2L;
                }
            }
            primeFactor.add(target);

            System.out.println(new HashSet<>(primeFactor).size() + 1);
        }
    }

    // 最大公約数を返す
    public static Long gcd(Long a, Long b) {
        Long temp;
        while ((temp = a % b) != 0) {
            a = b;
            b = temp;
        }
        return b;
    }
}