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.*;
 
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();
        HashSet<Long> al = getPrimeFactorization(a);
        HashSet<Long> bl = getPrimeFactorization(b);
        int prime = 1;
        for(long p : bl){
            if(al.contains(p)){
                prime++;
            }
        }
        System.out.println(prime);
    }
    
    static HashSet<Long> getPrimeFactorization(long x) {
        HashSet<Long> result = new HashSet<Long>();
        while (x >= 2 && x % 2 == 0) {
            result.add(2L);
            x /= 2;
        }
        long d = 3;
        long q = x / d;
        while (q >= d) {
            if (x % d == 0) {
                result.add(d);
                x = q;
            } else {
                d += 2;
            }
            q = x / d;
        }
        result.add(x);
        return result;
    }
}