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) {

        Scanner sc = new Scanner(System.in);
        long a = sc.nextLong();
        long b = sc.nextLong();
        HashSet<Long> primeA = new HashSet<>();
        HashSet<Long> primeB = new HashSet<>();
        prime(a,primeA);
        prime(b,primeB);
        int ans = 0;
        for(Long num:primeA)if(primeB.contains(num)){
            //System.out.println(num);
            ans++;
        }
        System.out.println(ans+1);
    }

    private static void prime(long a,HashSet<Long> primeA){
        for(long i=2;i<Math.sqrt(a)+1;i++){
            if(a%i==0){
                boolean ok = true;
                for(Long n:primeA)if(i%n==0){
                    ok = false;
                    break;
                }
                if(ok)primeA.add(i);
            }
        }
    }

}