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> aset = getPrimeSet(a);
        HashSet<Long> bset = getPrimeSet(b);
        long ans = 1;
        for(long l : aset){
            if(bset.contains(l)) ans++;
        }
        System.out.println(ans);
    }
    
    public static HashSet<Long> getPrimeSet(long l){
        long powl = (long)Math.pow(l,0.5);
        HashSet<Long> ret = new HashSet<Long>();
        boolean div2 = false;
        while(powl % 2 == 0){
            div2 = true;
            powl /= 2;
        }
        if(div2) ret.add((long)2);
        long i = 3;
        while(powl >= i){
            boolean divi = false;
            while(powl % i == 0){
                divi = true;
                powl /= i;
            }
            if(divi) ret.add(i);
            i += 2;
        }
        return ret;
    }
}