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();
        long gcd = GCD(A,B);
        long num = DIV(gcd);
        System.out.println(num+1);
    }
    public static long GCD(long num1,long num2){
        if(num1%num2==0) return num2;
        return GCD(num2,num1%num2);
    }
    public static long DIV(long A){
        long num = 0;
        Set<Long> set = new HashSet<>();
        for(int i=2;i<=Math.sqrt(A);i++){
            if(A%i==0) {
                set.add((long)i);
                A/=i;
            }
        }
        num+=set.size();
        if(A>1) num++;
        return num;
    }
}