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.*;
class Main {
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        long a = sc.nextLong();
        long b = sc.nextLong();
        long wk_no = a < b ? a : b;
        int cont = 0;
       
        for(long i = 1 ;i <= wk_no/2 ; i++){
            if(a%i == 0 && b%i == 0){
                if(is_prime(i)) {
                  if(i == 1){
                    cont ++;
                  }else{
                     cont += 2;
                   }
                  }
            }
        }
        
        System.out.println(cont);
        sc.close();
    }
    static boolean is_prime(long n) {
        if(n == 1) return true;
        for(long i = 2; i*i <= n; ++i) { // 2 <= i <= sqrt(n) に約数があれば、
            if(n % i == 0) return false; // n は素数ではない
        }
        return true;
    }
}