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: Accepted)

import java.util.*;
import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws Exception {
        // Your code here!
        
        BufferedReader stdR = new BufferedReader(new InputStreamReader(System.in));
        
        String[] x = stdR.readLine().split(" ");
        long A = Long.parseLong(x[0]);
        long B = Long.parseLong(x[1]);
        
        long gcd = GCD(A, B);
        System.out.println(primeFactorization(gcd) + 1);

    }
    
    public static long GCD(long x, long y) {
        if ( y == 0) return x;
        return GCD(y, x % y);
    }
    
    public static long primeFactorization(long n){
        
        int s = (int)(Math.sqrt(n));
        
        int[] primes = new int[s];

        int j = 0;
        for(int i = 2; i <= s; i++){
            
            if(n % i == 0){
                while(n % i  == 0){
                    n /= i;
                }
                primes[j++] = i;                
            }
        }
        
        if(n != 1){
            j++;
        }
        
        return j;
    }
    
}