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();
    //a,bの最大公約数を素因数分解して出てきた素数の個数+1が答え
    long c = gcd(a,b);
    long answer = 0;
//    System.out.println(c);
    answer = 1;
    long tmp =0;
    for(long prime :primes(c)){
      if(tmp!=prime){
        System.out.println(prime);
        answer = answer +1;
      }
      tmp=prime;
    }
    
    System.out.println(answer);
  }
  
    //最大公約数・最小公倍数（セットで使う）
    static long gcd (long a, long b) {return b>0?gcd(b,a%b):a;}
  
//素因数分解
  static List<Long> primes(long c){ 
    List<Long> primenums= new ArrayList<Long>();
            long x = c;
            long y = x;
          for (long i = 2; i<(long) Math.floor(Math.sqrt((double)x)+1)*2; i++) {
//            boolean divflg=false;
            while ((y % i) == 0 ) {
//               if(!divflg){
                primenums.add(i);
//               }
//                divflg=true;
                y = y/i;
            }
            
        }
    return primenums;
  }
  
}