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: Time Limit Exceeded)

import java.util.Scanner;
import java.util.HashSet;
import java.util.Set;

public class Main{
  public static void main(String[] args){
      Scanner sc = new Scanner(System.in);
      long a = sc.nextLong();
      long b = sc.nextLong();
    
	  if(a < b){
      	long c = a;
        a = b;
        b = c;
      }
      //最大公約数を求める
      while(a > 0 && b > 0){
        if(a < b){
      	long c = a;
        a = b;
        b = c;
      	}
      	a = a % b;
      }
      long gcd = a > b ? a : b;/*最大公約数*/
      Set<Integer> set = new HashSet<Integer>();      
      for(int i = 2; i <= gcd; i++){
       while(gcd % i == 0){
         gcd /= i;
         set.add(i);
       }
     }

	System.out.println(set.size()+1);
    }
}