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 c=gcd(a,b);
      int d=0;
      if(c==1){
        d=0;
      }else{
      for(int i=2; i<=c;i++){
        if(c%i!=0){continue;}
      	while(c%i==0){
        	c=c/i;
        }
        d++;
      }
      }
      System.out.println((d+1));
        
    }
  	
  	public static long gcd(long a, long b){
    	while(b%a!=0){
        	long c=b%a;
          	b=a;
          	a=c;
        }
      return b;
    }
  
}