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.*;

public class Main {

  public static void main (String[] args) {

    Scanner sc = new Scanner(System.in);
    
    long a = sc.nextLong();
    long b = sc.nextLong();

    sc.close();

    long v = gcd(a, b);

    long count = 1;

    if (v % 2 == 0){
      count++;
    }

    for(long i = 3; i <= v; i = i + 2){
      if(v % i == 0){
        if (isPrimeNum(i)){
          count++;
        }
      }
    }

    System.out.println(count);
    
  }

  public static long gcd(long x, long y){

    if (x < y) {
      long t = x;
      x = y;
      y = t;
    }
  
    while (y > 0) {
      long r = x % y;
      x = y;
      y = r;
    }
  
    return x;
  
  }

  static boolean isPrimeNum(long x) {

    if(x == 2) {
      return true;
    }

    if(x < 2 || x % 2 == 0) {
      return false;
    }

    for(long n = 3; n <= Math.sqrt((double)x); n += 2) {
      
      if(x % n == 0) {
        return false;
      }
    
    }

    return true;

  }

}