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();

    long min = gcd(a, b);

    List<Long> ky = new ArrayList<>();

    for (long i = 1; i <= min; i++) {
      if (a % i == 0 && b % i == 0) {
        boolean issosu = sosu(i);
        if (issosu) {
          ky.add(i);
        }
      }
    }

    System.out.println(ky.size());
    
  }

  private static boolean sosu(long m) {
    if (m == 0) return false;
    if (m <= 2) return true;
    for (long i = 2; i < m; i++) {
      if (m % i == 0) return false;
    }
    return true;
  }
  
  private static long gcd(long m, long n) {
    if(m < n) return gcd(n, m);
    if(n == 0) return m;
    return gcd(n, m % n);
  }
}