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.io.InputStream;
import java.io.PrintStream;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Main {

  public static void main(String[] args) {
    resolve(System.in, System.out);
  }

  static void resolve(InputStream is, PrintStream pw) {
    Scanner sc = new Scanner(is);
    Set<Integer> aPrimes = primes(Long.parseLong(sc.next()));
    Set<Integer> bPrimes = primes(Long.parseLong(sc.next()));
    aPrimes.retainAll(bPrimes);
    pw.println(aPrimes.size() + 1);
  }

  static Set<Integer> primes(long n) {
    Set<Integer> ans = new HashSet<>();
    for (int i = 2; i <= n; i++) {
      if (n % i == 0) {
        ans.add(i);
        n = n / i;
        i = 1;
      }
    }
    return ans;
  }
}