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 min = Math.min(a, b);
        List<Long> sets = new LinkedList<>();
        List<Long> sets2 = new LinkedList<>();
        long max = calc(a, b);
        System.out.println(math(max, 0));
    }
    private static long calc(long a, long b) {
        if(a == b ) return a;
        if(a > b) return calc(a -b, b);
        return calc(a, b -a);
    }
    private static long math(long m, long count) {
        if(m == 1.0) return count;
        if(m % 2 == 0) return math(m / 2, count + 1);
        for(long i = 3L; i <= m; i += 2) {
            if(m % i == 0) return math(m / i, count + 1);
        }
        return 0;
    }
}