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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
    static Scanner sc;
    public static void main(String[] args) {
        sc = new Scanner(System.in);
        long A = sc.nextLong();
        long B = sc.nextLong();
        long mi = Math.min(A,B);
        long ma = Math.max(A,B);

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

        for (long k = 1; k <= Math.sqrt(mi); k++) {
            if (mi % k == 0 && ma % k == 0) {
                //calculate the second and third factor
                long n1 = mi / k;
                long n2 = ma / k;

                boolean fits = true;
                boolean fits1 = true;
                boolean fits2 = true;

                for (long d : divisors) {
                    if (d != 1 && k % d == 0) {
                        fits = false;
                    }
                    if (d != 1 && n1 % d == 0) {
                        fits1 = false;
                    }
                    if (d != 1 && n2 % d == 0) {
                        fits2 = false;
                    }
                }
                if (fits) {
                    divisors.add(k);
                }
                boolean n1added = false;
                if(ma % n1 == 0 && fits1 && n1 % k != 0) {
                    divisors.add(n1);
                    n1added = true;
                }
                if(mi % n2 == 0 && fits2 && n2 % k != 0 && (n2 % n1 != 0 || !n1added)) {
                    divisors.add(n2);
                }
            }
        }


        if(ma % mi == 0 && mi != 1) {
            boolean fits = true;
            for (long d : divisors) {
                if (d != 1 && mi % d == 0) {
                    fits = false;
                }
            }
            if (fits) {
                divisors.add(mi);
            }
        }
        System.out.println(divisors.size());
    }
}