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();
        
        Set<Long> divA = new HashSet<Long>();
        Set<Long> divB = new HashSet<Long>();
        for (long i = 1; i*i < Math.max(A,B); i++) {
            if (A%i == 0) {
                divA.add((long)i);
                divA.add(A/i);
            }
            if (B%i == 0) {
                divB.add((long)i);
                divB.add(B/i);
            }
        }

        List<Long> divCommon = new ArrayList<Long>();
        for (Long l : divA) {
            if (divB.contains(l)) {
                divCommon.add(l);
            }
        }

        long ans = 0;
        Collections.sort(divCommon);
        for (Long i : divCommon) {
            if (i == 1) {
                ans++;
                continue;
            }
            boolean div = false;
            for (Long j : divCommon) {
                if (i <= j) {
                    break;
                }
                if (i%j == 0 && j != 1) {
                    div = true;
                    break;
                }
            }
            if (!div) {
                ans++;
            }
        }

        //System.out.println(divCommon.size());
        System.out.println(ans);
    }
}