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.BufferedReader;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.StringJoiner;

public class Main {
    public static void main(String[] args) {
        Main main = new Main();
        main.solve();
    }
    public void solve() {
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        long A = scan.nextLong();
        long B = scan.nextLong();
        long gcd = gcd(A, B);
        Map<Long, Long> map = divisors(gcd);
        System.out.println(map.size() + 1);
    }
    private Map<Long, Long> divisors(long N) {
        Map<Long, Long> result = new HashMap<>();
        for (long i = 2; i <= N; i++) {
            if (N % i != 0) {
                continue;
            }
            long count = 0;
            while (N % i == 0) {
                count += 1;
                N /= i;
            }
            result.put(i, count);
        }
        return result;
    }
    private long gcd(long A, long B) {
        if (B == 0) {
            return A;
        }
        return gcd(B,A % B);
    }
}