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: Accepted)

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

public class Main {
	InputStream in = System.in;
	PrintStream out = System.out;

	private Set<Long> divisors(long x) {
		Set<Long> result = new HashSet<>();

		for (int i = 1; i <= Math.sqrt(x) + 1; i++) {
			if (x % i == 0) {
				result.add((long) i);
				result.add(x / i);
			}
		}

		return result;
	}

	// 最大公約数
	private static long gcd(long a, long b) {
		if (b == 0) {
			return a;
		}
		return gcd(b, a % b);
	}

	private boolean isPrime(long x) {
		if (x < 2)
			return false;
		else if (x == 2)
			return true;
		else if (x % 2 == 0)
			return false;
		for (int i = 3; i <= Math.sqrt(x); i += 2) {
			if (x % i == 0) {
				return false;
			}
		}
		return true;
	}

	private List<Long> prime(List<Long> x) {
		List<Long> result = new ArrayList<>();
		result.add((long) 1);
		for (long n : x) {
			if (isPrime(n)) {
				result.add(n);
			}
		}
		return result;
	}

	public void _main(String[] args) {
		Scanner sc = new Scanner(in);

		long A = sc.nextLong();
		long B = sc.nextLong();

		sc.close();

		Set<Long> cd = divisors(A);
		cd.retainAll(divisors(B));

		List<Long> arr = new ArrayList<>(cd);

		arr = prime(arr);

		out.println(arr.size());

	}

	public static void main(String[] args) {
		new Main()._main(args);
	}
}