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.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Main {

	private static Set<Long> primeSet = new HashSet<Long>();
	public static void main(String[] args) {
		try (Scanner sc = new Scanner(System.in)) {
			long a = sc.nextLong();
			long b = sc.nextLong();

			long tmp = getGcd(a, b);
			makePrimeSet(tmp);

			Set<Long> resultSet = new HashSet<Long>();
			resultSet.add(1l);
			// 偶数なら2を入れとく
			if (tmp % 2 == 0) {
				resultSet.add(2l);
			}
			while(tmp % 2 == 0) {
				tmp /= 2;
			}

			long div = 3;
			while (tmp != 1) {
				if (tmp % div == 0) {
					resultSet.add(div);
					tmp /= div;
				} else {
					div += 2;
				}
			}
			System.out.println(resultSet.size());
		}
	}

	static long getGcd (long a, long b) {
		long temp;
		while((temp = a%b)!=0) {
			a = b;
			b = temp;
		}
		return b;
	}

	private static void makePrimeSet(long max) {
		if (max < 2) {
			return;
		}
		primeSet.add(2l);
		for (long i = 3; i <= max; i += 2) {
			if (isPrime(i)) {
				primeSet.add(i);
			}
		}
		return ;
	}

	private static boolean isPrime(long n) {
		for (long i : primeSet) {
			if (n % i == 0) {
				return false;
			}
		}
		return true;
	}
}