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.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long a = Long.parseLong(sc.next());
		long b = Long.parseLong(sc.next());
		long ans = 1;

		for (int i = 2; i <= 3; i++) {
			if (a % i == 0 && b % i == 0) {
				ans++;
			}
		}

		for (long i = 5; i <= getGcd(a, b); i += 2) {
			if (a % i == 0 && b % i == 0 && IsPrime(i)) {
				ans++;
			}
		}
		System.out.println(ans);
	}

	static boolean IsPrime(long n) {
		double sqrtNum = Math.sqrt(n);
		for (long i = 3; i <= sqrtNum; i += 2) {
			if (n % i == 0) {
				return false;
			}
		}
		return true;
	}

	//最大公約数(greatest common divisor)を返す関数
	static long getGcd(long a, long b) {
		long temp;
		while ((temp = a % b) != 0) {
			a = b;
			b = temp;
		}
		return b;
	}
}