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 Java (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);

		while (input.hasNext()) {
			long A = input.nextLong();
			long B = input.nextLong();
			long gcdAB = gcd(A, B);

			int countPrimeFactor = 0;
			for (long i = 2; i * i <= gcdAB; ++i) {
				if (gcdAB % i == 0) {
					++countPrimeFactor;
					
					while (gcdAB % i == 0) {
						gcdAB /= i;
					}
				}
			}
			
			if (gcdAB != 1) {
				++countPrimeFactor;
			}
			
			System.out.println(countPrimeFactor + 1);
		}
	}

	private static long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}
}