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 = sc.nextLong();
		long b = sc.nextLong();
		
		long gcd = gcd(a,b);
		
		System.out.println(prime(gcd));

	}
	
	static long gcd(long a ,long b) {
		if(b > a)
			return gcd(b,a);
		if(a % b == 0)
			return b;
		
		return gcd(b ,a%b);
	}

	static long prime(long n) {
		
		long d = n;
		long count = 1;
		
		for(long i = 2 ; i <= d ; i++) {
			if(n % i == 0) {
				count++;
				while(n % i == 0) {
					n = n/i;
				}
			}
			if(n == 1)
				break;
		}
		return count;
		}
}