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.util.Scanner;
 
public class Main {
 
	public static void main(String[] args) {
 
		try (Scanner scan = new Scanner(System.in)) {
 
			long x = scan.nextLong();
			long y = scan.nextLong();
 
			if(y > x) {
				long swap = x;
				x = y;
				y = swap;
			}
			
			Long s = 2L;
			long gcd = gcd(x, y);
       
			int ans = 1;
			while (s <= Math.sqrt(gcd)) {
				if ( gcd % s == 0 ) ans++;
				
				while ( gcd % s == 0 ) gcd /= s;
					s++;
				
			}
			if ( gcd != 1 ) ans++;
			System.out.println(ans);
		}
	}
  
  	private static long gcd(long x, long y) {
  		if(y == 0) return x;
     	return gcd(y, x%y);     
    }
}