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: Wrong Answer)

import java.util.Scanner;
public class Main{
	
	static long count = 0;
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		long a = sc.nextLong();
		long b = sc.nextLong();
		
		long n = prime1(a,b);
		
		System.out.println(n);
		
		
	}

	private static long prime1(long a,long b) {
		
		long p = (long)Math.floor(Math.sqrt(a));
		long q = (long)Math.floor(Math.sqrt(b));
		
		if(a == 1 || b == 1) {
			return 1;
		}
		for(long i = 1; i <= Math.min(p, q) ; i++) {
			if((a % i == 0) &&( b % i == 0)) {
				if(prime2(i)) {
					count++;
				}	
			}
		}
		return count;
		
	}
private static boolean prime2(long a) {
	
		long p = (long)Math.floor(Math.sqrt(a));
		
		if(a == 1) {
			return true;
		}
		if(a == 2) {
			return true;
		}
		
		int temp = -1;
		
		for(long i = 2; i <= p ; i++) {
			if(a % i == 0 ) {
				temp = 0;
				break;
			}	
		}
		if(temp == -1) {
			return true;
		}
		else {
			return false;
		}
		
}
	}