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.*;

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(Math.max(A, B), Math.min(A, B));
		
		int count = 0;
		for(long i = 2; i * i <= gcd; i++){
			if(gcd % i == 0){
				do{
					gcd /= i;
				}while(gcd % i == 0);
				count++;
			}
		}
		if(gcd != 1)count++;
		System.out.println(count + 1);
		
	}
	
	static private long GCD(long X, long Y) {
		while(Y != 0){
			long temp = X % Y;
			X = Y;
			Y = temp;
		}
		return X;
	}
}