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: Runtime Error)

import java.util.Scanner;

public class Main{
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		while (scan.hasNext()) {
			int a = scan.nextInt();
			int b = scan.nextInt();
			int temp,max;
			int[] arr = new int[1000];
			arr[0] = 1;
			if (a > b){
				temp = b;
				max=a;
			}
			else{
				temp = a;
				max=b;
			}
			int flag = 1;
			while (temp % 2 == 0) {
				flag = 0;
				temp /= 2;
				arr[1] = 2;
			}
			int sum = 1;
			int j;
			if (flag == 1)
				j = 1;
			else {
				j = 2;
				sum++;
			}

			for (int i = 3; i <= temp;) {
				if (temp % i == 0) {
					arr[j++] = i;
					temp /= i;
					sum++;
				} else
					i += 2;
			}
			int sum2=0;
			for(int i=0;i<sum;i++){
				if(max%arr[i]==0)
					sum2++;
			}
			System.out.println(sum2);
		}
	}
}