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.*;
import static java.lang.Integer.*;
import static java.lang.Long.*;
import static java.lang.Math.*;
import static java.lang.System.*;

public class Main {
	public static int MOD = 1000000007;
	public static void main(String[] args) {
		int i,j;
		Scanner sc = new Scanner(in);
		long n = parseLong(sc.next());
		long m = parseLong(sc.next());
		sc.close();
		long g = gcd(n,m);
		long cnt = 1;
		if(g%2L==0) {
			cnt++;
			g /= 2L;
			while(g%2L==0) {
				g /= 2L;
			}
		}
		long mx = (long)sqrt(g);
		for (long k = 3L; k <= mx; k+=2L) {
			if(g%k==0) {
				cnt++;
				g /= k;
				while(g%k==0) {
					g /= k;
				}
			}
		}
		if(g!=1) {
			cnt++;
		}
		out.println(cnt);
	}
	public static long gcd(long a, long b) {
		long tmp;
		while((tmp = a%b)!=0) {
			a = b;
			b = tmp;
		}
		return b;
	}

}