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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {


	public static void main(String[] args) throws IOException {
		BufferedReader bf = null;
		bf = new BufferedReader(new InputStreamReader(System.in));
		long a,b,ans = 0,temp,kouyaku,n;
		String str;
		String[] s;
		str = bf.readLine();
		s = str.split(" ");
		a = Long.parseLong(s[0]);
		b = Long.parseLong(s[1]);
		if(a < b) {
			temp = a;
			a = b;
			b = temp;
		}
		while(a%b != 0) {
			temp = a%b;
			a = b;
			b = temp;
		}
		kouyaku = b;
		ans = 1;
		if(kouyaku != 1) {
			if(kouyaku%2 == 0) {
				ans++;
				while(kouyaku%2 == 0) {
					kouyaku /= 2;
				}
			}
			n = 3;
			if(kouyaku != 1) {
				while(kouyaku >= n*n) {
					if(kouyaku%n == 0) {
						ans++;
						while(kouyaku%n == 0) {
							kouyaku /= n;
						}
					}
					n += 2;
				}
			}
			if(kouyaku != 1) {
				ans++;
			}
		}
		System.out.println(ans);
	}
}