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 C# (Mono 4.6.2.0) (judge verdict: Runtime Error)

using System;
using System.Collections.Generic;
using System.Linq;
using static System.Console;
using System.Runtime.CompilerServices;
using static MyUtil;

class MyUtil
{
    public static long[] ReadLongArray()
    {
	return ReadLine().Split().Select(x => long.Parse(x)).ToArray();
    }

    public static long Gcd(long a, long b)
    {
	if (a < b){ return Gcd(b, a); }
	while (b != 0)
	{
	    var r = a % b;
	    a = b;
	    b = r;
	}
	return a;
    }
}

class Program
{
    static long[] d;
    
    public static void Main()
    {
	var tmp = ReadLongArray();
	long a = tmp[0], b = tmp[1];
	long n = Gcd(a, b);

	int ct = 0;
	d = new long[n];

	for (long i = 1; i * i <= n; i++)
	{
	    if (n % i == 0 && (i == 2 || (i&1) == 1)) 
	    {
		if (i == 2)
		{
		    d[ct++] = i;
		    if (check(ct, n/i))
			d[ct++] = n / i;
		}
		else
		{
		    if(check(ct, i)){
			d[ct++] = i;
		    }
		}
	    }
	}

	WriteLine(ct);
    }

    static bool check(int ct, long x)
    {
	for (int i = 1; i < ct; i++) // skip d[0] == 1
	{
	    if (x%d[i] == 0) return false;
	}
	return true;
    }
}