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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;

class ABC142
{
    static void Main(string[] args)
    {
        var s = Console.ReadLine().Split().Select(long.Parse).ToArray();

        var a = s[0];
        var b = s[1];

        var m = Gcd(a, b);

        int c = 0;

        for (long i = 3; i <= m; i += 2)
        {
            long k = 0;
            for (long j = 3; j <= Math.Sqrt(i); j += 2)
            {
                if (i % j == 0)
                {
                    k = 1;
                    break;
                }
            }

            if (k == 0 && a % i == 0 && b % i == 0) c++;
        }

        Console.WriteLine(c+2);
    }

    // ユークリッドの互除法 
    public static long Gcd(long a, long b)
    {
        if (a < b)
            // 引数を入替えて自分を呼び出す
            return Gcd(b, a);
        while (b != 0)
        {
            var remainder = a % b;
            a = b;
            b = remainder;
        }
        return a;
    }
    
}