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: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApp11
{
    class Class6
    {
        public static void Main()
        {
            var I = Console.ReadLine().Split().Select(x => long.Parse(x)).ToArray();
            long gcd = GCD(I[0], I[1]);
            int count = 0;
            for(int i = 2; i <= gcd; i++)
            {
                if (gcd % i == 0)
                {
                    count += 1;
                    while (gcd % i == 0)
                        gcd /= i;
                }
                    
            }
            Console.WriteLine(count + 1);
        }
        public static long GCD(long a,long b)
        {
            var v = new[] { a, b };
            while (v[1] != 0) { v = new[] { v[1], v[0] % v[1] }; }
            return v[0];
        }
    }        
}