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: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using static System.Console;
using System.Text;
using System.IO;
using static System.Math;
namespace AtCoder
{
    class Program
    {
        static public long[] Sarray() { return ReadLine().Split().Select(long.Parse).ToArray(); }
        static public List<long> Slist() { return ReadLine().Split().Select(long.Parse).ToList(); }

        static long Mod = (long)1e9 + 7;
        static void Main(string[] args)
        {
            //SetOut(new StreamWriter(OpenStandardOutput()) { AutoFlush = false });
            var AB = Sarray();
            var A = AB[0];
            var B = AB[1];
            var gcd = Gcd(A, B);
            var ans = primefactorize(gcd);
            WriteLine(ans.Distinct().Count() + 1);
            //Out.Flush();
        }
        static long Gcd(long x, long y) => y == 0 ? x : Gcd(y, x % y);
        static List<long> primefactorize(long n)
        {
            var primes = new List<long>();
            for (long i = 2; i * i <= n; ++i)
            {
                if (n % i == 0)
                {
                    while (n % i == 0)
                    {
                        primes.Add(i);
                        n /= i;
                    }
                }
            }
            if (n != 1) primes.Add(n);
            return primes;
        }
    }
}