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.IO;
using System.Text;
using System.Diagnostics;
using static System.Console;
using static System.Math;

namespace CodeForces
{
    class CodeForcesss
    {
        static void Main(string[] args)
        {
            long[] s = ReadLine().Split(' ').Select(long.Parse).ToArray();
            long a = s[0];
            long b = s[1];
            var hoge = PrimeFactor(a).Distinct().ToList();
            var fuga = PrimeFactor(b).Distinct().ToList();
            WriteLine(
                hoge.Intersect(fuga).Count() + 1
                );

        }

        static IEnumerable<long> PrimeFactor(long n)
        {
            int i = 2;
            long tmp = n;

            while (i * i <=n)
            {
                if (tmp % i == 0)
                {
                    tmp /= i;
                    yield return i;
                }
                else i++;
            }
            if (tmp != 1) yield return tmp;
        }
    }
}