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# (.NET Core 3.1.201) (judge verdict: Accepted)

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


class Program
{
    static void Main(string[] args)
    {
        // 入力
        //var s = Console.ReadLine();
        //var n = int.Parse(Console.ReadLine());
        long[] ab = Console.ReadLine().Split().Select(long.Parse).ToArray();
        var a = ab[0];
        var b = ab[1];
        long ans = 0;

        IEnumerable<long> alist = PrimeFactors(a);
        IEnumerable<long> blist = PrimeFactors(b);
        var set1 = new HashSet<long>(alist);
        var set2 = new HashSet<long>(blist);

        ans = set1.Intersect(set2).Count() + 1;
        // 出力
        Console.WriteLine(ans);
    }

    //素因数分解
    public static IEnumerable<long> PrimeFactors(long n)
    {
        long 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;//最後の素数も返す
    }
}