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

class Myon
{
    public Myon() { }
    public static int Main()
    {
        cin = new Scanner();
        new Myon().calc();
        return 0;
    }

    static Scanner cin;

    // メンバ変数---
    long A;
    long B;
    // -------------

    void calc()
    {
        A = cin.nextLong();
        B = cin.nextLong();
        long gcd = Gcd(A, B);
        Dictionary<long, long> divisor = PrimeFactorization(gcd);
        Write(divisor.Count+1);

    }
    /// <summary>
    /// 最大公約数
    /// </summary>
    /// <param name="a"></param>
    /// <param name="b"></param>
    /// <returns></returns>
    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];
    }

    Dictionary<long, long> PrimeFactorization(long n)
    {
        var primes = new Dictionary<long, long>();
        for (long i = 2; i * i <= n; ++i)
        {
            while (n % i == 0)
            {
                if (primes.ContainsKey(i))
                    ++primes[i];
                else
                    primes[i] = 1;
                n /= i;
            }
        }
        if (n != 1) primes[n] = 1;
        return primes;
    }

    void Write(string v)
    {
        Console.WriteLine(v);
    }

    void Write(int v)
    {
        Console.WriteLine(v);
    }

    void Write(double v)
    {
        Console.WriteLine(v);
    }
    void Write(long v)
    {
        Console.WriteLine(v);
    }

    int Pow(int x, int y)
    {
        return (int)Math.Pow(double.Parse(x.ToString()), double.Parse(y.ToString()));
    }

    double Pow(double x, double y)
    {
        return Math.Pow(x, y);
    }

    class Scanner
    {
        string[] s;
        int i;

        char[] cs = new char[] { ' ' };

        public Scanner()
        {
            s = new string[0];
            i = 0;
        }

        public string next()
        {
            if (i < s.Length) return s[i++];
            string st = Console.ReadLine();
            while (st == "") st = Console.ReadLine();
            s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
            if (s.Length == 0) return next();
            i = 0;
            return s[i++];
        }

        public int nextInt()
        {
            return int.Parse(next());
        }
        public int[] ArrayInt(int N, int add = 0)
        {
            int[] Array = new int[N];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextInt() + add;
            }
            return Array;
        }

        public long nextLong()
        {
            return long.Parse(next());
        }

        public long[] ArrayLong(int N, long add = 0)
        {
            long[] Array = new long[N];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextLong() + add;
            }
            return Array;
        }

        public double nextDouble()
        {
            return double.Parse(next());
        }


        public double[] ArrayDouble(int N, double add = 0)
        {
            double[] Array = new double[N];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextDouble() + add;
            }
            return Array;
        }
    }
}