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;

public class Program
{
    public void Solve()
    {
        var sc = new Scanner();
        long a = sc.NextLong();
        long b = sc.NextLong();
        long gcd = MathEx.GCD(a, b);

        var cnt = 1;

        for (int i = 2; i*i <= gcd; i++)
        {
            if (gcd % i == 0)
            {
                cnt++;
                while (gcd % i==0)
                {
                    gcd /= i;
                }
            }
        }

        if (gcd != 1)
            cnt++;
        Console.WriteLine(cnt);
    }

    public static void Main(string[] args)
    {
        new Program().Solve();
    }
}

#region GCD LCM

/// <summary>
/// 様々な数学的関数の静的メソッドを提供します．
/// </summary>
public static partial class MathEx
{
    /// <summary>
    /// 2 つの整数の最大公約数を求めます．
    /// </summary>
    /// <param name="n">最初の値</param>
    /// <param name="m">2 番目の値</param>
    /// <returns>2 つの整数の最大公約数</returns>
    /// <remarks>ユークリッドの互除法に基づき最悪計算量 O(log N) で実行されます．</remarks>
    public static int GCD(int n, int m)
    {
        return (int) GCD((long) n, m);
    }


    /// <summary>
    /// 2 つの整数の最大公約数を求めます．
    /// </summary>
    /// <param name="n">最初の値</param>
    /// <param name="m">2 番目の値</param>
    /// <returns>2 つの整数の最大公約数</returns>
    /// <remarks>ユークリッドの互除法に基づき最悪計算量 O(log N) で実行されます．</remarks>
    public static long GCD(long n, long m)
    {
        n = Math.Abs(n);
        m = Math.Abs(m);
        while (n != 0)
        {
            m %= n;
            if (m == 0) return n;
            n %= m;
        }

        return m;
    }


    /// <summary>
    /// 2 つの整数の最小公倍数を求めます．
    /// </summary>
    /// <param name="n">最初の値</param>
    /// <param name="m">2 番目の値</param>
    /// <returns>2 つの整数の最小公倍数</returns>
    /// <remarks>最悪計算量 O(log N) で実行されます．</remarks>
    public static long LCM(long n, long m)
    {
        return (n / GCD(n, m)) * m;
    }
}

#endregion

#region PrimeSieve

public static partial class MathEx
{
    /// <summary>
    /// ある値までに素数表を構築します．
    /// </summary>
    /// <param name="max">最大の値</param>
    /// <param name="primes">素数のみを入れた数列が返される</param>
    /// <returns>0 から max までの素数表</returns>
    /// <remarks>エラトステネスの篩に基づき，最悪計算量 O(N loglog N) で実行されます．</remarks>
    public static bool[] Sieve(int max, List<int> primes = null)
    {
        var isPrime = new bool[max + 1];
        for (int i = 2; i < isPrime.Length; i++) isPrime[i] = true;
        for (int i = 2; i * i <= max; i++)
            if (!isPrime[i]) continue;
            else
                for (int j = i * i; j <= max; j += i)
                    isPrime[j] = false;
        if (primes != null)
            for (int i = 0; i <= max; i++)
                if (isPrime[i])
                    primes.Add(i);

        return isPrime;
    }
}

#endregion

class Scanner
{
    public Scanner()
    {
        _pos = 0;
        _line = new string[0];
    }

    const char Separator = ' ';
    private int _pos;
    private string[] _line;

    #region スペース区切りで取得

    public string Next()
    {
        if (_pos >= _line.Length)
        {
            _line = Console.ReadLine().Split(Separator);
            _pos = 0;
        }

        return _line[_pos++];
    }

    public int NextInt()
    {
        return int.Parse(Next());
    }

    public long NextLong()
    {
        return long.Parse(Next());
    }

    public double NextDouble()
    {
        return double.Parse(Next());
    }

    #endregion

    #region 型変換

    private int[] ToIntArray(string[] array)
    {
        var result = new int[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = int.Parse(array[i]);
        }

        return result;
    }

    private long[] ToLongArray(string[] array)
    {
        var result = new long[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = long.Parse(array[i]);
        }

        return result;
    }

    private double[] ToDoubleArray(string[] array)
    {
        var result = new double[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = double.Parse(array[i]);
        }

        return result;
    }

    #endregion

    #region 配列取得

    public string[] Array()
    {
        if (_pos >= _line.Length)
            _line = Console.ReadLine().Split(Separator);

        _pos = _line.Length;
        return _line;
    }

    public int[] IntArray()
    {
        return ToIntArray(Array());
    }

    public long[] LongArray()
    {
        return ToLongArray(Array());
    }

    public double[] DoubleArray()
    {
        return ToDoubleArray(Array());
    }

    #endregion
}