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

namespace debug
{
    class main
    {
        static void Main(string[] args)
        {
            //問題クラスを展開
            ProgramD a = new ProgramD();
            a.main();//実行する

        }

    }

    class ProgramA
    {
        public void main()
        {

            string input = Console.ReadLine();
            int a = int.Parse(input);
            double b = (double)((a / 2 + 1) / a);

            if (a % 2 == 0)
                Console.WriteLine("0.5");
            else
                Console.WriteLine(b);

        }
    }

    class ProgramB
    {
        public void main()
        {
            string[] input = Console.ReadLine().Split(' ');

            int n = int.Parse(input[0]);
            int k = int.Parse(input[1]);

            string[] a = Console.ReadLine().Split(' ');
            int count = 0;

            for (int i = 0; i < n; i++)
            {
                long ai = long.Parse(a[i]);
                if (k <= ai)
                    count++;

            }

            Console.WriteLine(count);
        }
    }

    class ProgramC
    {
        public void main()
        {

            string input = Console.ReadLine();
            int n = int.Parse(input);


            string[] input_b = Console.ReadLine().Split(' ');
            long[] b = new long[n];
            long[] c = new long[n];
            for (int i = 0; i < n; i++)
                b[i] = long.Parse(input_b[i]);


            for (int i = 1; i < n + 1; i++)
                c[b[i - 1] - 1] = i;

            for (int i = 0; i < n; i++)
                Console.Write(c[i] + " ");
        }
    }


    class ProgramD
    {
        public void main()
        {

            string[] input = Console.ReadLine().Split(' ');

            long a = long.Parse(input[0]);
            long b = long.Parse(input[1]);

            long c = gcd(a, b);//最大公約数を求める
            long count = 1;//1はあらかじめカウント
            int flag = 0;

            //最大公約数が1なら答は1
            if(c == 1)
            {
                Console.WriteLine("1");
                return;

            }

            //素因数分解してみる（例えば、2で割った後は2より大きい素数で2つの数で表せるかどうかを随時判定
            for (long i = 2; i <= Math.Sqrt(c); i++)
            {
                while (c % i == 0)
                {
                    c /= i;
                    flag = 1;
                }

                if (flag == 1)
                {
                    count++;
                    flag = 0;
                }

            }

            //ただし、計算結果でcが1以外ということは最後の素数が計上されていない。
            if (c != 1)
                count++;

            Console.WriteLine(count);


        }

        /*最大公約数を求める関数*/
        static long gcd(long p, long q)
        {
            if (p < q)
            {
                long temp = p;
                p = q;
                q = temp;
            }

            long r = 1;

            while (r != 0)
            {
                r = p % q;
                p = q;
                q = r;
            }

            return p;
        }

    }

    class ProgramE
    {
        public void main()
        {

            string[] input = Console.ReadLine().Split(' ');

            int a = int.Parse(input[0]);
            int b = int.Parse(input[1]);
        }
    }


    
}