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: Runtime Error)

using System;
using System.Collections.Generic;

class Program
{
    static void Main(string[] args)
    {
        //// 整数の入力
        //int a = int.Parse(Console.ReadLine());
        //// スペース区切りの整数の入力
        //string[] input = Console.ReadLine().Split(' ');
        //int b = int.Parse(input[0]);
        //int c = int.Parse(input[1]);
        //// 文字列の入力
        //string s = Console.ReadLine();
        ////出力
        //Console.WriteLine((a + b + c) + " " + s);

        string[] input = Console.ReadLine().Split(' ');
        int A = int.Parse(input[0]);
        int B = int.Parse(input[1]);

        int answer = 0;

        List<int> H = new List<int>();
        List<int> C = new List<int>();
        for(int i = 1;i <= A; i++)
        {
            if (i > B)
            {
                break;
            }
            if ( A%i == 0 && B % i == 0)
            {
                H.Add(i);
            }
        }
        C.Add(H[0]);
        H.RemoveAt(0);
        while(H.Count != 0)
        {
            int r = H[0];
            
            H.RemoveAt(0);

            if (r != 0)
            {
                C.Add(r);

                for(int i = 0; i < H.Count; i++)
                {
                    if (H[i] % r == 0)
                    {
                        H[i] = 0;
                    }
                }
            }
        }

        Console.WriteLine(C.Count);

    }
}