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;
using System.Linq;

namespace test
{
    class Program
    {
        static void Main(string[] args)
        {
            var val = Console.ReadLine().Split(' ');
            var a = int.Parse(val[0]);
            var b = int.Parse(val[1]);

            int max = a > b ? b : a;
            //var h = new int[max];
            int v = 0;
            for (int num = 1; num < max; num++)
            {
                if (a % num == 0 && b % num == 0)
                {
                    if (num < 2) continue;
                    else if (num == 2)
                    {
                        //h[v] = num;
                        v++;
                        continue;
                    }
                    else if (num % 2 == 0) continue;

                    double sqrtNum = Math.Sqrt(num);
                    bool t = false;
                    for (int i = 3; i <= sqrtNum; i += 2)
                    {
                        if (num % i == 0)
                        {
                            // 素数ではない
                            t = true;
                            break;
                        }
                    }
                    if(t) continue;

                    // 素数である
                    //h[v] = num;
                    v++;
                }
            }


            Console.Write(v+1);

        }
    }
}