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.Linq;
using System.Collections;
using System.Collections.Generic;

namespace AtCoder
{
    class Program
    {
        static void Add(ref List<int> res, ref List<int> res2, int value)
        {
            if (res.Contains(value))
            {
                res2.Add(value);
            }
            else
            {
                res.Add(value);
            }
        }

        static List<int> GetDivisor(int a, int b)
        {
            int minValue = Math.Min(a, b);
            List<int> divisors = new List<int>();
            for (int i = 1; i <= minValue; i++)
            {
                if (a % i == 0 && b % i == 0)
                {
                    divisors.Add(i);
                }
            }

            return divisors;
        }

        static void Main(string[] args)
        {
            int[] numbers = Console.ReadLine().Split().Select(int.Parse).ToArray();

            List<int> divisors = GetDivisor(numbers[0], numbers[1]);
            List<int> getDivisor = new List<int>();
            List<int> result = new List<int>();
            List<int> result2 = new List<int>();

            int div0 = divisors[divisors.Count - 1];
            int div1 = divisors[0];

            getDivisor = GetDivisor(div0, div1);

            if (getDivisor.Count == 1 && getDivisor[0] == 1)
            {
                Add(ref result, ref result2, div0);
                Add(ref result, ref result2, div1);
            }

            for (int i = 0;i < divisors.Count;i++)
            {
                if(i < divisors.Count - 1)
                {
                    div0 = divisors[i];
                    div1 = divisors[i + 1];

                    getDivisor = GetDivisor(div0, div1);
                    if (getDivisor.Count == 1 && getDivisor[0] == 1)
                    {
                        Add(ref result, ref result2, div0);
                        Add(ref result, ref result2, div1);
                    }
                }
            }

            Console.WriteLine(result2.Count);

            /*
            System.Text.StringBuilder sb = new System.Text.StringBuilder();
            for (int j = 0;j < result2.Count;j++)
            {
                sb.Append( result2[j].ToString());
                if(j < result2.Count - 1)
                {
                    sb.Append(" ");
                }
            }

            Console.WriteLine(sb);
            */

        }
    }
}