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: Wrong Answer)

using System;
using System.Diagnostics;
using System.IO;
using System.Text;
using System.Linq;
using System.Collections.Generic;

class Program
{
    static void Main(string[] args) {
#if DEBUG
        //MyDebugger.MakeTestCase("../TextFile2.txt"); return;
        MyDebugger.Test(process, "../TextFile1.txt");
#else
        object ret = process(new StreamReader(Console.OpenStandardInput()));
        Console.WriteLine(ret);
#endif
    }

    static object process(TextReader input) {
        var spl = input.ReadLine().Split(' ');
        var a = long.Parse(spl[0]);
        var b = long.Parse(spl[1]);

        List<long> list = new List<long>();
        long max = (long)Math.Min(Math.Sqrt(a),Math.Sqrt(b))+1;

        for (long i = 3; i <= max; i+=2) {
            if (a % i == 0 && b % i == 0) {
                bool isok = true;
                foreach (var v in list) {
                    if (i % v == 0) {
                        isok = false;
                        break;
                    }
                }
                if (isok) {
                    list.Add(i);
                }
            }
        }

        {
            long i = Math.Min(a, b);
            if (a % i == 0 && b % i == 0) {
                bool isok = true;
                foreach (var v in list) {
                    if (i % v == 0) {
                        isok = false;
                        break;
                    }
                }
                if (isok) {
                    list.Add(i);
                }
            }
        }

        list.Add(1);
        if(a%2==0 && b%2==0){
            list.Add(2);
        }

        return list.Count;
    }
}