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.Collections;
using System.Collections.Generic;
using System.Text;
using System.Linq;
using static MyIO;


public class D
{
	public static void Main() => (new Solver()).Solve();
}


public class Solver
{
	public void Solve()
	{
		long A = GetLong();
		long B = GetLong();

		var divsA = new List<long>();
		for(long i = 1; i * i <= A; i++)
		{
			if(A % i == 0)
			{
				divsA.Add(i);
				if(i * i < A)
					divsA.Add(A / i);
			}
		}

		var comDivs = new List<long>();
		for(int i = 0; i < divsA.Count; i++)
			if(B % divsA[i] == 0)
				comDivs.Add(divsA[i]);

		var primes = sieve(1000000);

		int ans = 0;
		var set = new HashSet<long>();
		for(int i = 0; i < comDivs.Count; i++)
		{
			bool isOK = true;
			long rest = comDivs[i];
			for(int j = 0; j < primes.Count; j++)
			{
				if(rest % primes[j] != 0)
					continue;

				if(set.Contains(primes[j]))
				{
					isOK = false;
					break;
				}
				set.Add(primes[j]);

				while(rest % primes[j] == 0)
					rest /= primes[j];
			}
			if(rest > 1)
			{
				if(set.Contains(rest))
				{
					isOK = false;
				}
				else
				{
					set.Add(rest);
				}
			}
			if(isOK)
				ans++;
		}

		Console.WriteLine(ans);
	}

	private static List<long> sieve(long N)
	{
		List<long> primeList = new List<long>();
		bool[] isPrime = new bool[N+1];
		for(long i = 2; i <= N; i++)
			isPrime[i] = true;

		for(long i = 2; i <= N; i++)
		{
			if(isPrime[i])
			{
				primeList.Add(i);
				for(long j = 2 * i; j <= N; j += i)
					isPrime[j] = false;
			}
		}

		return primeList;
	}
}


public static class MyIO
{
	private static string[] args = null;
	private static int num = -1;
	private static int used = -1;

	private static string getArg()
	{
		if(used == num)
		{
			args = Console.ReadLine().Split(' ');
			num = args.Length;
			used = 0;
		}
		return args[used++];
	}

	public static int GetInt() => int.Parse(getArg());
	public static long GetLong() => long.Parse(getArg());
	public static double GetDouble() => double.Parse(getArg());
	public static string GetString() => getArg();
	public static char GetChar() => getArg()[0];
	public static int[] GetInts(int N) => Enumerable.Range(0, N).Select(_ => GetInt()).ToArray();
	public static long[] GetLongs(int N) => Enumerable.Range(0, N).Select(_ => GetLong()).ToArray();
	public static double[] GetDoubles(int N) => Enumerable.Range(0, N).Select(_ => GetDouble()).ToArray();
	public static string[] GetStrings(int N) => Enumerable.Range(0, N).Select(_ => GetString()).ToArray();
}