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.IO;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCorder
{
	public class Program
	{
		static void Main(string[] args)
		{
			Scanner cin = new Scanner();
			long a = cin.Long();
			long b = cin.Long();

			long gcd = GCD(a, b);
			var gp = GetPrimeFactor(gcd);
			if (gp.Count == 0) {
				gp.Add(gcd);
			}

			gp.Add(1);

			Console.WriteLine(gp.Count());
		}

		public static HashSet<long> GetPrimeFactor(long value)
		{
			long max = (long)Math.Ceiling(Math.Sqrt(value));
			var factors = new HashSet<long>();
			for (int i = 2; i <= max; ++i) {
				if (value % i == 0) {
					factors.Add(i);
					while (value % i == 0) {
						value /= i;
					}
				}
			}

			return factors;
		}

		public static long GCD(long a, long b)
		{
			if (a > b) {
				if (b == 0) {
					return a;
				}

				return GCD(b, a % b);
			} else {
				if (a == 0) {
					return b;
				}

				return GCD(a, b % a);
			}
		}
	}

	public class Scanner
	{
		private readonly char[] delimiter_ = new char[] { ' ' };
		private string[] buf_;
		private int index_;

		public Scanner()
		{
			buf_ = new string[0];
			index_ = 0;
		}

		public string Next()
		{
			if (index_ < buf_.Length) {
				return buf_[index_++];
			}

			string st = Console.ReadLine();
			while (st == "") {
				st = Console.ReadLine();
			}

			buf_ = st.Split(delimiter_, StringSplitOptions.RemoveEmptyEntries);
			if (buf_.Length == 0) {
				return Next();
			}

			index_ = 0;
			return buf_[index_++];
		}

		public int Int() => int.Parse(Next());
		public long Long() => long.Parse(Next());
		public double Double() => double.Parse(Next());

		public int[] ArrayInt(int N, int add = 0)
		{
			int[] Array = new int[N];
			for (int i = 0; i < N; i++) {
				Array[i] = Int() + add;
			}
			return Array;
		}

		public long[] ArrayLong(int N, long add = 0)
		{
			long[] Array = new long[N];
			for (int i = 0; i < N; i++) {
				Array[i] = Long() + add;
			}
			return Array;
		}

		public double[] ArrayDouble(int N, double add = 0)
		{
			double[] Array = new double[N];
			for (int i = 0; i < N; i++) {
				Array[i] = Double() + add;
			}
			return Array;
		}
	}
}