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.Generic;
using System.Linq;

namespace AtCoder
{
	class Program
	{
		static void Main(string[] args)
		{
			long[] arr = Console.ReadLine().Trim().Split(' ').Select(hogehoge => long.Parse(hogehoge)).ToArray<long>();

			long a = arr[0];
			long b = arr[1];



			long mod = a;

			while (true)
			{
				mod = a % b;
				if (mod == 0) break;
				a = b;
				b = mod;
			}
			int cnt = 0;

			for (int i = 1; i < Math.Sqrt(b) + 1; i++)
			{
				if (b % i == 0)
				{
					cnt++;
					b /= i;
				}
			}
			if (b != 0) cnt++;

			Console.WriteLine(cnt);
		}



	}

}