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: Accepted)

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

class D
{
	static void Main()
	{
		var h = Console.ReadLine().Split().Select(long.Parse).ToArray();
		Console.WriteLine(Factorize(Gcd(h[0], h[1])).Distinct().Count() + 1);
	}

	static long Gcd(long x, long y) { for (long r; (r = x % y) > 0; x = y, y = r) ; return y; }

	static long[] Factorize(long n)
	{
		long x = 2;
		var r = new List<long>();
		while (n % x == 0) { r.Add(x); n /= x; }
		for (x++; x * x <= n && n > 1; x += 2)
			while (n % x == 0) { r.Add(x); n /= x; }
		if (n > 1) r.Add(n);
		return r.ToArray();
	}
}