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: Runtime Error)

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

class D
{
	static void Main()
	{
		var h = Console.ReadLine().Split().Select(long.Parse).ToArray();
		var g = Gcd(h[0], h[1]);

		Console.WriteLine(g == 1 ? 1 : PrimesF(2, g).Count(p => g % p == 0) + 1);
	}

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

	static IEnumerable<long> Range(long m, long M) { for (var i = m; i <= M; i++) yield return i; }
	static long[] PrimesF(long m, long M)
	{
		var b = PrimeFlags(M);
		return Range(m, M).Where(i => b[i]).ToArray();
	}
	static bool[] PrimeFlags(long M)
	{
		var rM = (long)Math.Sqrt(M);
		var b = new bool[M + 1]; b[2] = true;
		for (long i = 3; i <= M; i += 2) b[i] = true;
		for (long p = 3; p <= rM; p++) if (b[p]) for (var i = p * p; i <= M; i += 2 * p) b[i] = false;
		return b;
	}
}