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.IO;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;

namespace AtCoder
{
	// ABC138D X
	public class ABC
	{
		static void Main(string[] args)
		{
			Console.SetOut(new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false });


			long[] vs = Console.ReadLine().Split().Select(long.Parse).ToArray();
			long A = vs[0];
			long B = vs[1];

			long[] fa = PrimeFactorization(A);
			long[] fb = PrimeFactorization(B);

			long tmp;
			List<long> pa = new List<long>();
			pa.Add(fa[0]);
			tmp = fa[0];
			for (int i = 1; i < fa.Length; i++) {
				if (tmp != fa[i]) {
					pa.Add(fa[i]);
					tmp = fa[i];
				}
			}
			List<long> pb = new List<long>();
			pb.Add(fb[0]);
			tmp = fb[0];
			for (int i = 1; i < fb.Length; i++) {
				if (tmp != fb[i]) {
					pb.Add(fb[i]);
					tmp = fb[i];
				}
			}
			int ans = 1;
			int ia = 0, ib = 0;
			while (ia < pa.Count && ib < pb.Count) {
				if (pa[ia] == pb[ib]) {
					ans++;
					ia++;
					ib++;
				} else if (pa[ia] < pb[ib]) {
					ia++;
				} else {
					ib++;
				}
			}
			Console.WriteLine(ans);


			Console.Out.Flush();
		}

		#region 素因数分解
		static long[] PrimeFactorization(long x)
		{
			if (x < 2) { return new long[] { x }; }

			List<long> ans = new List<long>();
			while (x % 2 == 0) {
				ans.Add(2);
				x /= 2;
			}
			for (int i = 3; i <= Math.Sqrt(x); i += 2) {
				while (x % i == 0) {
					ans.Add(i);
					x /= i;
				}
			}
			if (x > 1) {
				ans.Add(x);
			}
			return ans.ToArray();
		}
		#endregion

	}
}