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;
using System.Text;
using System.Threading.Tasks;

namespace ABC142_D
{
    class Program
    {
        static void Main(string[] args)
        {
            var ab = Console.ReadLine().Split().Select(long.Parse).ToArray();
            var a = ab[0];
            var b = ab[1];

            var ap = Factors(a).Distinct();
            var bp = Factors(b).Distinct();

            var ans = 1;
            foreach (var i in ap)
            {
                if (bp.Contains(i) && IsPrime(i)) ans++;
            }

            Console.WriteLine(ans);
        }

        public static IEnumerable<long> Factors(long n)
        {
            long i = 2;
            long tmp = n;

            while (i * i <= n)
            {
                if (tmp % i == 0)
                {
                    tmp /= i;
                    yield return i;
                }
                else
                {
                    i++;
                }
            }
            if (tmp != 1) yield return tmp;
        }

        public static bool IsPrime(long num)
        {
            if (num < 2) return false;
            else if (num == 2) return true;
            else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

            double sqrtNum = Math.Sqrt(num);
            for (int i = 3; i <= sqrtNum; i += 2)
            {
                if (num % i == 0)
                {
                    // 素数ではない
                    return false;
                }
            }

            // 素数である
            return true;
        }
    }
}