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 AtCoder
{
    class Abc142D
    {
        public static void Main()
        {
            var ab = Console.ReadLine().Split(' ').Select(s => long.Parse(s)).ToArray();
            var a = ab[0];
            var b = ab[1];

            var c = Gcd(a, b);

            var primeFactors = new List<long>();

            if (c % 2 == 0)
            {
                primeFactors.Add(2);
                while (c % 2 == 0)
                    c /= 2;
            }

            for (long i = 3; i * i <= c; i += 2)
            {
                if (c % i != 0)
                    continue;
                primeFactors.Add(i);
                while (c % i == 0)
                    c /= i;
            }

            if (c != 1)
                primeFactors.Add(c);

            Console.WriteLine(primeFactors.Count + 1);
        }

        public static long Gcd(long a, long b)
        {
            if (a > b)
            {
                var c = a;
                a = b;
                b = c;
            }

            while (a > 0)
            {
                var r = b % a;
                b = a;
                a = r;
            }
            return b;
        }
    }
}