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# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ABC142D
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] AB = Console.ReadLine().Split(' ');
            long A = long.Parse(AB[0]);
            long B = long.Parse(AB[1]);
            long[] Ayaku = new long[1000001];
            long[] Byaku = new long[1000001];
            Ayaku[1] += 1;
            Byaku[1] += 1;
            long a = 2;
            long Atemp = A;
            while (a * a <= A)
            {
                while (Atemp % a == 0)
                {
                    Ayaku[a] += 1;
                    Atemp /= a;
                }
                a += 1;
            }
            long b = 2;
            long Btemp = B;
            while (b * b <= B)
            {
                while (Btemp % b == 0)
                {
                    Byaku[b] += 1;
                    Btemp /= b;
                }
                b += 1;
            }
            long count = 0;
            for (int i = 0; i < 1000001; i++)
            {
                if (Ayaku[i] > 0 && Byaku[i] > 0)
                {
                    count += 1;
                }
            }
            if (Atemp > 1)
            {
                if (Atemp == Btemp)
                {
                    count += 1;
                }
            }
            Console.WriteLine(count);
        }
    }
}