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;

namespace D
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] AB = Console.ReadLine().Split(' ');
            int A = int.Parse(AB[0]);int B = int.Parse(AB[1]);
            int c;
            if (A < B) { c = B;B = A;A = c; }
            int wa=A,wb=B,wc;
            while(true)
            {
                wc = wa % wb;
                if (wc == 1)
                {
                    Console.WriteLine('1');
                    return;
                }
                else
                {
                    if(wc==0)
                    {
                        c = wb;
                        break;
                    }
                }
                wa = wb;wb = wc;
            }
            int cnt = 1;
            if (c % 2 == 0) { cnt++; }
            if (c % 3 == 0) { cnt++; }
            if (c > 3)
            {
                for (int i = 4; i <= c; i++)
                {
                    if (c % i == 0)
                    {
                        if (i % 2 == 0 | i % 3 == 0)
                        {
                            continue;
                        }
                        else
                        {
                            cnt++;
                        }
                    }
                }
            }
            Console.WriteLine(cnt);
        }
    }
}