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: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Numerics;

namespace atcoder
{
    class Program
    {
        static void Main(string[] args)
        {
            var ab = Console.ReadLine().Split(' ').Select(num => long.Parse(num)).ToArray();
            var gcd = GCD(ab[0], ab[1]);

            var pCnt = 1;
            for (var i = 2; i <= Math.Sqrt(gcd); i++)
            {
                if (gcd % i == 0)
                {
                    pCnt++;
                    while (gcd % i == 0) gcd /= i;
                }
            }

            if (gcd != 1) pCnt++;

            Console.WriteLine(pCnt);

            long GCD(long x, long y)
            {
                var max = Math.Max(x, y);
                var min = Math.Min(x, y);

                var remain = max % min;
                if (remain == 0)
                {
                    return min;
                }
                else
                {
                    return GCD(min, remain);
                }
            }
        }
    }
}