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: Time Limit Exceeded)

using System;
using System.Text;
using System.Collections.Generic;
using System.Linq;
using E = System.Linq.Enumerable;
 
class Program
 {
  
        static long[] a;
        static long N;
        static long K;
        
        static void Main(string[] args)
        {
          //入力を受け取る
          //long N = int.Parse(Console.ReadLine());
          long[] AB = Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToArray();
          long A = AB[0];
          long B = AB[1];
          long mld = Gcd(A, B);;
          


          var count = 1;
          if(mld%2 == 0){
            while(mld%2 == 0){
              mld = mld/2;
            }
            count++;
          }
          
          for(int i= 3;i < (mld+1)/2 ;i+=2){
            if(mld%i == 0){
              while(mld%i == 0){
                mld = mld/i;
              }
              count++;
            }
          }
          
          if(mld!=1){
            count++;
          }
          
          //結果を出力
          Console.WriteLine(count);
        }
  
          // ユークリッドの互除法 
        static long Gcd(long a, long b)
        {
            return b == 0 ? a : Gcd(b, a % b);
        }
 }