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;//list用
using System.Text;//StringBuilder用
using System.Collections;//ハッシュテーブル用
using System.Linq;

class Program
{
    //Hashtable ht = new Hashtable();
    static List<int> s = new List<int>();
    static List<int>[] sl;
    static decimal result;
    static decimal N,K;
    //StringBuilder strresult= new StringBuilder("");//Append(str ○○)関数で文字を追加していく
    static List<decimal> res =new List<decimal>();

    static void Main()
    {
        //Modular mo = new Modular(0);//素数で割った余りを出せで使う。()内が初期値。★使い方：Modular型で宣言すると、その時点でのその素数で割った余りを保持する変数になる（powべき乗、Fac累乗、nCr組み合わせも関数として存在）
        //Node<> no = new Node<>();//UnionFind★Find関数で根っこの値を返す（根っこの値が一緒なら同グループ）、Unite関数で二つの木をくっつける
        //Permutation<> pe = new Permutation<>();//配列＆リストを並べ替えた場合の全ケースを出力★All関数でリスト形式で返す。
        //Treap<> tr = new Treap<>();//平衡二分木★Insert()関数で挿入、FindIndex()関数で該当する大きさの物が入るべき配列番号を返す。
        //PriorityQueue<int,int> pr = new PriorityQueue<int,int>();//優先度付キュー★Add(key,value)で格納、Dequeue()で取りだし。keyが小さいものから取り出す。
Read4();
//最大公約数出して、それの素な約数の数を……
      decimal tmp=Gcd((long)N,(long)K);
PF(tmp);
      result++;
        Console.WriteLine(result.ToString());

    }
public static void PF(decimal n)
    {
        decimal i = 2;
        decimal tmp = n;

        while (i * i <= n) //※1
        {
          a1:
            if(tmp % i == 0){
                tmp /= i;
              for(int j=0;j<res.Count;j++){if(res[j]==i){goto a1;}}
              res.Add(i);
                result++;
            }else{
                i++;
            }
        }
        if(tmp != 1) result++;//最後の素数も返す
    }
  
static long Gcd(long a, long b)
{
    var v = new[] { a, b };
    while (v[1] != 0) { v = new[] { v[1], v[0] % v[1] }; }
    return v[0];
}
  
    static void Read1()//1項
    {
        string tmp = Console.ReadLine();
        N = int.Parse(tmp);
    }

    static void Read2()//複数項（項目数Ｎ）
    {
        string[] tmp = Console.ReadLine().Split(' ');
        for (int i = 0; i < N; i++) { s.Add(int.Parse(tmp[i])); }
    }
    static void Read3()//1行複数項
    {
        string[] tmp = Console.ReadLine().Split(' ');
        for (int i = 0; i < tmp.Length; i++) { s.Add(int.Parse(tmp[i])); }
    }
    static void Read4()//NK１行
    {
        string[] tmp = Console.ReadLine().Split(' ');
        N = decimal.Parse(tmp[0]);
        K = decimal.Parse(tmp[1]);
    }
    static void Read5()//N行K項
    {


    }
}