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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <cmath>

using namespace std;
#define inputInt(a) int a; cin >> a; 
#define inputInt2(a,b) int a; int b; cin >> a >> b; 
#define inputInt3(a,b,c) int a; int b; int c; cin >> a >> b >> c;
#define inputLong(a) long a;cin >> a; 
#define inputIntArray(a,N) int a[N];for(int i=0;i<N;i++){cin >> a[i];}
#define inputIntArray2(a,b,N) int a[N]; int b[N]; for(int i=0;i<N;i++){cin >> a[i] >> b[i];}
#define output(answer) cout << answer << endl;
#define fN(i,N)    for(int i=0; i<N; i++)
#define fSE(i,s,e) for(int i=s; i<=e; i++)
#define ll long long
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

//互除法
ll GCD(ll a, ll b){
  return b ? GCD(b,a%b) : a;
}

ll LCM(ll a, ll b){
  return (a / GCD(a,b))*b;
}

int main()
{
  //方針
  //最大公約数を素因数分解して素因数の種類+1を求める。
  
  //input
  ll a, b;
  cin >> a >> b;
  
  //main
  int ans = 0;
  ll g = GCD(a,b);
  ll f = g;
  
  if(f%2 == 0){
    while(f%2 == 0){
      f = f/2; 
    }
    ans ++;
  }
  
  for(int i=3; i*i<=f; i+=2){//1と2は別に扱っても問題ない。
    if(f%i == 0){
      while(f%i == 0){
        f /= i; 
      }
      ans ++;
    }
  }
  if(f != 1){//f自身が素数の時
    ans ++; 
  }
  
  // output
  output(ans+1);//1の分
    
  return 0;
}