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: Accepted)

#include<bits/stdc++.h>
using namespace std;
//long longをllと省略
#define ll long long
//ユークリッドの互除法
ll gcd(ll a, ll b){
  ll t = a%b;
  while(t > 0){
    a = b;
    b = t;
    t  = a%b;
  }
  return b;
}
    
int main(){
  ll A,B;
  cin >> A >>B;
  //ansは素因数分解で出た数字の数＋1
  int ans=1;
  ll N =gcd(A,B);
  //素因数分解を実施(i*iは、素因数の小さい方は√N以下であるといういう性質から)
  for(ll i = 2;i*i <= N;i++){
    if(N%i==0){
      ans++;
      while(N%i==0){
        N/=i;
      }
    }
  }
  //素因数の大きいほうが√N以上だった場合
  if(N>1)ans++;
  cout << ans << endl;
  
  
  
}