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<iostream>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;


#if 1
ll gcd(ll x,ll y){
  if(x<y) swap(x,y);
  //xの方が常に大きい
  while(y>0){
    ll r=x%y;
    x=y;
    y=r;
  }
  return x;
}

ll lcm(ll x,ll y){
  return ll(x*y/gcd(x,y));
}

//vに素因数分解した要素突っ込む
void Prime_factorize(vector<ll>& v,ll n){
  if(n<=1) return;
  ll l=ll(sqrt(n));
  for(ll i=2;i<l+1;i++){
    if(n%i==0){
      Prime_factorize(v,i);Prime_factorize(v,ll(n/i));return;
    }
  }
  v.push_back(n);return;
}

void erase_same(vector<ll>& v){
  sort(v.begin(),v.end());
  //隣接している重複した要素を除いた要素を先頭に集めるのが、unique関数
  //集めた後の返り値は重複を除いた範囲の末尾の次のイテレータになる(つまり、このイテレータ以降の要素はゴミ)
  v.erase(unique(v.begin(),v.end()),v.end());
}

#endif
//n個の場合のlcmも2個で比べて順に適用していくだけ
int main(){
  ll a,b;cin >> a >> b;
  ll k=gcd(a,b);
  vector<ll> fac;
  Prime_factorize(fac,k);
  erase_same(fac);
  cout << fac.size()+1 << endl;
}