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>
typedef long long ll;
using namespace std;
using pint = pair<int, int>;
using pll = pair<ll,ll>;

const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<(n);++i)
#define COUT(x) cout<<(x)<<"\n"

map<ll,int> divisor;
void test(ll x){
    ll num = sqrt(x);
    for(int i=1;i<num+1;i++){
        if(x%i==0){
            divisor[x/i]++;
            divisor[i]++;
        }
    }
}
bool prime(ll n){
  int n_root = sqrt(n);
  if(n==1)return 1;
  else if(n==2)return true;
  else if (n%2==0)return false;
  for(int i=3;i<n_root+1;i+=2){
    if(n%i==0)return false;
  }
  return true;
}

int main(){
    ll a,b;
    cin >> a >> b;
    if(a==1||b==1){
        COUT(1);
        return 0;
    }
    test(a);test(b);
    vector<ll>vec;
    for(const auto &x:divisor){
        if(x.second==2){
            vec.push_back(x.first);
        }
    }
    int count = 0;
    for(ll x:vec){
        if(prime(x)==1){
            count++;
        }
    }
    COUT(count);
    return 0;
}