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 <bits/stdc++.h>
#define REP(i, n) for(int i = 0;i < n;i++)
#define VSORT(v) sort(v.begin(), v.end())
#define VRSORT(v) sort(v.rbegin(), v.rend())
#define ll long long
using namespace std;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> PP;
typedef pair<ll, LP> LPP;
//typedef vector<unsigned int>vec;
typedef vector<ll>vec;
typedef vector<vec> mat;
typedef vector<vector<int>> Graph;

const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
const int INF = 1000000000;
const ll LINF = 1000000000000000000;//1e18
const ll  MOD = 1000000007;
const double PI = acos(-1.0);
const double EPS = 1e-10;

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; }
template<class T> inline void add(T &a, T b){a = ((a+b) % MOD + MOD) % MOD;};

int gcd(int a, int b){
	if(!a) return b;
	return gcd(b%a,a);
}
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int A, B;
    cin >> A >> B;
    
    int C = gcd(A, B);
    
    vector<int> yakusu(100000);
    int count = 0;
    for(int i = 1; i <= C; i++ ){
        if( C % i == 0 ){
        yakusu.at(count) = i;
        count++;
        }
    }
    
    for( int i = 1; i < count; i++ ){
        if( yakusu.at(i) == -1 ) continue;
        for( int j = i+1; j < count; j++ ){
        if( yakusu.at(j) == -1 ) continue;
        if( yakusu.at(j) % yakusu.at(i) == 0 ){
        yakusu.at(j) = -1;
        }
        }
    }
  
  int ans = 0;
  for( int i = 0; i < count; i++ ){
    if( yakusu.at(i) != -1 ){
      ans++;
    }
  }
  cout << ans << endl;

}