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 <cstdio>
#include <queue>
#include <algorithm>
#include <functional>
#include <string>
#include <queue>
#include <cmath>
#include <iomanip>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <random>
#include <unordered_map>
#include <unordered_set>
#include <numeric>
#define ll long long
#define fr(i,k,N) for(ll i=k; i<N;i++)
#define rep(i,N) for(ll i=0; i<N;i++)
#define ALL(v) v.begin(),v.end()
#define mp make_pair
using namespace std;
template <class X> void pr(X test){cout<<test<<endl;}
template <class X> void prr(X test){for(auto it:test){cout<<it<<endl;}}
template <class X> void prrr(X test){int f=0;for(auto it:test){cout<<(f++!=0?" ":"")<<it;}cout<<endl;}
template<typename X>
istream&operator>>(istream&i,vector<X>&v){for(X&x:v)i>>x;return i;}

class UF { // union find from https://github.com/kartikkukreja/blog-codes (MIT)
    // [usage]
    // auto uf = new UF(n);
    // uf->merge(l, r); // 0 <= l, r <= n-1
    // ...
public:
    int *id; // inner id
    int cnt; // number of all vartices
    int *sz; // sz[i] means size of i's set
    UF(int N)   {
        cnt = N;
        id = new int[N];
        sz = new int[N];
        for(int i=0; i<N; i++) {
            id[i] = i;
            sz[i] = 1;
        }
    }
    ~UF() {
        delete [] id;
        delete [] sz;
    }
    int find(int p) {
        int root = p;
        while (root != id[root]) root = id[root];
        while (p != root) {
            int newp = id[p];
            id[p] = root;
            p = newp;
        }
        return root;
    }
    void merge(int x, int y) {
        int i = find(x);
        int j = find(y);
        if (i == j) return;
        
        if (sz[i] < sz[j]) {
            id[i] = j;
            sz[j] += sz[i];
        } else {
            id[j] = i;
            sz[i] += sz[j];
        }
        cnt--;
    }
    bool connected(int x, int y) {
        return find(x) == find(y);
    }
    int count() {
        return cnt;
    }
};

// return 約数set 1 <= x <= n
auto divv = [](auto n) {
  unordered_set<decltype(n)> v;
  for(decltype(n) i = 1; i*i <= n; i++) {
    if(n%i == 0) {
      v.insert(i);
      if(i*i != n) v.insert(n/i);
    }
  }
  return v;
};

auto isprime = [](auto n){
  bool isPrime = true;
  for(decltype(n) i = 2; i <= sqrt(n); ++i) {
      if(n % i == 0) {
          return false;
      }
  }
  return isPrime;
};

int main() {
    ll n,k;
    cin>>n>>k;


    unordered_set<ll> divn,divk;
    divn = divv(n);
    divk = divv(k);
    ll sum=0;
    for(auto it:divk) if(divn.count(it)) if(isprime(it)) sum++;
    pr(sum);
}