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>

#include <iostream>
#include <algorithm>

#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <stack>
#include <queue>
#include <deque>
#include <cstring>
#include <string>
#include <utility>
#include <array>
#include <complex>
#include <valarray>

#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <chrono>
#include <random>
#include <numeric>

#include<memory>


using namespace std;
#define int long long

typedef long long ll;
typedef unsigned long long ull;
//typedef unsigned __int128 HASH;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef pair<ull, ull> pullull;
typedef pair<ll,int> plli;
typedef pair<double,int> pdi;
typedef pair<long double, int> pdbi;
typedef pair<int,pii> pipii;
typedef pair<ll,pll> plpll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<pii> vpii;
typedef vector<vector<int>> mat;

#define rep(i,n) for (int i=0;i<(n);i++)
#define rep2(i,a,b) for (int i=(a);i<(b);i++)
#define rrep(i,n) for (int i=(n);i>0;i--)
#define rrep2(i,a,b) for (int i=(a);i>b;i--)
#define pb push_back
#define fi first
#define se second
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()

const ll hmod1 = 999999937;
const ll hmod2 = 1000000000 + 9;
const int INF = 1<<30;
const ll INFLL = 1LL<<62;
const long double EPS = 1e-12;
const ll mod = 1000000000 + 7;
//const ll mod = 998244353;
const int dx4[4] = {1, 0, -1, 0};
const int dy4[4] = {0, 1, 0, -1};
const int dx8[8] = {1, 1, 1, 0, 0, -1, -1, -1};
const int dy8[8] = {0, 1, -1, 1, -1, 0, 1, -1};
const long double pi = 3.141592653589793;

#define addm(X, Y) (X) = ((X) + ((Y) % mod) + mod) % mod
#define inside(y, x, h, w) (0 <= (y) && (y) < (h) && 0 <= (x) && (x) < (w)) ? true : false

//debug
#define DEBUG

#define DUMPOUT cerr

#ifdef DEBUG
#define dump(...) DUMPOUT<<#__VA_ARGS__<<" :["<<__FUNCTION__<<":"<<__LINE__<<"]"<<endl; DUMPOUT<<"    "; dump_func(__VA_ARGS__)
#else
#define dump(...)
#endif

void dump_func() {DUMPOUT << endl;};

template <class Head, class... Tail> void dump_func(Head&& head, Tail&&... tail) {
    DUMPOUT << head;
    if (sizeof...(Tail) == 0) DUMPOUT << " ";
    else DUMPOUT << ", ";
    dump_func(std::move(tail)...);
}

//ostream
template<typename T> ostream& operator << (ostream& os, vector<T>& vec) {
    os << "["; for (int i = 0; i<vec.size(); i++) os << vec[i] << (i + 1 == vec.size() ? "" : ", "); os << "]";
    return os;
}

template<typename T, typename U> ostream& operator << (ostream& os, pair<T, U>& pair_var) {
    os << "(" << pair_var.first << ", " << pair_var.second << ")";
    return os;
}

template<typename T, typename U> ostream& operator << (ostream& os, map<T, U>& map_var) {
    os << "[";
    for (auto itr = map_var.begin(); itr != map_var.end(); itr++) {
        os << "(" << itr->first << ", " << itr->second << ")"; itr++;  if(itr != map_var.end()) os << ", "; itr--;
    }
    os << "]";
    return os;
}

template<typename T> ostream& operator << (ostream& os, set<T>& set_var) {
    os << "[";
    for (auto itr = set_var.begin(); itr != set_var.end(); itr++) {
        os << *itr; ++itr; if(itr != set_var.end()) os << ", "; itr--;
    }
    os << "]";
    return os;
}

int a, b;

long long  gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
}

map<int, int> primefactor(int x) {
    map<int, int> fac;
    for (int i = 2; i * i <= x; i++){
        while (x % i == 0) {
            fac[i]++;
            x = x / i;
        }
    }
    if (x != 1) fac[x] += 1;
    return fac;
}


signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> a >> b;
  int g = gcd(a, b);
  map<int, int> p = primefactor(g);
  int ans = 1 + p.size();
  cout << ans << endl;
}