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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pll = pair<ll, ll>;
using vl = vector<ll>;

/* short */
#define pb push_back
#define mp make_pair
#define Fi first
#define Se second
#define ALL(v) begin(v), end(v)
#define RALL(v) rbegin(v), rend(v)

/* REPmacro */
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define FORR(i, a, b) for(int i = (a); i >= (b); i--)
#define REP(i, n) for(int i = 0; i < (n); i++)
#define FOREACH(x, a) for(auto x : a)

/* exchange */
#define CHMIN(a, b) (a) = min((ll)(a), (ll)(b))
#define CHMAX(a, b) (a) = max((ll)(a), (ll)(b))

/* function */
#define IN(x) cin >> x
#define DEBUG(x) cerr << (x) << " "
#define LN() cerr << "\n"
#define PRINT(x) cout << (x) << endl
#define BR cout << endl

/* const */
const int ARRAY = 100005;
const int INF = 1001001001; // 10^9
const ll LINF = 1001001001001001001; // 10^18
const int MOD = 1e9 + 7;

ll N = 0;
ll A, B;
ll t;
ll ret = 0;

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

void input() {
  IN(A);
  IN(B);
}

void solve() {
  ll G = gcd(A, B);
  vector<ll> arr;
  arr.pb(-1);
  for (ll i = 1; i*i <= G; i++) {
    arr.pb(i);
  }
  // DEBUG(arr.size());
  FOR(i, 2, arr.size()) {
    if (arr[i] == -1) continue;
    if (G % i == 0) {
      // DEBUG(i);
      ret += 1;
      for (ll j = i; j < arr.size(); j += i) {
        arr[j] = -1;
      }
    }
  }
  PRINT(ret+1);
}

int main(void){
  input();
  solve();
}