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 <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <unordered_map>
#include <vector>
#include <string.h>
#include <set>

using namespace std;

#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl

#define rep(i, n) for(int i = 0; i < n; ++i)
#define all(s) s.begin(), s.end()

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef tuple<ll, ll, ll> triple;
typedef double D;

const ll INF = 1e9;
const ll MOD = 1000000007;  // 1e9 + 7

ll powmod(ll x, ll n) { // like pow(x, n)
  ll r = 1;
  while (n) {
    if (n & 1) {
      r = r * x % MOD;
    }
    x = x * x % MOD;
    n >>= 1;
  }
  return r;
}

int gcd(int a, int b) {
  if (a < b) { swap(a, b); }
  if (b == 0) { return a; }
  return gcd(b, a % b);
}

int main(int argc, char** argv) {
  ll A, B;
  cin >> A >> B;

  set<ll> a_candidates;
  a_candidates.insert(1);
  for (ll i = 2; i * i <= A; ++i) { // i is the divisor
    while ((A % i) == 0) {
      a_candidates.insert(i);
      A /= i;
    }
  }
  if (A > 1) {
    a_candidates.insert(A);
  }
  // cout << "a_candidates: " << endl;
  // for (auto x : a_candidates) {
  //   cout << x << endl;
  // }
  set<ll> b_candidates;
  b_candidates.insert(1);
  for (ll i = 2; i * i <= B; ++i) { // i is the divisor
    while ((B % i) == 0) {
      b_candidates.insert(i);
      B /= i;
    }
  }
  if (B > 1) {
    b_candidates.insert(B);
  }
  // cout << "b_candidates: " << endl;
  // for (auto x : b_candidates) {
  //   cout << x << endl;
  // }

  set<ll> candidates;
  for (auto x : a_candidates) {
    if (b_candidates.find(x) != b_candidates.end()) {
      candidates.insert(x);
    }
  }
  // cout << "candidates: " << endl;
  // for (auto x : candidates) {
  //   cout << x << endl;
  // }
  // cout << "answer:" << endl;

  cout << candidates.size() << endl;
}