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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include<iostream>
#include<cmath>
#include<vector>
using namespace std;
#define ll long long

int main() {
  ll A,B;
  cin >> A >> B;
  std::vector<ll> avec={1};
  ll na = A;
  //ll asqrt = std::floor(std::sqrt(A));
  //for (int i = 2; i<=asqrt; ++i) {
  for (int i = 2; i<=A; ++i) {
    if (na%i==0) avec.push_back(i);
    while (na%i==0) na/=i;
  }
  std::vector<ll> bvec;
  for (int j = 0; j<avec.size(); ++j)
    if (B%avec[j]==0) bvec.push_back(avec[j]);
  cout << bvec.size() << endl;
}