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 <cmath>

#define rep(i, n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;

ll gcd(ll a, ll b)
{
  if( a < b ) 
  {
    a ^= b;
    b ^= a;
    a ^= b;
  }
  return b ? gcd(b, a % b) : a;
}

void calcCd(ll x, vector<ll>& cd)
{
  cd.emplace_back(1);
  if(x == 1)
  {
    return;
  }
  for(ll i = 2; i <= sqrt(x); ++i)
  {
    bool push = false;
    while(x%i == 0)
    {
      x /= i;
      push = true;
    }
    if(push) cd.emplace_back(i);
  }
  if(x != 1)cd.emplace_back(x);
}

int main()
{
  ll a, b;
  cin >> a >> b;
  ll gcdab = gcd(a, b);
  vector<ll> cd;
  calcCd(gcdab, cd);

  cout << cd.size() << endl;
  return 0;
}