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: Time Limit Exceeded)

#include<iostream>
#include<vector>
#include<cmath>
#include<numeric>
#include <unistd.h>
using namespace std;

long gcd(long x,long y)
{
  if(y<1)
    return x;
  else
    return gcd(y, x%y);
}

int main()
{
  int A, B;
  cin >> A >> B;

  int i,j;
  long div = gcd(A,B);
  int count=0;
  for(i=2;i*i<=div;i++)
    {
      if(div % i == 0)
	{
	  count++; 
	  while(div%i==0)
	      div/=i;
        }
    }

  if(count>0)
    count++;
  cout << count+1 << endl;
  
  return 0;
}