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<bits/stdc++.h>
#define ll long long int
using namespace std;
int main()
  {
  ll a,b,cd[100000],cnt=1,j=0;
  cin>>a>>b;
  ll m=min(a,b);
  for(ll i=1;i<=m;i++){
    if(a%i==0 && b%i==0){
        cd[j]=i;
        ++j;
    }
  }
  for(ll i=1;i<j;i++){
    for(ll k=1;k<j;k++){
        if(i==k)
            continue;
        if(cd[i]%cd[k]==0){
            cnt--;
            break;
        }
    }
    cnt++;
  }
  cout<<cnt<<endl;
  return 0;
  }