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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define MOD 1000000007
#define INF 1000000010
#define EPS 1e-9
#define fst first
#define scd second

#define debug(x) cout<<x<<endl;
#define repi(i,x,n) for(int i=x;i<n;i++)
#define rep(i,n) repi(i,0,n)
#define lp(i,n) repi(i,0,n)
#define repn(i,n) for(int i=n;i>=0;i--)
#define int long long
#define endl "\n"

#define N 10000000

int prime[N+1];

void eratosthenes(){
  rep(i,N+1) prime[i]=1;
  prime[0]=prime[1]=0;
  rep(i,N+1){
    if(prime[i]){
      for(int j=i+i;j<N+1;j+=i) prime[j]=0;
    }
  }
}

signed main(){
  cin.tie(0);	
  ios::sync_with_stdio(false);
  eratosthenes();
  int a,b;
  cin>>a>>b;
  int ans=1;
  rep(i,N+1){
    if(prime[i]){
      if(a%i==0 && b%i==0){
	ans++;
	a/=i;
	b/=i;
	//	cout<<i<<endl;
      }
    }
  }
  if(ans==1 && a==b && a!=1) ans++;
  cout<<ans<<endl;

  return 0;
}