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 rep(i,j,n) for(int i=(int)(j);i<(int)(n);i++)
#define REP(i,j,n) for(int i=(int)(j);i<=(int)(n);i++)
#define MOD 1000000007
#define int long long
#define ALL(a) (a).begin(),(a).end()
#define vi vector<int>
#define vii vector<vi>
#define pii pair<int,int>
#define priq priority_queue<int>
#define disup(A,key) distance(A.begin(),upper_bound(ALL(A),(int)(key)))
#define dislow(A,key) distance(A.begin(),lower_bound(ALL(A),(int)(key)))
#define tii tuple<int,int,int>
int gcd(int A,int B){
  if(A<B)
    swap(A,B);
  while(1){
    if(A%B==0)
      return B;
    int C=A%B;
    A=B;
    B=C;
  }
}
signed main(){
  int A,B; cin>>A>>B;
  int memo=1;
  int K=gcd(A,B);
  int ans=1;
  if(K==1)
    ans=1;
  else{
  while(K>1&&K%2==0){
    if(memo==1){
      memo=2;
      ans++;
    }
    K/=2;
  }
  int i=3;
  while(K>1&&K>=i*i){
    if(K%i==0){
      if(memo<i){
        ans++;
        memo=i;
      }
      K/=i;
    }
    else
      i+=2;
  }
  if(memo<2)
    ans++;
  if(K>1)
    ans++;
  }
  cout<<ans<<endl;
  }