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)))
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 G=gcd(a,b);
  bitset<1000000> A;
  int ans=1;
  if(G!=1){
    while(1){
      if(G%2==0){
        G/=2;
        A.set(1);
        }
      else
        break;
      }
    int i=3;
    int g=G;
    while(i<=G){
      if(G%i==0){
        G/=i;
        A.set(i-1);
        }
      else{
        i+=2;
        if(g<i*i)
          break;
        }
      }
    ans+=A.count();
    if(ans==1)
      ans++;
    }
  cout<<ans<<endl;
  }