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: Memory Limit Exceeded)

#include <iostream>
#include <cstring>
#include <map>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <set>
#include <vector>
#include <queue>
#include <list>
#include <numeric>
#include <stdio.h>
#include <string>
#include <cstdlib>
#include <math.h>
#include <stack>
#include <climits>
#include <bitset>
#include <utility>


using namespace std;
typedef long long ll;

//最小公倍数
ll gcd(ll x, ll y) {
  ll tmp = 0;
  if (x < y){
    tmp=x;
    x=y;
    y=tmp;
  }
  while (y > 0) {
    ll r = x % y;
    x = y;
    y = r;
  }
  return x;
}

//最大公倍数
ll lcm(ll x,ll y){
  return x/gcd(x,y)*y;
}

//階乗
ll kaijo(ll k){
    ll sum = 1;
    for (ll i = 1; i <= k; ++i)
    {
        sum *= i;
        sum%=1000000000+7;
    }
    return sum;
}

//for(int i = ; i < ; i++){}

ll lmax(ll s,ll t){
  if(s>t){
    return s;
  }
  else{
    return t;
  }
}

ll lmin(ll s,ll t){
  if(s<t){
    return s;
  }
  else{
    return t;
  }
}
// ここから開始

int main(){
  ll a,b;
  cin>>a>>b;
  if(a>b){
    ll tmp = a;
    a = b;
    b = tmp;
  }
  //素因数分解
  vector<ll> v;
  vector<bool> prime_count(10000000000,false);
  v.push_back(1);
  ll n = a;
  for(ll i = 2; i < sqrt(n)+2; i++){
    if(a%i!=0){
      continue;
    }
    else{
      while(a%i==0){
        if(!prime_count[i]){
          prime_count[i]=true;
          v.push_back(i);
          a/=i;
        }
        else{
          a/=i;
        }
      }
    }
    
    if(a==1) break;  
  }
  if(a>1)v.push_back(a);
  // for(int i = 0; i<v.size(); i++) cout<<v[i]<<endl;
  ll cnt = 0;
  for(int i = 0; i < v.size(); i++){
    if(b%v[i]==0){
      cnt++;
    }
  }
  //debug
  
  cout<<cnt<<endl;
  return 0;
}