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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#include<numeric>
#define int long long
#define ld long double
#define all(x) x.begin(),x.end()
const int maxn=1000000;
#define what_is(x) cerr << #x << " is " << x << endl;
const int MOD = 1e9 + 7;
using namespace std;
string abc="abcdefghijklmnopqrstuvwxyz";
int powmod(int a, int b, int MOD = MOD){
    if(!a and !b)    return 1ll;
    if(!a)    return a;
    int res = 1ll;
    b %= (MOD - 1ll), a %= MOD;
    while(b){
        if(b%2 == 1)    res = (res * a) % MOD;
        a = (a * a) % MOD;
        b /= 2;
    }
    return res;
}
int popcount(uint32_t n)
{
    n = n - ((n >> 1) & 0x55555555);
    n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
    return ((n + (n >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
}
bool check(int n) {return (((double)sqrt(n)-floor((double)sqrt(n)))==0);}
vector<bool> sieve(maxn + 1, true);
void fillprime() {
    sieve[0] = sieve[1] = false;
    for (int i = 2; i <= maxn; ++i) {
        if (sieve[i] && (int) i * i <= maxn) {
            for (int j = i * i; j <= maxn; j += i) {
                sieve[j] = false;
            }
        }
    }
}
const int nn=1e5+5;
int result[100005];
void solve() {
  
  
  int x,y;
  cin>>x>>y;
  int cnt=1;
  //int xx=min(x,y);
  for(int i=1;i*i<min(x,y);++i){
      if(x%i==0 and y%i==0 and sieve[i]) cnt++;
  }
  cout<<cnt;
   
    
}
int32_t main() {
    ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
    fillprime();
    //int t;cin>>t;while(t--)
    solve();
}