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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define me BOSS
#define bug() cout<<"AIkahne aise"<<endl
ll A[5000001],bin[5000001];
vector<ll>v;
map<int,int>myMap;
set<int>mySet;
vector<pair<ll,ll> >vp;
/* Precode Start Here *\


    Integer  to Binary

void bti(ll n)
{
ll k=1;
  while(n!=0)
    {
    bin[k++]=n%2;
    n/=2;
    }
}

     Sieve
void sieve()
{
    for(ll i=3; i<=2000; i+=2)
    {
        for(ll j=i*i; j<=4000000; j+=i)
            A[j]=1;
    }
    v.push_back(2);
    for(ll i=3; i<=4000000; i+=2)
        if(A[i]==0)
            v.push_back(i);
}

     Power Function Modulo P

ll modPow(ll a, ll x, ll p) {
    //calculates a^x mod p in logarithmic time.
    long res = 1;
    while(x > 0) {
        if( x % 2 != 0) {
            res = (res * a) % p;
        }
        a = (a * a) % p;
        x /= 2;
    }
    return res;
}


   factorial

ll fact(ll n)
{
    ll ans=1;
    for(ll i=1;i<=n;i++)
    {
        ans=(ans*i)%mod;
    }
    return ans;
}
     Calculating nCr using Fermats law Theorem

ll modInverse(ll a, ll p) {
    //calculates the modular multiplicative of a mod m.
    //(assuming p is prime).
    return modPow(a, p-2, p);
}


ll modBinomial(ll n, ll k, ll p) {
// calculates C(n,k) mod p (assuming p is prime).

    ll numerator = 1; // n * (n-1) * ... * (n-k+1)
    for (ll i=0; i<k; i++) {
        numerator = (numerator * (n-i) ) % p;
    }

    ll denominator = 1; // k!
    for (ll i=1; i<=k; i++) {
        denominator = (denominator * i) % p;
    }

    // numerator / denominator mod p.
    return ( numerator* modInverse(denominator,p) ) % p;
}


/* Precode Ends */
void sieve()
{
    for(ll i=3; i<=2000; i+=2)
    {
        for(ll j=i*i; j<=4000000; j+=i)
            A[j]=1;
    }
    v.push_back(2);
    for(ll i=3; i<=4000000; i+=2)
        if(A[i]==0)
            v.push_back(i);
}
ll gcd(ll a,ll b)
{
    while(b!=0)
    {
        ll t=b;
        b=a%b;
        a=t;
    }
    return a;
}
int main()
{
    sieve();
   ll a,b,cnt=0;
   cin>>a>>b;
   ll d=gcd(max(a,b),min(a,b));
   for(ll i=0;v[i]<=sqrt(d);i++)
   {
       if(d%v[i]==0)
       {
           while(d%v[i]==0)
             {
                 d/=v[i];
             }
           cnt++;
       }
   }
   if(d!=1)
   {
       cnt++;
   }
   cout<<cnt+1<<endl;
return 0;
}