Problem:
Score : 400 points

Problem Statement

Given is a positive integer N. Consider repeatedly applying the operation below on N:

First, choose a positive integer z satisfying all of the conditions below:

z can be represented as z=p^e, where p is a prime number and e is a positive integer;

z divides N;

z is different from all integers chosen in previous operations.

Then, replace N with N/z.

Find the maximum number of times the operation can be applied.

Constraints

All values in input are integers.

1 \leq N \leq 10^{12}

Input

Input is given from Standard Input in the following format:

N

Output

Print the maximum number of times the operation can be applied.

Sample Input 1

24

Sample Output 1

3

We can apply the operation three times by, for example, making the following choices:

Choose z=2 (=2^1). (Now we have N=12.)

Choose z=3 (=3^1). (Now we have N=4.)

Choose z=4 (=2^2). (Now we have N=1.)

Sample Input 2

1

Sample Output 2

0

We cannot apply the operation at all.

Sample Input 3

64

Sample Output 3

3

We can apply the operation three times by, for example, making the following choices:

Choose z=2 (=2^1). (Now we have N=32.)

Choose z=4 (=2^2). (Now we have N=8.)

Choose z=8 (=2^3). (Now we have N=1.)

Sample Input 4

1000000007

Sample Output 4

1

We can apply the operation once by, for example, making the following choice:

z=1000000007 (=1000000007^1). (Now we have N=1.)

Sample Input 5

997764507000

Sample Output 5

7

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

class Eratosthenes{ //NMAX変更しておｋ
    private:
        const static int NMAX = 2000010;
        vector<bool> prime;
    
    public:
        Eratosthenes(){
            prime.resize(NMAX,true);//一旦全部trueとする(true=素数)
            prime[0] = false;
            prime[1] = false;
            for(int i=2;i*i<=NMAX;i++){
                if(prime[i]){
                    for(int j=2;i*j<=NMAX;j++) prime[i*j] = false;
                }
            }
        }
        inline vector<bool> isPrime(){ return prime; }

};
#define debug(x) cerr << x << " = "(x) << endl;
#define ll long long
#define rep(i,x,y) for(int i=x;i<y;i++)
#define rel(i,x,y) for(int i=x-1;i>=y;i--)
#define all(x) x.begin(),x.end()

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    ll n; cin >> n;
    ll ni = n;
    ll ans = 0;
    vector<int> sosu;
    bool sos = false;
    rep(i,2,sqrt(n)+100){
        if(ni<i) break;
        bool ok = true;
        rep(j,0,sosu.size()){
            if(i % sosu[j]==0) ok = false;
        }
        if(!ok) continue;
        sosu.push_back(i);
        ll ti = i;
        while(ni%ti==0){
            sos = true;
            ans++;
            ni /= ti;
            ti *= i;
            //cout << "aa" << endl;
        }
    }
    if(n == 1) cout << 0 << endl;
    else if(sos)cout << ans << endl;
    else cout << 1 << endl;
}