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>
#define REP(i, n) for(long long i=0; i<n; i++)
#define REPR(i, n) for(long long i=n-1; i>=0; i--)
#define FOR(i, m, n) for(long long i=m; i<=n; i++)
#define FORR(i, m, n) for(long long i=m; i>=n; i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define VSORTR(v) sort(v.rbegin(), v.rend());
#define ALL(v) (v).begin(),(v).end()
#define FIN ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

using namespace std;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using P = pair<ll, ll>;
const ll mod = 1e9+7;
const ll inf = 1e15;

map<ll,ll> primes(ll n){
    map<ll, ll> r;
    for(int i=2; i*i<=n; i++){ 
        while(n % i ==0){
            r[i]++;
            n /= i;
        }
    }

    if(n != 1) r[n] = 1;
    return r;
}

int main(){FIN
    
    // double a,b ;cin>>a>>b;
    // cout<<ll(a*b)<<endl;
    ll n; cin>>n;
    ll ans=0;
    map<ll,ll> mp = primes(n);
    for(auto p: mp){
        ll q = p.second;
        ll idx=1;
        while(1){
            q-= idx;
            ans ++ ;
            if(q<=idx) break;
            idx++;
        }
    }
    cout<<ans<<endl;
    return 0;
}