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: Accepted)

#include<iostream>
#include<iomanip>
#include<cmath>
#include<string>
#include<vector>
#include<list>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
#define fi first
#define se second
#define mp make_pair
#define rep(i, n) for(int i=0;i<n;++i)
#define rrep(i, n) for(int i=n;i>=0;--i)
const int inf=1e9+7;
const ll mod=1e9+7;
const ll big=1e18;
const double PI=2*asin(1);

int main() {
    ll N;
    cin>>N;
    map<ll, ll> prime;
    vector<ll> parr;
    for(ll i=2;i*i<=N;++i){
        if(N%i==0){
            parr.push_back(i);
        }
        while(N%i==0) {
            prime[i]++;
            N /= i;
        }
    }
    if(N>1){
        parr.push_back(N);
        prime[N]++;
    }
    /**
    for(int i=0;i<parr.size();++i){
        cout<<parr[i]<<" "<<prime[parr[i]]<<endl;
    }
    **/
    ll ans = 0;
    ll tmp;
    for(int i=0;i<parr.size();++i){
        for(int j=0;j<=prime[parr[i]];++j){
            if(j*(j+1)<=2*prime[parr[i]]){
                tmp = j;
            }
            else break;
        }
        ans += tmp;
    }
    cout<<ans<<endl;
}