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: Wrong Answer)

// Hail god Yato
 
#include <bits/stdc++.h> 
using namespace std;
 
#define hs ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
typedef long long ll;
const ll mod = 1000000007;
const ll INF = 1e18;
const ll MAX = 100001;
//
//
void solve(){
    ll n;
    cin>>n;
    ll ans = 0;
    for(ll i = 2; i*i <= n; i++){
        if(n%i == 0){
            // ++ans;
            ll cnt = 0;
            while(n%i == 0){
                n /= i;
                ++cnt;
            }
            ll maybe = sqrt(cnt);
            if((maybe+2)*(maybe+1) <= cnt*2)
                ans += maybe+1;
            else if(maybe*(maybe+1) <= cnt*2)
                ans += maybe;
            else
                ans += maybe-1;
        }
    }
    if(n != 1)
        ++ans;
    cout<<ans;
}
int main(){ 
        hs;
        ll t;
        t=1;
        // cin>>t;
        for (int i=1; i<=t; i++){
                //cout<<"Case #"<<i<<": ";
                solve();
         }
        return 0; 
}