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)

#include <bits/stdc++.h>
#define PREP(i, s, x) for(ll i = (s); i < (x); i++)
#define MREP(i, s, x) for(ll i = (s); i >= (x); i--)
#define REP(i, x) PREP(i, 0, x)
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
template<typename T> inline bool chmax(T &a, T b) { return ((a < b) ? (a = b, true) : (false)); }
template<typename T> inline bool chmin(T &a, T b) { return ((a > b) ? (a = b, true) : (false)); }

// variable -------------------------------------------

constexpr ll INF = (1ll << 60);
constexpr ll MOD = 1000000007ll;

// struct ---------------------------------------------

// function -------------------------------------------


// main -----------------------------------------------

int main() {
    ll n;
    cin >> n;

    vector<bool> is_prime(1000001, true);
    is_prime[0] = is_prime[1] = false;

    vector<ll> prime_table;
    REP(i, 1000001){
        if(is_prime[i] == true){
            prime_table.push_back(i);
            for(ll k = i * i; k < 1000001; k += i){
                is_prime[k] = false;
            }
        }
    }

    ll cnt = 0;
    for(ll i = 0; i < (ll)(prime_table.size()); i ++){
        ll ppp = prime_table[i];
        while(n % ppp == 0){
            n /= ppp;
            cnt ++;
            ppp *= prime_table[i];
        }
    }
    if(n != 1){
        cnt ++;
    }
    cout << cnt << endl;

    return 0;
}