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 <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (int)(n); i++)
#define REP(i,n) for (int i = 1; i < (int)(n); i++)
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define vout(x) rep(i,x.size()) cout << x[i] << " "
template<class T> bool chmin(T &a, T b) {if(a>b) {a=b;return 1;}return 0;}
template<class T> bool chmax(T &a, T b) {if(a<b) {a=b;return 1;}return 0;}
using namespace std;
using vint = vector<int>;
using vvint = vector<vector<int>>;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using P = pair<int,int>;
const int inf = 1e9;
const ll inf_l = 1e18;
const int MAX = 1e5;
const int mod = 1e9+7;

//素因数分解
vector<pair<ll,int>> factorize(ll n) {
    vector<pair<ll,int>> res;
    ll N = n;
    for (ll i = 2; i * i <= N; i++) {
        if (n % i != 0) continue;
        res.emplace_back(i,0);
        while (n % i == 0) {
            n /= i;
            res.back().second++;
        } 
    }
    if (n != 1) res.emplace_back(n,1);
    return res;
}

int main() {
    ll n; cin >> n;
    auto p = factorize(n);
    int ans = 0;
    for (auto x : p) {
        int a = x.second;
        int n = 1;
        while (n <= a) {
            a -= n;
            ans++; n++;
        }
    }
    cout << ans << endl;
}