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<iostream>
#include<vector>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<queue>
#include<map>

using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vector<int> >;
using pii = pair<int, int>;
using ppi = pair<pii, int>;

int f_dir[2][4] = {{-1, 0, 1, 0}, {0, 1, 0, -1}}; //↑→↓←
int e_dir[2][8] = {{-1, -1, -1, 0, 0, 1, 1, 1}, {-1, 0, 1, -1, 1, -1, 0, 1}};
bool is_in_field(int y, int x, int h, int w){
  return (y >= 0 && y < h && x >= 0 && x < w);
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int ti = clock();
  // start-----------------------------------------------
  ll n;
  cin >> n;
  vi isp(sqrt(n)+1, 1);
  for(int i = 2; i <= sqrt(n); i++){
    if(!isp[i]) continue;
    int k = 2;
    while(i * k <= sqrt(n)){
      isp[i * k] = 0;
      k++;
    }
  }
  map<ll, ll> mp;
  while(n > 1){
    bool f = false;
    for(int i = 2; i < sqrt(n); i++){
      if(!(n%i)){
        f = true;
        mp[i]++;
        n /= i;
        i--;
      }
    }
    if(!f && n != 1){
      mp[n]++;
      n /= n;
    }
  }
  ll ans = 0;
  for(auto i : mp){
    // cout << i.first << ' ' << i.second << endl;
    ans += (-1 + sqrt(1 + 8 * i.second)) / 2;
  }
  cout << ans << endl;

  // end-----------------------------------------------
  // cerr << 1.0 * (clock() - ti) / CLOCKS_PER_SEC << endl;
}