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>
#include <algorithm>

using namespace std;
typedef long long ll;
typedef pair<long long,long long> P;
#define rep(n) for(ll i=0; i<n; i++)
#define MAX_V 100001
const ll INF = 1e+9;
const ll MOD = 1000000007;

vector<ll> v;

void init()
{
  for (ll i = 0; i < 41; i++)
  {
    v.push_back(i*(i+1)/2);
  }
}

P exponent(ll n, ll p)
{
  if(n%p==0)
  {
    P next = exponent(n/p, p);
    return make_pair(next.first, ++next.second);
  } else {
    return make_pair(n, 0);
  }
}


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

  init();
  /*
  for(auto p:v){
    cout << p << " ";
  }
  cout << endl;
  */
  
  if(n==1){
    cout << 0 << endl;
    return 0;
  }


  ll cur = 1;
  ll ans = 0;
  while(n>1)
  {
    cur++;
    if(n%cur!=0) {
      if(cur*cur>n) 
      {
        ans++;
        break;
      }
      continue;
    }
    P tmp = exponent(n, cur);
    n = tmp.first;
    auto itr = upper_bound(v.begin(), v.end(), tmp.second);
    /*
    cout << n << " ";
    cout << cur << " ";
    cout << tmp.second << " " << itr-v.begin()-1 << endl;
    */
    ans += itr-v.begin()-1;
  }

  cout << ans << endl;
  

  return 0;
}