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)

#pragma target("avx")
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> P;
typedef vector<ll> V;
typedef unordered_map<ll, ll> U_MAP;
typedef priority_queue<ll> pq;
typedef priority_queue<ll, vector<ll>, greater<ll>> rpq;
constexpr ll INF = 1e9, MOD = 1e9 + 7, ohara = 1e6 + 10;
constexpr ll LINF = 1e18;

#define rep(i, n) for (ll(i) = 0; (i) < (int)(n); (i)++)
#define rrep(i, a, b) for (ll i = (a); i < (b); i++)
#define rrrep(i, a, b) for (ll i = (a); i >= (b); i--)
#define all(v) (v).begin(), (v).end()
#define Size(n) (n).size()
#define Cout(x) cout << (x) << endl
#define doublecout(a) cout << fixed << setprecision(15) << a << endl;
#define fi first
#define se second
#define m_p make_pair
#define p_b push_back
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)

//------ Believe yourself as a genius!!!!!! ------

int dy[] = {1, 0, -1, 0};
int dx[] = {0, 1, 0, -1};
// int dy[]={-1,0,1,-1,1,-1,0,1};int dx[]={-1,-1,-1,0,0,1,1,1};
string alph("abcdefghijklmnopqrstuvwxyz"), s;
ll n, cnt, ans, a, b, c, d, tmp, m, h, w, x, y, sum, k, q;

map<int64_t, int> prime_factor(int64_t n) {
  map<int64_t, int> ret;
  for (int64_t i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if (n != 1) ret[n] = 1;
  return ret;
}

int main(void) {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(false);

  cin >> n;
  for (auto p : prime_factor(n)) {
    sum = 0;
    cnt = 1;
    while (1) {
      if (sum + cnt > p.se) break;
      sum += cnt;
      ans++;
      cnt++;
    }
  }
  Cout(ans);
  return 0;
}