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>
using namespace std;
typedef long long ll;
#define FOR(var, begin, end) for (int var = (begin); var <= (end); var++)
#define RFOR(var, begin, end) for (int var = (begin); var >= (end); var--)
#define REP(var, length) FOR(var, 0, length - 1)
#define RREP(var, length) RFOR(var, length - 1, 0)
#define EACH(value, var) for (auto value : var)
#define SORT(var) sort(var.begin(), var.end())
#define REVERSE(var) reverse(var.begin(), var.end())
#define RSORT(var) SORT(var); REVERSE(var)
#define OPTIMIZE_STDIO ios::sync_with_stdio(false); cin.tie(0); cout.precision(10); cout << fixed
#define endl '\n'

const int INF = 1e9;
const int MOD = 1e9 + 7;
const ll LINF = 1e18;

bool is_prime(ll x) {
  for (ll i = sqrt(x); i >= 2; i--) {
    if (x % i == 0) return false;
  }
  return true;
}

bool is_prime_pow(ll j, ll i) {
  if (i == j) return true;
  ll k = 1;
  while (k <= j) {
    k *= i;
    if (k == j) return true;
  }
  return false;
}

bool is_z(ll x) {
  if (is_prime(x)) return true;
  for (ll i = sqrt(x); i >= 2; i--) {
    if (x % i == 0) {
      ll j = x / i;
      if (is_prime(i) && is_prime_pow(j, i)) return true;
    }
  }
  return false;
}

void solve(istream& cin, ostream& cout) {
  ll n;
  cin >> n;
  if (n == 1) {
    cout << 0 << endl;
    return;
  }
  bool flag = false;
  priority_queue<int, vector<int>, greater<int>> pq;
  for (ll i = sqrt(n); i >= 2; i--) {
    if (n % i == 0) {
      flag = true;
      pq.push(i);
    }
  }
  if (!flag) {
    cout << 1 << endl;
    return;
  }
  map<int, bool> mp;
  ll ans = 0;
  while (!pq.empty()) {
    ll x = pq.top();
    pq.pop();
    if (mp.count(x)) {
      if (mp[x]) continue;
    }
    if (n % x != 0) continue;
    if (n < x) break;
    if (!is_z(x)) continue;
    n /= x;
    mp[x] = true;
    ans++;
    if (n == 1) break;
  }
  cout << ans << endl;
}

#ifndef TEST
int main() {
  OPTIMIZE_STDIO;
  solve(cin, cout);
  return 0;
}
#endif