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>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
#define repA(i,n) for(int i=n; i>=0; i--)
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define sz(a) (int)(a.size())
typedef long long ll;
const ll mod = 1e9+7;//998244353
int t = 1;
/*
int rooms,psgs;
vector<int> a[(int)1e5 + 3];
vector<int> p((int)1e5 + 3 , 0);
vector<bool> v((int)1e5 + 3, false);

void dfs(int cur) {
  vector<int>
  for(auto &i:a[cur]) {
    if(v[i])  continue;
    p[i] = cur;
    v[i]=true;
  }
}*/

int calc(int x) {
  long double ans = 0.25;
  ans = sqrt((long double)x*2 + ans);
  ans -=0.5;
  return (int)ans;
}

void solve(int BYasD) {
  ll n; cin>>n;
  int cnt = 0;
  while(n%2 == 0) {
    cnt++;
    n = n/2;
  }
  int ans = 0;
  ans += calc(cnt);
  for(int i=3; i<=sqrt(n); i+=2) {
    cnt = 0;
    while(n%i == 0) {
      cnt++;n=n/i;
    }
    ans += calc(cnt);
  }
  if(n!=1)  ans++;
  cout<<ans<<'\n';
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0); cout.tie(0);
  #ifndef ONLINE_JUDGE
  freopen("input.txt", "r", stdin);
  #endif
  //cin>>t;
  rep(i,t)  {
  //  cout<<"Case #"<<(i+1)<<'\n';
    solve(i);
  //  cout<<'\n';
  }
  return 0;
}