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;
using ll=long long;
const ll MOD=1000000007;
const ll MOD2=998244353;
const double PI=acos(-1);
const ll INF=1e18;
typedef pair<ll,ll> P;
typedef vector<ll> vl;
typedef vector<vl> vvl;
#define FOR(i,a,b) for(ll i=a;i<b;i++)
#define rep(i,n) FOR(i,0,n)
string abc="abcdefghijklmnopqrstuvwxyz";
string ABC="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
struct edge{ll to,cost;};

int main() {
  ll n;
  cin >> n;
  if(n==1){
    cout << 0 << endl;
  }
  else{
    ll ans=0;
    ll k=sqrt(n)+1;
    FOR(i,2,k+2){
      ll cnt=0;
      while(n%i==0){
        n/=i;
        cnt++;
      }
      ll z=0;
      ll add=1;
      while(z+add<=cnt){
        ans++;
        z++;
        add++;
      }
    }
    if(n!=1){
      ans+=1;
    }
    cout << ans << endl;
  }
}