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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(v) v.begin(), v.end()
#define ll int64_t
#define _GLIBCXX_DEBUG
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
int main(){
  ll n;
  cin >> n;
  vector<int> v(1000004,1);
  int ans=0;
  for(int i=1;i<1000000;i++){
    if(i+1>n)break;
    if(v.at(i)==0)continue;
    int a=i+1;
    if((n%a)==0){
      ans++;
      n=n/a;
      if(v.at(i)==2)continue;
      for(int j=1;j<=1000000/a;j++){
        if(a*j-1<1000000)v.at(a*j-1)=0;
      }
      ll h=a*a;
      while(h<1000000){
        v.at(h-1)=2;
        h*=a;
      }
    }
  }
  if(n>1000000)ans++;
  cout << ans << endl;
}