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 <algorithm>
#include <iostream>
#include <queue>
#include <cmath>
#include <string>
#include <iterator>
#include <map>
#include <set>
#include <iomanip>
#include <vector>
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<int, int>;
using Graph = vector<vector<int>>;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
#define INF 100000000
#define MAX 200001
#define MOD 1000000007
ll fac[MAX], finv[MAX], inv[MAX];
const int MX = 1000005;

int main(){
    ll n;cin>>n;
    vector<ll> r;
    int ans=0;
    for(int i=2;i*i<=n;i++){
        if(n%i==0){
            n/=i;
            r.push_back(i);
            ans++;
        }
    }
    if(n!=1){
        for(int i=0;i<r.size();i++){
            if(r[i]==n) break;
            if(i==r.size()-1) ans++;
        }
    }
    if(ans==0) ans=1;
    if(n==1) ans=0;
    cout<<ans<<endl;
}