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;
#define rep(i, n) for (int64_t i=0; i<(int64_t)(n); i++)
#define all(v) v.begin(), v.end()
//d
int main(){

    int64_t N; cin >> N;
    int64_t n=N;

    if(N==1){
        cout << 0 << endl;
        return 0;
    }

    vector<int> x(1000000);
    rep(i, 1000000){
        if(i==0) continue;
        while((int64_t)N%(i+1)==0){
            x[i]++;
            N/=(int64_t)(i+1);
        }
    }

    if((int64_t) n>1000000 && N==n){
        cout << 1 << endl;
        return 0;
    }

    int ans=0;
    rep(i, 1000000){
        if(i==0) continue;
        int cnt=1;
        while(x[i]>=cnt){
            ans++;
            x[i]-=cnt;
            cnt++;
        }
    }

    cout << ans << endl;

}