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(int i=0; i<(n); i++)
#define rep2(i,x,n) for(int i=x; i<(n); i++)
#define all(n) begin(n),end(n)
struct cww{cww(){ios::sync_with_stdio(false);cin.tie(0);}}star;
const long long INF = numeric_limits<long long>::max();
int main(){
    long long N;
    cin>>N;
    if(N==1){
        cout << 0 << endl;
        return 0;
    }
    if(N==0){
        cout << 0 << endl;
        return 0;
    }
    long long tmp=N;
    map<long long,int> divcnt;//素因数、個数
    bool over=false;
        for(long long i=2;i<10000000;i++){
            if(over){
                break;
            }
            bool flag=true;
            while(flag){
                if(tmp%i==0){
                    if(divcnt.count(i)){
                        divcnt[i]++;
                    }else{
                        divcnt[i]=1;
                    }
                    tmp=tmp/i;
                    if(tmp==1){
                        over=true;
                    }
                }else{
                    flag=false;
                }
            }
        }
    if(tmp!=1){
        // cout << "koko" << endl;
        if(divcnt.count(tmp)){
            divcnt[tmp]++;
        }else{
        divcnt[tmp]=1;
        }
    }
    int ans =0;
    // cout << divcnt[2] << divcnt[3] << endl;
    //mapに素数の個数が入った
    for (auto iter = divcnt.begin(); iter != divcnt.end(); iter++) {
        // first: key, second: value
        int k =iter->first;
        // cout << "koko" << k << endl;
        for(int j=1;j<100;j++){
            //  cout <<"soko"<< j << "i" <<divcnt[k] << endl;
             
            if(j<=divcnt[k]){
                // cout << "asoko" << endl;
                ans++;
                divcnt[k]=divcnt[k]-j;
            }else{
                break;
            }

        }
    }

    cout << ans << endl;
    return 0;
}