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++ (Clang 10.0.0) (judge verdict: Runtime Error)

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <cmath>
#include <stack>
#include <queue>
#include <functional>
#include <set>
#include <map>
#include <tuple>
#include <bitset>
#include <random>
#define REP(i,n) for(int i=0;i<n;i++)
#define INF 100000000000007
using namespace std;
using pii=pair<int,int>;

vector<bool> IsPrime;


void sieve(int max){
    IsPrime=vector<bool>(max,true);
    IsPrime[0] = false; // 0は素数ではない
    IsPrime[1] = false; // 1は素数ではない

    for(int i=2; i*i<=max; ++i) // 0からsqrt(max)まで調べる
        if(IsPrime[i]) // iが素数ならば
            for(int j=2; i*j<=max; ++j) // (max以下の)iの倍数は
                IsPrime[i*j] = false;      // 素数ではない
}


vector<pair<int64_t, int64_t>> fs;//素因数分解の素数，指数のペア
void factorization(int64_t base){//baseを素因数分解
    //base以下のiが約数か
    for(int i=2;i*i<=base;i++){
        //iで割り切れるなら限界まで割る
        if(!IsPrime.at(i)){
            continue;
        }
        if(base%i==0){
            int cnt=0;
            while(base%i==0){
                base/=i;
                cnt++;
            }
            fs.push_back(make_pair(i,cnt));
        }
    }
    //baseが素数
    if(base>1){
        fs.push_back(make_pair(base,1));
    }
}
int main(){
    int64_t n;
    cin>>n;
    
    sieve(1000000+10);
    factorization(n);
    int64_t ans=0;
    for(auto p : fs){
        int64_t e;
        e=p.second;
        int k=1;
        while(e-k>=0){
            e-=k;
            k++;
            ans++;
        }
    }

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