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 FOR(i,n,j) for(int i=(j);i<(n);++i)
#define rrep(i,n) for(int i=(n);i>=0;--i)
#define ssort(n) sort((n).begin(),(n).end())
using ll=long long;
using ld=long double;
typedef pair<int,int> P;

int findFactors(int num){
    int i;
    
    while (num % 2 == 0){ // 偶数の素数は2のみ
        
        num /= 2;
    }
    for (i = 3; i <= num; i++){ // 他の素数で割っていく
        while (num % i == 0){
            num /= i;
        }
    }
    return 0;
}

int main()
{
    ll N;
    cin >> N;
    map<ll,int> num_count;
    queue<ll> num_list;
    ll n_=N;
    while (N % 2 == 0){ // 偶数の素数は2のみ
        if(!num_count.count(2)){
            num_list.push(2);
            num_count[2]=1;
        }
        num_count[2]++;
        N /= 2;
    }
    for (int i = 3; i <=sqrt(N); i+=2){ // 他の素数で割っていく
        while (N % i == 0){
            if(!num_count.count(i)){
                num_list.push(i);
                num_count[i]=1;
            }
            num_count[i]++;
            N /= i;
        }
    }
    if(N%n_==0){
        num_list.push(n_);
        num_count[n_]=1;
    }

    ll ans=0;
    while(!num_list.empty()){

        int n_count;
        int num=num_list.front();
        n_count=num_count[num];
        
        num_list.pop();
        for(int i=1;n_count>=i;++i){
            n_count-=i;
            ans++;
        }
    }
    cout << ans << endl;
}