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: Runtime Error)

#pragma region 宣言部分
#include <iostream>
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <iterator>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <utility>
#include <vector>
typedef long long ll;
#define REP(i,n) for(int i = 0;i < (n);i++)
#define rep(i,m,n) for(int i = (m);i < (n);i++)
#define P pair<int,int>
#define pb push_back
#define mk make_pair
using namespace std;
#define Vec(a) vector <int> a
const int INF = 1 << 20;
const int MOD = 1e9+7;

#pragma endregion



int main(){
    ll n;
    cin >> n;
    if(n == 1){
        cout << 0 << endl;
        return 0;
    }
    ll m = n;
    
    int num[1000010] = {0};
    //cout << "!" << endl;
    bool flag = false;
    for(ll i = 2;i*i <= n;i++){
        while(m%i == 0){
            flag = true;
            m /= i;
            num[i]++;
        }
    }
    if(m > 1)
        num[m]++;
    if(!flag){
        cout << 1 << endl;
        return 0;
    }
    ll res = 0;
    for(ll i = 2;i*i <= n;i++){
        //cout << num[i] << endl;
        
        if(num[i] != 0){
            //cout << i << " " << num[i] << endl;
            res += (-1 + sqrt(1+8*num[i]))/2;
        }
    }
    cout << res << endl;
    return 0;
}