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: Accepted)

/**
*    author:  Dooloper
*    created: 06.06.2020 00:59:13
**/

#include <iostream>
#include <vector>
#include <map>
#include <string>
#include <utility>
#include <algorithm>
#include <queue>
#include <deque>
#include <stack>
#include <cmath>
#include <functional>
#include <cassert>


using namespace std;
typedef unsigned long long ull;
typedef long long ll;

vector<pair<ll, int>> primeFactor(ll N){
    vector<pair<ll, int>> ret;
    ll target = N;

    for (ll i = 2; i * i < target; i++)
    {
        int facotrNum = 0;
        while (N % i == 0)
        {
            facotrNum++;
            N /= i;
        }
        if (facotrNum)
            ret.emplace_back(make_pair(i, facotrNum));
    }
    if (N > 1)
        ret.emplace_back(make_pair(N, 1));
    return ret;
}

int main() {
    cin.tie(0);
    ll N;
    cin >> N;
    vector<pair<ll, int>> factor = primeFactor(N);
    int ans = 0;
    for (pair<ll,int> P : factor)
    {
        for (int j = 1; j <= P.second; j++){
            ans++;
            P.second -= j;
        }
    }

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