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;

typedef long long ll;

int main() {

#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    ll m = 0;
    cin >> m;

    vector<ll> v(1e6 + 5, 0);
    ll n = m;
    while(n % 2 == 0) {
        n = n/2;
        v[2]++;
    }

    for(int i = 3; i < sqrt(m); i+= 2) {
        while(n % i == 0) {
            n = n /i;
            v[i]++;
        }
    }

    // for(int i = 0; i < 5; i++) {
    //     cout << v[i] << " ";
    // }
    // cout << endl;


    ll ans = 0; 
    for(int i = 0; i < v.size(); i++) {
        if(v[i] != 0) {
            int temp = 1;
            while(v[i] - temp >= 0) {
                v[i] -= temp;
                temp++;
                ans++;
            }
        }
    }

    cout << ans << endl;


    
    return 0;
}