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)

#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <numeric>
#include <chrono>
#include <bitset>
#include <tuple>
#include <queue>
#include <map>
#include <cstring>
#include <cassert>
#include <math.h>
using namespace std;

const int N = 2e5 + 10, mod = 1e9 + 7;


int main() {
    // ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int tc = 1;
    // scanf("%d", &tc);
    for(int cn = 1; cn <= tc; cn++) {
        long long n; scanf("%lld", &n);
        vector<pair<long long, int>> fac;
        for(int i = 2; 1ll * i * i <= n; i++) {
            if(n % i == 0) {
                fac.push_back({i, 0});
                while(n % i == 0) {
                    fac.back().second++;
                    n /= i;
                }
            }
        }
        if(n > 1) fac.push_back({n, 1});
        int ans = 0;
        for(int i = 0; i < fac.size(); i++) {
            int cnt = fac[i].second;
            for(int j = 1; j <= cnt; j++) {
                ans ++;
                cnt -= j;
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}