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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<string> vs;
typedef pair<int, int> P;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()

void YN(bool flg){if(flg) cout << "YES" << endl; else cout << "NO" << endl;}
void Yn(bool flg){if(flg) cout << "Yes" << endl; else cout << "No" << endl;}
void yn(bool flg){if(flg) cout << "yes" << endl; else cout << "no" << endl;}

//素因数分解
vector<pair<ll, int>> factorize(ll n) {
    vector<pair<ll, int>> res;
    for(ll i = 2; i * i <= n; i++) {
        if(n % i) continue;
        res.emplace_back(i, 0);
        while(n % i == 0) {
            n /= i;
            res.back().second++;
        }
    }
    if(n != 1) res.emplace_back(n, 1);
    return res;
}

int main()
{
    ll n;
    cin >> n;

    vector<pair<ll, int>> f = factorize(n);
    
    ll ans = 0;
    for(auto u : f) {
        int num = u.second;
        int i = 1;
        while(i <= num) {
            num -= i;
            ans++;
            i++;
        }
    }

    cout << ans << endl;

    return 0;
}