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;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const ll INF = 1LL << 60;
const int MAX = 1000000010;
const ll MOD = 1000000007;

int isPrime(ll x) {
    ll i;
    if (x < 2) return 0;
    else if (x == 2) return 1;
    if (x % 2 == 0) return 0;
    for (i = 3; i * i <= x; i += 2) {
        if (x % i == 0) return 0;
    }
    return 1;
}

int main() {
    ll N; cin >> N;
    map<int, int> m;
    ll x = N;
    for (ll i = 2; i * i <= x; i++) {
        while (N % i == 0 && N >= i) {
            N /= i;
            m[i]++;
        }
        if (N == 1) break;
    }

    ll ans = 0;
    for (auto p : m) {
        int cnt = p.second;
        int sum = 1;
        while (cnt - sum >= 0) {
            cnt -= sum;
            sum++;
            ans++;
        }
    }

    if (x == 1) {
        cout << 0 << endl;
    } else {
        if (isPrime(x)) {
            cout << 1 << endl;
        } else {
            cout << ans << endl;
        }
    }
    return 0;
}