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;
using ll = long long;
using VI = vector<int>;
using VL = vector<ll>;
using VS = vector<string>;
using VB = vector<bool>;
using VVI = vector<VI>;
using VVL = vector<VL>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;
using TI3 = std::tuple<int, int, int>;
using TI4 = std::tuple<int, int, int, int>;
using TL3 = std::tuple<ll, ll, ll>;
using TL4 = std::tuple<ll, ll, ll, ll>;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repr(i, n) for (int i = (int)(n)-1; i >= 0; i--)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
#define rep3(i, s, n, d) for (int i = (s); i < (int)(n); i += (d))
#define allpt(v) (v).begin(), (v).end()
#define allpt_c(v) (v).cbegin(), (v).cend()
#define allpt_r(v) (v).rbegin(), (v).rend()

const int mod1 = 1e9 + 7, mod2 = 998244353, mod3 = 1e9 + 9;
const int mod = mod2;
const ll inf = 1e18;

const string wsp = " ";
const string tb = "\t";
const string rt = "\n";

template <typename T>
void show1dvec(const vector<T> &v)
{
    if (v.size() == 0)
        return;
    int n = v.size() - 1;
    rep(i, n) cout << v[i] << wsp;
    cout << v[n] << rt;

    return;
}

template <typename T>
void show2dvec(const vector<vector<T>> &v)
{
    int n = v.size();
    rep(i, n) show1dvec(v[i]);
}

template <typename T, typename S>
void show1dpair(const vector<pair<T, S>> &v)
{
    int n = v.size();
    rep(i, n) cout << v[i].first << wsp << v[i].second << rt;
    return;
}

template <typename T, typename S>
void pairzip(const vector<pair<T, S>> &v, vector<T> &t, vector<T> &s)
{
    int n = v.size();
    rep(i, n)
    {
        t.push_back(v[i].first);
        s.push_back(v[i].second);
    }
    return;
}

template <typename T>
void maxvec(vector<T> &v)
{
    T s = v[0];
    int n = v.size();
    rep(i, n - 1)
    {
        if (s > v[i + 1])
        {
            v[i + 1] = s;
        }
        s = v[i + 1];
    }
}

template <typename T, typename S>
bool myfind(T t, S s)
{
    return find(t.cbegin(), t.cend(), s) != t.cend();
}

bool check(int y, int x, int h, int w)
{
    return 0 <= y && y < h && 0 <= x && x < w;
}

template <typename T>
vector<T> cumsum(const vector<T> &v)
{
    T s = 0;
    vector<T> ret;

    rep(i, v.size())
    {
        s += v[i];
        s %= mod;
        ret.emplace_back(s);
    }
    return ret;
}

bool iskadomatsu(int a, int b, int c)
{
    return (a != b && b != c && c != a) && ((a > b && b < c) || (a < b && b > c));
}

VS split(string s, char c)
{
    VS ret;
    string part;
    s += c;
    rep(i, s.length())
    {
        if (s[i] == c)
        {
            ret.emplace_back(part);
            part = "";
        }
        else if (s[i] != c)
        {
            part += s[i];
        }
    }
    return ret;
}

string removezero(string &s)
{
    string ret;
    for (auto z : s)
        if (z != '0')
            ret += z;
    return ret;
}

ll sumdigit(ll x)
{
    ll ans{0};
    while (x > 0)
    {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}

template <typename T, typename S, typename R>
ll pow_mod(T p, S q, R mod = 1ll)
{
    ll ret = 1, r = p;
    while (q)
    {
        if (q % 2)
            ret *= r, ret %= mod;
        r = (r * r) % mod, q /= 2;
    }
    return ret % mod;
}

void make_frac_tables(VL &frac_list, VL &frac_inv_list)
{
    rep(i, frac_list.size() - 1)
    {
        frac_list[i + 1] *= frac_list[i] * (i + 1);
        frac_list[i + 1] %= mod;
        frac_inv_list[i + 1] *= frac_inv_list[i] * pow_mod(i + 1, mod - 2, mod);
        frac_inv_list[i + 1] %= mod;
    }
}

ll comb(int a, int b, const VL &frac_list, const VL &frac_inv_list)
{
    if (a < b)
        return 0;
    ll ret = frac_list[a];
    ret *= frac_inv_list[b];
    ret %= mod;
    ret *= frac_inv_list[a - b];
    ret %= mod;
    return ret;
}



int main()
{

    cin.tie(0);
    ios::sync_with_stdio(false);

#ifdef DEBUG
    cout << "DEBUG MODE" << endl;
    ifstream in("input.txt"); //for debug
    cin.rdbuf(in.rdbuf());    //for debug
#endif

    ll n;
    int ans = 0;
    cin >> n;
    map<ll, int> x;
    rep2(i, 2, 100000)
    {
        while (n % i == 0)
        {
            x[i]++;
            n /= i;
        }
    }
    if (n != 1)
    {
        x[n] = 1;
    }

    VI v{0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66};

    for (auto i : x)
    {
        rep(j, 11)
        {
            if (v[j] <= i.second && i.second < v[j + 1])
            {
                ans += j;
            }
        }
    }
    cout << ans << rt;    

    return 0;
}