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>
#include <unistd.h>
#include <time.h>
#include <sys/time.h>

using namespace std;

///////////////////////////////////////////////////////////////////////////////

#define DEBUG 0

#define pb push_back
#define V vector
#define M unordered_map
#define S static
#define PI  (3.1415926535)

#define rep(i,n) for(ll i=0LL; i<n; ++i)
#define srep(i,s,n) for(ll i=s; i<n; ++i)
#define rrep(i,n) for(ll i=n-1LL; i>=0LL; --i)
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define CIN(x) do { \
        assert(!cin.eof()); \
        cin >> x; \
        assert(!cin.fail()); \
} while(0);
#if DEBUG
#define debug_print(...) _debug_print(__VA_ARGS__)
#define debug_printf(...) printf(__VA_ARGS__)
#define debug_print_time _debug_print_time
#else // DEBUG
#define debug_print(...)
#define debug_printf(...)
#define debug_print_time
#endif // DEBUG

typedef long long ll;
typedef unsigned long long ull;
typedef tuple<ll, ll> t2;
typedef tuple<ll, ll, ll> t3;
typedef tuple<ll, ll, ll, ll> t4;
typedef tuple<ll, ll, ll, ll, ll> t5;
template <typename T>
using priority_queue_incr = priority_queue<T, V<T>, greater<T> >;

///////////////////////////////////////////////////////////////////////////////

template <typename TT, typename T>
T get_m(M<TT, T> &m, TT k, T default_value)
{
    if (m.find(k) == m.end()) return m[k] = default_value;
    return m[k];
}

template <typename TT, typename T>
void incr_m(M<TT, T> &m, TT k)
{
    if (m.find(k) == m.end()) m[k] = 0;
    m[k]++;
}

void llin(ll &a)
{
    CIN(a);
}

void llinl1(auto &v, ll count)
{
    for (ll i = 0LL; i < count; ++i) {
        ll a;
        CIN(a);
        v.push_back(a);
    }
}

void llinl2(auto &v, ll count)
{
    for (ll i = 0LL; i < count; ++i) {
        ll a, b;
        CIN(a >> b);
        v.push_back(t2(a, b));
    }
}

void llinl3(auto &v, ll count)
{
    for (ll i = 0LL; i < count; ++i) {
        ll a, b, c;
        CIN(a >> b >> c);
        v.push_back(t3(a, b, c));
    }
}

void llinl4(auto &v, ll count)
{
    for (ll i = 0LL; i < count; ++i) {
        ll a, b, c, d;
        CIN(a >> b >> c >> d);
        v.push_back(t4(a, b, c, d));
    }
}

void llina(auto &v, ll count)
{
    llinl1(v, count);
}

template <typename T>
void sort(V<T> &v)
{
    sort(v.begin(), v.end());
}

template <typename T>
void sort_reverse(V<T> &v)
{
    sort(v.begin(), v.end(), greater<T>());
}

void _debug_print(auto x)
{
    cout << x << endl;
}

void _debug_print(const t2 &x)
{
    ll x1 = get<0>(x);
    ll x2 = get<1>(x);
    cout << "-- " << x1 << " -- " << x2 << endl;
}

void _debug_print(const t3 &x)
{
    ll x1 = get<0>(x);
    ll x2 = get<1>(x);
    ll x3 = get<2>(x);
    cout << "-- " << x1 << " -- " << x2 << " -- " << x3 << endl;
}

void _debug_print(const t4 &x)
{
    ll x1 = get<0>(x);
    ll x2 = get<1>(x);
    ll x3 = get<2>(x);
    ll x4 = get<3>(x);
    cout << "-- " << x1 << " -- " << x2
         << " -- " << x3 << " -- " << x4 << endl;
}

template <typename T>
void _debug_print(T xarray[], ll n)
{
    rep (i, n) _debug_print(xarray[i]);
}

template <typename T>
void _debug_print(const V<T> &xlist)
{
    for (auto x : xlist) {
        cout << "-- ";
        _debug_print(x);
    }
}

template <typename T>
void _debug_print(const set<T> &xset)
{
    for (auto x : xset) {
        cout << "-- ";
        _debug_print(x);
    }
}

template <typename TT, typename T>
void _debug_print(const M<TT, T> &xlist)
{
    for (auto x : xlist) {
        TT k = x.first;
        T v = x.second;
        cout << "====" << endl;
        cout << "K=";
        _debug_print(k);
        cout << "V=";
        _debug_print(v);
    }
}

void _debug_print_time(const char *prefix)
{
    struct timeval tv;
    gettimeofday(&tv, NULL);
    struct tm *tm = localtime(&tv.tv_sec);
    printf("-- %s %02d:%02d:%02d.%06ld\n",
           prefix, tm->tm_hour, tm->tm_min, tm->tm_sec, tv.tv_usec);
}



void get_factors(V<ll> &retlist, ll x)
{
    retlist.pb(1LL);
    for (ll i = 2LL; i < (ll)(sqrt(x)) + 3LL; ++i) {
        while (x % i == 0LL) {
            retlist.pb(i);
            x /= i;
        }
    }
    retlist.pb(x);
}



///////////////////////////////////////////////////////////////////////////////

void _main();
int main()
{
    cout << setprecision(12);
#if !DEBUG
    ios::sync_with_stdio(false);
    cin.tie(0);
#endif

    _main();
    return 0;
}

void _main()
{
    ll n; llin(n);
    V<ll> flist;
    get_factors(flist, n);

    M<ll, ll> cnts;
    for (auto f : flist) incr_m(cnts, f);

    ll ans = 0;

    for (auto cntm : cnts) {
        ll f = cntm.first;
        if (f < 2) continue;
        ll cnt = cntm.second;
        ll incr = 0;
        srep (i, 1, 100000) {
            if (i * (i+1LL) / 2LL > cnt) break;
            incr++;
        }
        ans += incr;
    }

    cout << ans << endl;
}

///////////////////////////////////////////////////////////////////////////////