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)

/*
これを入れて実行
g++ code.cpp
./a.out
 */
 
#include <iostream>
#include <stdio.h>
#include <vector>
#include <string>
#include <cstring>
#include <queue>
#include <deque>
#include <stack>
#include <algorithm>
#include <utility>
#include <set>
#include <map>
#include <unordered_map>
#include <cmath>
#include <math.h>
#include <tuple>
#include <iomanip>
#include <bitset>
#include <functional>
#include <cassert>

#define all(x) (x).begin(),(x).end()
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
 
using namespace std;
typedef long long ll;
typedef long double ld;
 
int dy4[4] = {-1, 0, +1, 0};
int dx4[4] = {0, +1, 0, -1};
int dy8[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
int dx8[8] = {0, 1, 1, 1, 0, -1, -1, -1};
 
const long long INF = 1LL << 62;
//const ll MOD = 1e9 + 7;
 
bool greaterSecond(const pair<int, int>& f, const pair<int, int>& s){
    return f.second > s.second;
}
 
ll gcd(ll a, ll b){
	if (b == 0)return a;
	return gcd(b, a % b);
}
 
ll lcm(ll a, ll b){
    return a / gcd(a, b) * b;
}

ll conbinationMemo[100][100];

void cmemoInit(){
    rep(i, 100){
        rep(j, 100){
            conbinationMemo[i][j] = -1;
        }
    }
}
 
ll nCr(ll n, ll r){
    if(conbinationMemo[n][r] != -1) return conbinationMemo[n][r];
    if(r == 0 || r == n){
        return 1;
    } else if(r == 1){
        return n;
    }
    return conbinationMemo[n][r] = (nCr(n - 1, r) + nCr(n - 1, r - 1));
}
 
ll nPr(ll n, ll r){
    r = n - r;
    ll ret = 1;
    for (ll i = n; i >= r + 1; i--) ret *= i;
    return ret;
}
//-----------------------ここから-----------
template<class T> T fastpow(T a, T b){
    T res = 1;
    while(b > 0){
        if(b & 1){
            res *= a;
        }
        a *= a;
        b >>= 1;
    }
    return res;
}
int main(void){
    ll n;
    cin >> n;
    vector<bool> prime(1001001, false);

    for(ll i = 0; i <= 1001000; i++){
        prime[i] = true;
    }

    vector<ll> p;

    for(ll i = 2; i <= 1001000; i++){
        if(prime[i]){
            for(ll j = 2 * i; j <= 1001000; j += i){
                prime[j] = false;
            }
            p.push_back(i);
        }
    }
    vector<ll> li;
    for(ll i = 0; i < p.size(); i++){
        ll now = p[i];
        while(now <= (ll)1e13){
            li.push_back(now);
            now *= p[i];
        }
    }

    bool np = true;

    for(ll i = 2; i * i <= n; i++){
        if(n % i == 0){
            np = false;
        }
    }
    ll plus = 0;
    if(n >= 1001000){
        if(np){
            cout << 1 << endl;
            return 0;
        }
    }
    sort(all(li));
    ll ans = 0;
    ll num = n;
    for(ll i = 0; i < li.size(); i++){
        if(num % li[i] == 0){
            num /= li[i];
            ans++; 
        }
    }

    cout << ans << endl;
}