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>
#pragma GCC target ("avx2")
#pragma GCC optimization ("Ofast")
#pragma GCC optimization ("unroll-loops")
#pragma comment(linker, "/stack:200000000")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
#define int long long 
typedef pair<int,int>pi;
typedef vector<int>vi;

const int MAXN=1000005;
const int MOD=1000000007;

#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define DEC(i,a,b) for(int i=(a);i>=(b);i--)
#define pb push_back
#define mp make_pair
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(),(v).end()
#define db(x) cerr<<#x<<" = "<<x<<"\n"
#define LOW(v,x) (lower_bound(all(v),(x))-(v).begin())
#define UP(v,x) (upper_bound(all(v),(x))-(v).begin())

bitset<MAXN+1> isPrime;
vector<int> primeList;
void sieve(){
    isPrime.reset();
    isPrime.flip();
    isPrime[0] = isPrime[1] = 0;
    primeList.push_back(2);
    for (int i = 4; i < MAXN; i+=2) isPrime[i] = 0;
    for (int i = 3; i < MAXN; i+=2) {
        if (!isPrime[i]) continue;
        primeList.push_back(i);
        for (int j = i; j <= MAXN/i; ++j) isPrime[i*j] = 0;
    }
}

vector<pi> primeFactorize(int X) {
    vector<pi> result;
    for (auto i:primeList) {
        if (i > X/i) break;
        int cnt = 0;
        while (X % i == 0) {
            X /= i;
            cnt++;
        }
        if (cnt > 0) result.emplace_back(i, cnt);
    }
    if (X != 1) result.emplace_back(X, 1);
    return result;
}

main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	sieve();
	
	int N; cin>>N;
	
	vector<pi> vec = primeFactorize(N);
	int ans = 0;
	for (auto i:vec){
		int lim = i.second;
		int s = 0, e = 1e9;
		while (e-s>1){
			int m = (s+e)/2;
			int tmp = m*(m+1)/2;
			if (tmp > lim) e = m;
			else s = m;
		}
		ans += s;
	}
	cout<<ans;
}