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>
using namespace std;

typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pi;
typedef pair<ll,ll> pll;

#define endl '\n'
#define ff first
#define ss second
#define sz(v) (int)v.size()
#define int long long
#define pb push_back
#define f(i,l,r) for(ll i=l;i<=r;i++)
#define rf(i,r,l) for(ll i=r;i>=l;i--)
#define Max 5000001
#define inf 2147483647
#define llinf 9223372036854775807
#define all(v) v.begin(),v.end()
#define bp(n) __builtin_popcountll(n)
#define fast ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)

#define d0(a) cout<<a<<' ';
#define d1(a) cout<<a<<endl;
#define d2(a,b) cout<<a<<' '<<b<<endl;
#define d3(a,b,c) cout<<a<<' '<<b<<' '<<c<<endl;
#define d4(a,b,c,d) cout<<a<<' '<<b<<' '<<c<<' '<<d<<endl;
#define d5(a,b,c,d,e) cout<<a<<' '<<b<<' '<<c<<' '<<d<<' '<<e<<endl;
#define d6(a,b,c,d,e,f) cout<<a<<' '<<b<<' '<<c<<' '<<d<<' '<<e<<' '<<f<<endl;

const ll N=1e5+5,mod=1e9+7,bit=61;

int solve(int x)
{
    int l=1,r=1e9;
    int ans=1;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        int res=(mid*(mid+1))>>1;
        if(res<=x)
        {
            ans=mid;
            l=mid+1;
        }
        else
        {
            r=mid-1;
        }
    }
    return ans;
}

signed main()
{
    fast;

    ll t=1;
    //cin>>t;
    while(t--)
    {
        ll n;
        cin>>n;
        ll ans=0;
        map<int,int> mp;
        f(i,2,sqrt(n))
        {
            int pw=0;
            while(n%i==0)
            {
                n/=i;
                pw++;
            }
            if(pw>=1)
            {
                mp[i]=pw;
            }
        }
        if(n>1)
        {
            mp[n]+=1;
        }
        for(auto x:mp)
        {
            ans+=solve(x.ss);
        }
        cout<<ans<<endl;
    }
    return 0;
}