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<stdio.h>
#include<iostream>
#include<vector>
#include<math.h>
#include<queue>
#include<map>
#include<algorithm>
#include<string.h>
#include<functional>
#include<limits.h>
#include<stdlib.h>
#include<string>
#include<unordered_map>
#include<set>
using namespace std;

#define intmax INT_MAX
#define lmax LONG_MAX
#define uintmax UINT_MAX
#define ulmax ULONG_MAX
#define llmax LLONG_MAX
#define ll long long
#define rep(i,a,N) for((i)=(a);(i)<(N);(i)++)
#define rrp(i,N,a) for((i)=(N)-1;(i)>=(a);(i)--)
#define llfor ll i,j,k
#define sc(a) cin>>a
#define pr(a) cout<<a<<endl
#define pY puts("YES")
#define pN puts("NO")
#define py puts("Yes")
#define pn puts("No")
#define pnn printf("\n")
#define all(a) a.begin(),a.end()
#define push(a,b) (a).push_back(b)
#define llvec vector<vector<ll>>
#define charvec vector<vector<char>>
#define sizeoof(a,b) (a,vector<ll>(b))
#define llpvec vector<pair<ll,ll>>
/*繰り上げ除算*/ll cei(ll x,ll y){ll ans=x/y;if(x%y!=0)ans++;return ans;}
/*最大公約数*/ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
/*最小公倍数*/ll lcm(ll x,ll y){return x/gcd(x,y)*y;}
/*n乗*/ll llpow(ll x,ll n){ll i,ans=1;rep(i,0,n)ans*=x;return ans;}
/*階乗*/ll fact(ll x){ll i,ans=1;rep(i,0,x)ans*=(x-i);return ans;}
/*nCr*/ll ncr(ll n,ll r){return fact(n)/fact(r)/fact(n-r);}
/*nPr*/ll npr(ll n,ll r){return fact(n)/fact(n-r);}
/*primejudge*/bool prime(ll a){if(a<=1)return false;ll i;for(i=2;i*i<=a;i++){if(a%i==0)return false;}return true;}
/*約数*/vector<ll>div(ll a){vector<ll>b;ll i;rep(i,1,a){if(!a%i)push(b,a);}return b;}
llfor;///////////////////////////////////////////////////////////
int main(){
 ll n;
 sc(n);
 ll b;
 ll tmp=n;
 ll ans=1;
 ll before;
 bool jud=prime(tmp);
rep(i,2,n){
 if(i>tmp)break;
  if(tmp%i==0){
   before=tmp;
   ans++;tmp/=i;
  // cout<<i<<" "<<tmp<<endl;
  }
  jud=prime(tmp);
  if(jud==true){
   if(i==tmp)ans--;
   //cout<<i<<" "<<tmp<<endl;
   break;
  }
 if(tmp==1)ans--;
 }
 pr(ans);
return 0;}