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 ll;
typedef unsigned long long int ull;
#define endl "\n"
#define pb push_back
#define sq(a) (a)*(a)
#define all(x) (x).begin(), (x).end()
#define debug(x) cerr<<#x<<'='<<(x)<<endl;
#define debugv(v) cerr<<#v<<" : ";for (auto x:v) cerr<<x<<' ';cerr<<endl;
#define MOD 1000000007
#define PI 3.141592653589793238
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);

ll gcd(ll a, ll b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

ll exp(ll x,ll n,ll mod)
{
    ll result=1;
    while(n)
    {
        if(n & 1)
            result=(result * x)%mod;
        x=(x*x)%mod;
        n=n>>1;
    }
    return result;
}



int32_t main()
{
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif
    IOS

    ll n;
    cin>>n;
    ll ans=0;

    for(ll i=2;i<=1000005;i++)
    {
    	ll temp=0;
    	while(n%i==0)
    	{
    		temp++;
    		n/=i;
    	}
    	for(int j=1;j<1010;j++)
    	{
    		if(j>temp)
    			break;
    		ans++;
    		temp-=j;
    	}
    }

    if(n>1)
    	ans++;

    cout<<ans<<endl;
    
    return 0;
}