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*/

 // #include <iostream>
#include <bits/stdc++.h>
/*#include <string>
#include <vector>
#include <algorithm>
#include <sstream>
#include <queue>
#include <deque>
#include <bitset>
#include <iterator>
#include <list>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <numeric>
#include <utility>
#include <limits>
#include <time.h>
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>*/
#define ll long long
#define dl double
#define rep(i,n)  for(int i = 0; i < n; i++)
#define all(cont) cont.begin(), cont.end()
#define rall(cont) cont.rbegin(), cont.rend()
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define unique(a)       sort((a).begin(), a.end()), (a).erase(unique((a).begin(), (a).end()),(a).end())
#define fastio          ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define Assert(x)       {if(!(x)){cerr<<"Assertion failed at line "<<__LINE__<<": "<<#x<<" = "<<(x)<<"\n";exit(1);}}
#define MP make_pair
#define PB push_back
#define INF (int)1e9
#define mod 1000000007

using namespace std;

typedef long long LL;
   bool isPrime(int n)
{
    // Corner cases
    if (n <= 1)  return false;
    if (n <= 3)  return true;

    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n%2 == 0 || n%3 == 0) return false;

    for (int i=5; i*i<=n; i=i+6)
        if (n%i == 0 || n%(i+2) == 0)
           return false;

    return true;
}
// A utility function to get the middle index from corner indexes.


/*  A recursive function to get the xor of values in given range
    of the array. The following are parameters for this function.

    st    --> Pointer to segment tree
    si    --> Index of current node in the segment tree. Initially
              0 is passed as root is always at index 0
    ss & se  --> Starting and ending indexes of the segment
                 represented by current node, i.e., st[si]
    qs & qe  --> Starting and ending indexes of query range */


/* A recursive function to update the nodes which have the given
   index in their range. The following are parameters
    st, si, ss and se are same as getXorUtil()
    i    --> index of the element to be updated. This index is
             in input array.
   diff --> Value to be added to all nodes which have i in range */

//     bool prime[1000010];
// void SieveOfEratosthenes(int n)
// {
//     // Create a boolean array "prime[0..n]" and initialize
//     // all entries it as true. A value in prime[i] will
//     // finally be false if i is Not a prime, else true.

//     memset(prime, true, sizeof(prime));

//     for (int p=2; p*p<=n; p++)
//     {
//         // If prime[p] is not changed, then it is a prime
//         if (prime[p] == true)
//         {
//             // Update all multiples of p greater than or
//             // equal to the square of it
//             // numbers which are multiple of p and are
//             // less than p^2 are already been marked.
//             for (int i=p*p; i<=n; i += p)
//                 prime[i] = false;
//         }
//     }

//     // Print all prime numbers
//     // for (int p=2; p<=n; p++)
//        // if (prime[p])
//        //    cout << p << " ";
// }

// The function to update a value in input array and segment tree.
// It uses updateValueUtil() to update the value in segment tree

const long long MAX_SIZE = 1000001;

// isPrime[] : isPrime[i] is true if number is prime
// prime[] : stores all prime number less than N
// SPF[] that store smallest prime factor of number
// [for Exp : smallest prime factor of '8' and '16'
// is '2' so we put SPF[8] = 2 , SPF[16] = 2 ]
vector<long long >isprime(MAX_SIZE , true);
vector<long long >prime;
vector<long long >SPF(MAX_SIZE);

// function generate all prime number less then N in O(n)
void manipulated_seive(int N)
{
    // 0 and 1 are not prime
    isprime[0] = isprime[1] = false ;

    // Fill rest of the entries
    for (long long int i=2; i<N ; i++)
    {
        // If isPrime[i] == True then i is
        // prime number
        if (isprime[i])
        {
            // put i into prime[] vector
            prime.push_back(i);

            // A prime number is its own smallest
            // prime factor
            SPF[i] = i;
        }

        // Remove all multiples of  i*prime[j] which are
        // not prime by making isPrime[i*prime[j]] = false
        // and put smallest prime factor of i*Prime[j] as prime[j]
        // [ for exp :let  i = 5 , j = 0 , prime[j] = 2 [ i*prime[j] = 10 ]
        // so smallest prime factor of '10' is '2' that is prime[j] ]
        // this loop run only one time for number which are not prime
        for (long long int j=0;
             j < (int)prime.size() &&
             i*prime[j] < N && prime[j] <= SPF[i];
             j++)
        {
            isprime[i*prime[j]]=false;

            // put smallest prime factor of i*prime[j]
            SPF[i*prime[j]] = prime[j] ;
        }
    }
}

ll power(ll x,ll y,ll p)
{
    ll res = 1;      // Initialize result

    x = x % p;  // Update x if it is more than or
                // equal to p

    while (y > 0)
    {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res*x) % p;

        // y must be even now
        y = y>>1; // y = y/2
        x = (x*x) % p;
    }
    return res%p;
}

ll solve(ll n)
{
    // Print the number of 2s that divide n
    ll ans= 0,sum = 0;
    while (n % 2 == 0)
    {
        sum++;
        n = n/2;
    }

    ll temp = sqrt(1 + 8*sum);
    temp = temp-1;
    temp = temp/2;

    ans += temp;
    // n must be odd at this point. So we can skip
    // one element (Note i = i +2)
    for (ll i = 3; i <= sqrt(n); i = i + 2)
    {
        // While i divides n, print i and divide n
        sum = 0;
        while (n % i == 0)
        {
            sum++;
            n = n/i;
        }

        temp = sqrt(1 + 8*sum);
        temp = temp-1;
        temp = temp/2;
        ans += temp;
    }

    // This condition is to handle the case when n
    // is a prime number greater than 2
    sum = 0;
    if (n > 2)
        ans++;

    return ans;

}

int main()
{
    ll i,j,n,m,b,t,k;

    cin>>n;

    cout<<solve(n);
    return 0;
}



Measured memory_bytes: 18684