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>
      #include <ext/pb_ds/assoc_container.hpp>
      #include <ext/pb_ds/tree_policy.hpp>
      using namespace __gnu_pbds;
      using namespace std;
      #define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
      #define ll            long long
      #define pb            push_back
      #define ppb           pop_back
      #define si            set <ll>
      #define endl          '\n'
      #define fr            first
      #define sc            second
      #define mii           map<ll,ll>
      #define msi           map<string,ll>
      #define mis           map<ll,string>
      #define rep(i,a,b)    for(ll  i=a;i<b;i++)
      #define all(v)        v.begin(),v.end()
      #define pii           pair<ll,ll>
      #define vi            vector<ll>
      #define vii           vector<pair<ll,ll>>
      #define vs            vector<string>
      #define sz(x)         (ll)x.size()
      #define rt            return
      #define M             1000000007
      #define bs            binary_search
      #define rev(a)        reverse(all(a));
      #define sp(n)         setprecision(n)
      #define spl           " "
      #define rt            return
      #define arr(a,n)      rep(i,0,n) cin>>a[i]
      #define mod           998244353
      #define time          cout << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
      #define INF           1ll<<31
      void __print(int x) {cerr << x;}
      void __print(long x) {cerr << x;}
      void __print(long long x) {cerr << x;}
      void __print(unsigned x) {cerr << x;}
      void __print(unsigned long x) {cerr << x;}
      void __print(unsigned long long x) {cerr << x;}
      void __print(float x) {cerr << x;}
      void __print(double x) {cerr << x;}
      void __print(long double x) {cerr << x;}
      void __print(char x) {cerr << '\'' << x << '\'';}
      void __print(const char *x) {cerr << '\"' << x << '\"';}
      void __print(const string &x) {cerr << '\"' << x << '\"';}
      void __print(bool x) {cerr << (x ? "true" : "false");}
      template<typename T, typename V>
      void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
      template<typename T>
      void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
      void _print() {cerr << "]\n";}
      template <typename T, typename... V>
      void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
      #ifndef ONLINE_JUDGE
      #define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
      #else
      #define debug(x...)
      #endif
      ll bpow(ll a, ll b, ll mm = M)
      {
          ll res = 1;

          while(b)
          {
              if(b & 1)
                  res = (res * a) % mm;
              a = (a * a) % mm;

              b >>= 1;
          }

          return res;
      }
      ll modInverse(ll A,ll mm)
      {
          return bpow(A,mm-2,mm);
      }
      ll nCrModPFermat(ll n, ll r, ll p)
      {
         if (r==0)
            return 1;
          ll fac[n+1];
          fac[0] = 1;
          for (ll i=1 ; i<=n; i++)
              fac[i] = fac[i-1]*i%p;

          return (fac[n]*modInverse(fac[r],p)%p*modInverse(fac[n-r], p)%p)%p;
      }
       vector<ll> primeFactors(ll n)
    {

    vi v;
    while (n % 2 == 0)
    {
        v.pb(2);
        n = n/2;
    }
    for (ll i = 3; i <= sqrt(n); i = i + 2)
    {
        while (n % i == 0)
        {
            v.pb(i);
            n = n/i;
        }
    }

    if (n > 2)
        v.pb(n);
    return v;
}
      ll ok(ll n)
      {
          return (sqrt(1+8*n)-1)/2;
      }
      void solve()
      {
      ll n;
      cin>>n;
      vector<ll> v=primeFactors(n);
      map<ll,ll> mp;
      for(auto e:v) mp[e]++;
      vi ans;
      for(auto e:mp) ans.pb(e.sc);
      ll res=0;
      for(auto e:ans) res+=ok(e);
      cout<<res<<endl;
      

      } 
      signed main()
      { 
          ios_base::sync_with_stdio(false);
          cin.tie(0);
          ll t=1;
          //cin>>t;
          while(t--) solve();
      }