You are given an integer number n. The following algorithm is applied to it:

  1. if n = 0, then end algorithm; 
  2. find the smallest prime divisor d of n; 
  3. subtract d from n and go to step 1. 



Determine the number of subtrations the algorithm will make.

Input

The only line contains a single integer n (2 ≤ n ≤ 10^{10}).

Output

Print a single integer — the number of subtractions the algorithm will make.

Examples

Input


5


Output


1


Input


4


Output


2

Note

In the first example 5 is the smallest prime divisor, thus it gets subtracted right away to make a 0.

In the second example 2 is the smallest prime divisor at both steps.
#include <bits/stdc++.h>
using namespace std;
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, 1, -1, 0, 0, -1, 1};
const double PI = acos(-1.0);
const double EPS = 1e-6;
const int MOD = (int)1e9 + 7;
const int maxn = (int)2e5 + 5;
const int LOGN = 20;
const int MX = 1000010;
bool is_composite[MX];
int sp[MX];
vector<int> primes;
void Sieve() {
  primes.push_back(2);
  for (int i = 2; i < MX; i += 2) sp[i] = 2;
  for (int i = 3; i < MX; i += 2) {
    if (is_composite[i]) continue;
    primes.push_back(i);
    sp[i] = i;
    for (long long j = (long long)i; (j * i) < MX; j += 2) {
      if (!is_composite[j * i]) {
        is_composite[j * i] = true;
        sp[j * i] = i;
      }
    }
  }
}
long long factorize(long long n) {
  long long ans = 0LL;
  for (int p : primes) {
    long long tem = n;
    while (tem % p == 0) {
      ans += (tem - tem / p) / p;
      tem /= p;
    }
    n = tem;
  }
  if (n > 0) ans++;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  Sieve();
  long long n;
  cin >> n;
  long long ans = factorize(n);
  cout << ans;
  return 0;
}
