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;
double const EPS = 1e-8, PI = acos(-1);
const int N = 30, M = 1e6 + 1, OO = 1e6 + 1;
const long long MOD = 1e9 + 7;
void INPUT() {
  cout << fixed << setprecision(0);
  ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
}
long long cnt, x;
void solve(long long n) {
  int i = 1, d = 1;
  for (i = 2; i * i <= n; i += d, d = 2) {
    if (n % i == 0) x = n / i;
    cnt += x;
    n -= x * i;
  }
  if (n > 1) cnt++;
}
int main() {
  INPUT();
  long long n;
  cin >> n;
  solve(n);
  cout << cnt;
  return 0;
}
