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;
long long prime[1000010];
bool vis[1000010];
long long cnt = 0;
void Euler_peime(long long n) {
  for (long long i = 2; i <= n; ++i) {
    if (!vis[i]) {
      prime[cnt++] = i;
      vis[i] = true;
    }
    for (long long j = 0; j < cnt; ++j) {
      if (i * prime[j] > n) break;
      vis[i * prime[j]] = true;
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  Euler_peime(1000010);
  long long n, k = 0;
  scanf("%lld", &n);
  while (n != 0) {
    if (n % 2 == 0) break;
    int p = 0;
    int m = sqrt(n);
    for (int i = 0; i < cnt && prime[i] <= m; i++) {
      if (n % prime[i] == 0) {
        n -= prime[i];
        k++;
        p = 1;
        break;
      }
    }
    if (p == 0) {
      k++;
      break;
    }
  }
  if (n % 2 == 0) k += n / 2;
  cout << k << endl;
  return 0;
}
