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;
const int N = 3e5 + 7;
int vis[N], zhi[N];
int cnt;
void ol() {
  memset(vis, 0, sizeof(vis));
  cnt = 0;
  for (int i = 2; i <= N; i++) {
    if (!vis[i]) zhi[++cnt] = i;
    for (int j = 1; j <= cnt && zhi[j] * i <= N; ++j) {
      vis[zhi[j] * i] = 1;
      if (i % zhi[j] == 0) break;
    }
  }
}
int main() {
  ol();
  long long n;
  scanf("%lld", &n);
  if (n % 2 == 0) {
    printf("%lld\n", n / 2);
    return 0;
  }
  int flag = 0;
  for (int i = 2; i <= cnt; i++) {
    if (n % zhi[i] == 0) {
      n = n - zhi[i];
      flag = 1;
      break;
    }
  }
  if (!flag)
    printf("1\n");
  else
    printf("%lld\n", n / 2 + 1);
  return 0;
}
