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 MAX = 1e7 + 5;
long long n;
int prim[MAX];
void prime() {
  memset(prim, 0, sizeof(prim));
  for (int i = 2; i < MAX; i++) {
    if (!prim[i]) prim[++prim[0]] = i;
    for (int j = 1; j <= prim[0] && prim[j] <= MAX / i; j++) {
      prim[prim[j] * i] = 1;
      if (i % prim[j] == 0) break;
    }
  }
}
int main() {
  prime();
  while (~scanf("%lld", &n)) {
    long long ans = 0;
    for (int i = 1; i <= prim[0]; ++i) {
      if (n % prim[i] == 0) {
        ans++;
        n -= prim[i];
        break;
      }
    }
    if (n % 2 == 0) ans += n / 2;
    if (ans == 0) ans = 1;
    printf("%lld\n", ans);
  }
  return 0;
}
