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 Size = 300005;
const int INF = 0x3f3f3f3f;
long long n, sn;
bool check(int x) {
  if (x == 2) return true;
  if (!(x & 1)) return false;
  int sx = sqrt(x);
  for (register int i = 3; i <= sx; i += 2) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
}
int main() {
  int A[10001];
  cin >> n;
  sn = sqrt(n);
  bool flag = false;
  for (int i = 2; i <= sn; i++) {
    if (n % i == 0 && check(i)) {
      flag = true;
      n -= i;
      break;
    }
  }
  if (!flag) {
    printf("1");
  } else {
    printf("%I64d", (long long)(n >> 1) + 1ll);
  }
  return 0;
}
