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 n;
int m = 0, a[100010], b[100010], cnt = 0;
int prime(long long x) {
  a[0] = 1;
  a[1] = 1;
  for (long long i = 2; i <= x; i++)
    if (!a[i])
      for (long long j = i * 2; j <= x; j += i) a[j] = 1;
  for (long long i = 2; i <= x; i++)
    if (!a[i]) b[++m] = i;
}
bool isp(long long x) {
  long long d = sqrt(x);
  for (int i = 2; i <= d; i++)
    if (x % i == 0) return 0;
  return 1;
}
int main() {
  cin >> n;
  while (n) {
    if (n % 2 == 0) {
      cnt += (n / 2);
      break;
    }
    if (isp(n)) {
      cnt++;
      break;
    }
    int x = 1;
    prime(sqrt(n));
    while (n % b[x] != 0) x++;
    n -= b[x];
    cnt++;
  }
  cout << cnt << endl;
  return 0;
}
