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;
double pi = acos(-1);
inline long long lowbit(long long x) { return x & (-x); }
inline long long min(long long a, long long b) { return a < b ? a : b; }
inline long long max(long long a, long long b) { return a > b ? a : b; }
const int INF = 0x3f3f3f3f, max_n = 200005, mod = 998244353;
inline long long gcd(long long a, long long b) {
  return b == 0 ? a : gcd(b, a % b);
}
inline long long pow(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long n;
int main() {
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2 << '\n';
  } else {
    int s = sqrt(n);
    int ans = 0;
    int flag = 0;
    for (int i = 3; i <= s; i++) {
      if (n % i == 0) {
        n -= i;
        ans++;
        flag = 1;
        break;
      }
    }
    if (!flag) {
      cout << 1 << '\n';
    } else
      cout << ans + n / 2 << '\n';
  }
}
