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 long long N = 200002;
long long MOD = 1e9 + 7;
long long gcd(long long a, long long b) {
  while (b) {
    long long t = a % b;
    a = b;
    b = t;
  }
  return a;
}
long long mod(long long n) { return (n % MOD + MOD) % MOD; }
long long mexp(long long a, long long b) {
  if (b == 0) return 1;
  long long t = mexp(a, b / 2);
  if (b % 2 == 0)
    return mod(t * t);
  else
    return mod(mod(t * t) * a);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, i, ans = 0;
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2;
    return 0;
  }
  for (i = 3; i * i <= n; i++) {
    if (n % i == 0) {
      cout << (1 + (n - i) / 2);
      return 0;
    }
  }
  cout << 1;
  return 0;
}
