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 multi(long long a, long long b, long long mod) {
  return (a * b) % mod;
}
long long power(long long a, long long b, long long mod) {
  long long powans = 1;
  for (; b > 0; a = multi(a, a, mod), b /= 2)
    if (b % 2 == 1) powans = multi(powans, a, mod);
  return powans;
}
void fastscan(int &x) {
  bool neg = false;
  register int c;
  x = 0;
  c = getchar();
  if (c == '-') {
    neg = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) x = (x << 1) + (x << 3) + c - 48;
  if (neg) x *= -1;
}
int gcd(int a, int b) {
  if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
void solve() {
  long long int n;
  cin >> n;
  int count = 0;
  if (n % 2 == 0)
    cout << n / 2;
  else {
    for (int i = 3; i <= sqrt(n); i++) {
      if (n % i == 0) {
        count = i;
        break;
      }
    }
    if (count == 0)
      cout << 1;
    else {
      n = n - count;
      cout << n / 2 + 1;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
