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;
bool is(long long n) {
  for (long long p = 2; p * p <= n; p++) {
    if (n % p == 0) return false;
  }
  return true;
}
long long go2(long long n) {
  if (!(n & 1)) {
    return 2;
  }
  for (long long p = 3; p * p <= n; p++) {
    if (n % p == 0) {
      return p;
    }
  }
  return n;
}
long long go1(long long n) {
  long long ans = 0;
  while (n > 0) {
    long long x = go2(n);
    ans++;
    n -= x;
  }
  return ans;
}
long long go(long long n) {
  if (!(n & 1)) {
    return (n / 2);
  }
  for (long long p = 3; p * p <= n; p++) {
    if (n % p == 0) {
      return (n / p);
    }
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  if (is(n))
    cout << "1\n";
  else {
    if (n < 1000000)
      cout << go1(n) << "\n";
    else
      cout << max(n / 2, go(n)) << "\n";
  }
}
