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 int maxn = 100010;
int prime[maxn], vis[maxn];
int cnt = 0;
void Prime() {
  for (int i = 2; i <= maxn; i++) {
    if (!vis[i]) {
      prime[++cnt] = i;
      for (int j = 1; j <= cnt && i * prime[j] <= maxn; j++) {
        vis[i * prime[j]] = 1;
        if (i % prime[j] == 0) break;
      }
    }
  }
}
int main() {
  long long n;
  Prime();
  bool flag = true;
  cin >> n;
  for (int i = 1; i <= cnt; i++) {
    if (n % prime[i] == 0 && (n - prime[i]) % 2 == 0) {
      cout << (n - prime[i]) / 2 + 1 << endl;
      return 0;
    } else if (n % prime[i] == 0) {
      cout << n / prime[i] << endl;
      return 0;
    }
  }
  puts("1");
  return 0;
}
