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 isPrime(long long x) {
  for (long long i = 2; i <= x / 2; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
int main() {
  long long n;
  int cnt = 0;
  int prime[100001];
  bool vis[100001];
  memset(vis, false, sizeof(vis));
  memset(prime, 0, sizeof(prime));
  for (int i = 2; i <= 90000; i++) {
    if (!vis[i]) prime[cnt++] = i;
    for (int j = 0; j < cnt && i * prime[j] <= 90000; j++) {
      vis[i * prime[j]] = true;
      if (i % prime[j] == 0) break;
    }
  }
  while (~scanf("%lld", &n)) {
    if (isPrime(n)) {
      printf("1\n");
    } else {
      for (int i = 0; i < cnt; i++) {
        if (n % prime[i] == 0) {
          printf("%d\n", n / prime[i]);
          break;
        }
      }
    }
  }
  return 0;
}
