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 N = 1e5 + 10;
const int K = 200;
const int mod = 998244353;
int power(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) {
      res = 1LL * res * a % mod;
    }
    a = 1LL * a * a % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  long long n, cnt = 0;
  scanf("%lld", &n);
  if (n % 2 == 0) {
    printf("%lld\n", n / 2);
  } else {
    long long idx = n;
    for (long long i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        idx = i;
        break;
      }
    }
    if (idx == n)
      printf("1\n");
    else {
      printf("%lld\n", n / 2 + 1);
    }
  }
  return 0;
}
