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 INF = 1000000005;
const long long INFLL = 1000000000000000002ll;
const long long MOD = 998244353;
inline long long min(long long a, long long b, long long c) {
  return min(min(a, b), c);
}
inline long long min(long long a, long long b, long long c, long long d) {
  return min(min(min(a, b), c), d);
}
inline long long max(long long a, long long b, long long c) {
  return max(max(a, b), c);
}
inline long long max(long long a, long long b, long long c, long long d) {
  return max(max(max(a, b), c), d);
}
bool U[100005];
vector<long long> Primes;
long long make(long long n) {
  for (long long i = 2; i <= 100000; i++)
    if (n % i == 0) return n / i;
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  cout << make(n);
}
