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 = 100005;
const int mod = 1000000007;
const int N = 100005;
bool mark[N];
vector<long long> prime;
void sieve() {
  mark[1] = true;
  prime.push_back(2);
  long long limit = sqrt(N) + 5;
  for (long long i = 4; i < N; i += 2) mark[i] = true;
  for (long long i = 3; i < N; i += 2) {
    if (!mark[i]) {
      prime.push_back(i);
      if (i <= limit) {
        for (long long j = i * i; j < N; j += 2 * i) {
          mark[j] = true;
        }
      }
    }
  }
}
int main() {
  sieve();
  long long n;
  scanf("%lld", &n);
  long long ans = 0;
  if (n % 2 == 0)
    ans = n / 2;
  else {
    for (int i = 0; i < prime.size(); i++) {
      if (n % prime[i] == 0) {
        ans++;
        n -= prime[i];
        break;
      }
    }
    ans += n / 2;
  }
  printf("%lld\n", ans);
}
