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 int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
std::vector<long long int> primes_upto(long long int N) {
  const long long int M = N / 2;
  std::vector<bool> sieve(M, true);
  for (long long int i = 1; i < M; ++i) {
    const long long int L = (M - i) / (2 * i + 1);
    for (long long int j = i; j <= L; ++j) sieve[i + j + 2 * i * j] = false;
  }
  std::vector<long long int> primes;
  primes.push_back(2);
  for (long long int i = 1; i < M; ++i)
    if (sieve[i]) primes.push_back(i * 2 + 1);
  return primes;
}
long long int smallest_prime_factor(long long int N) {
  if (N < 2) return N;
  std::vector<long long int> primes = primes_upto(std::sqrt(N) + 1);
  for (std::vector<long long int>::size_type i = 0; i < primes.size(); ++i)
    if ((N % primes[i]) == 0) return primes[i];
  return N;
}
int main() {
  long long int n;
  cin >> n;
  long long int ans;
  if (isPrime(n))
    cout << "1" << endl;
  else {
    if (n % 2 == 0) {
      cout << n / 2 << endl;
    } else {
      cout << n / smallest_prime_factor(n) << endl;
    }
  }
}
