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 = 2e5 + 5;
bool not_prime[maxn];
vector<int> prime;
void init() {
  for (int i = 2; i < maxn; i++) {
    if (!not_prime[i]) prime.emplace_back(i);
    int sz = prime.size();
    for (int j = 0; j < sz && i * prime[j] < maxn; j++) {
      not_prime[i * prime[j]] = true;
      if (i % prime[j] == 0) break;
    }
  }
}
bool checkp(long long x) {
  for (auto i : prime) {
    if (1LL * i * i > x) break;
    if (x % i == 0) return false;
  }
  return true;
}
int getp(long long x) {
  int p = upper_bound(prime.begin(), prime.end(), int(sqrt(1.0 * x) + 0.5)) -
          prime.begin();
  for (int i = p; i >= 0; i--) {
    if (x % prime[i] == 0) return prime[i];
  }
  cout << "error\n";
}
long long eular(long long x) {
  long long ret = 1;
  for (int i = 2; 1LL * i * i <= x; i++) {
    if (x % i == 0) {
      x /= i, ret *= i - 1;
      while (x % i == 0) {
        x /= i;
        ret *= i;
      }
    }
  }
  if (x > 1) ret *= x - 1;
  return ret;
}
int main() {
  long long n;
  cin >> n;
  cout << n - eular(n) << endl;
  return 0;
}
