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;
vector<long long> p;
long long vis[1000010];
void kb() {
  long long n = 1000000;
  long long m = sqrt(n + 0.5);
  for (long long i = 2; i <= m; i++)
    if (!vis[i])
      for (long long j = i * i; j <= n; j += i) vis[j] = 1;
  for (long long i = 2; i <= n; i++) {
    if (vis[i] == 0) p.push_back(i);
  }
}
long long is(long long n) {
  long long m = sqrt(n);
  long long i;
  for (i = 2; i <= m; i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
int main() {
  kb();
  long long n;
  cin >> n;
  long long num = 0, i;
  for (i = 0; i < p.size(); i++) {
    if (n % p[i] == 0) {
      num += n / p[i];
      n = n - num * p[i];
      if (n == 0) break;
    }
  }
  if (num == 0)
    printf("1\n");
  else
    cout << num << endl;
}
