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;
inline void read(int &x) {
  char ch;
  bool flag = false;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') flag = true;
  for (x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  x = flag ? -x : x;
}
int f(long long x) {
  for (int i = 2; i <= sqrt(x) + 1; i++) {
    if (x % i == 0) return 1;
  }
  return 0;
}
int main() {
  long long n;
  cin >> n;
  if (f(n) == 0) {
    cout << 1 << endl;
  } else {
    for (int i = 2; i <= sqrt(n) + 1; i++) {
      if (n % i == 0) {
        cout << i << endl;
        break;
      }
    }
  }
  return 0;
}
