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;
int a[100100];
void sieve() {
  int i, j, s, x = 0;
  int n = 100100;
  a[1] = a[0] = 1;
  s = sqrt(n);
  for (i = 2; i <= s; i++) {
    if (a[i] == 0) {
      for (j = i * i; j <= n; j += i) a[j] = 1;
    }
  }
  for (i = 0; i < 100; i++) {
  }
}
int main() {
  long long int n, d, c = 0, sq, f, i = 2, g;
  sieve();
  cin >> n;
  for (;;) {
    f = 0;
    sq = sqrt(n);
    for (;;) {
      g = 0;
      if (a[i] == 0) {
        if (n % i == 0) {
          d = i;
          f = 1;
          g = 1;
          break;
        }
      }
      if (g == 0) {
        i++;
        if (i > sq) break;
      }
    }
    if (f == 0) {
      cout << "1" << endl;
      break;
    } else {
      cout << n / d << endl;
      break;
    }
  }
  return 0;
}
