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>
#pragma GCC optimize(3)
using namespace std;
const int maxn = 1;
inline char gc() {
  return getchar();
  static const int maxs = 1 << 16;
  static char buf[maxs], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, maxs, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long rd() {
  long long x = 0;
  char c = gc();
  bool neg = 0;
  while (c < '0' || c > '9') {
    if (c == '-') neg = 1;
    c = gc();
  }
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = gc();
  return neg ? (~x + 1) : x;
}
int main() {
  long long n = rd(), p = n;
  if (n % 2 == 0)
    printf("%I64d\n", n / 2);
  else {
    for (long long i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        p = i;
        break;
      }
    }
    printf("%I64d\n", (n - p) / 2 + 1);
  }
  return 0;
}
