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;
long long n;
vector<long long> prime;
vector<int> vis;
long long ep(long long n) {
  long long i, j;
  for (i = 2; i <= n; i++) {
    for (j = 2; j <= sqrt(i); j++) {
      if (i % j == 0) break;
    }
    if (j > sqrt(i)) {
      if (n % i == 0) return i;
    }
  }
  return -1;
}
bool isPrime(unsigned long n) {
  if (n == 2 || n == 3) return true;
  if (n % 6 != 1 && n % 6 != 5) return false;
  float nsqrt = floor(sqrt((float)n));
  for (int i = 6; i <= nsqrt; i += 6) {
    if (n % (i - 1) == 0 || n % (i + 1) == 0) return false;
  }
  return true;
}
int main() {
  long long sum = 0;
  scanf("%lld", &n);
  if (n % 2 == 0) {
    printf("%lld", n / 2);
    return 0;
  }
  if (isPrime(n)) {
    printf("1");
    return 0;
  }
  long long p = ep(n);
  if (p > 0)
    printf("%lld", (n - p) / 2 + 1);
  else
    printf("1");
  return 0;
}
