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 modulo(long long num, long long MOD = 1000000007) {
  return ((num % MOD) + MOD) % MOD;
}
long long power(long long b, long long e, long long MOD = 1000000007) {
  long long ans = 1;
  while (e) {
    if (e % 2) ans = (ans * b) % MOD;
    b = (b * b) % MOD;
    e /= 2;
  }
  return ans;
}
long long inv(long long num, long long MOD = 1000000007) {
  return power(modulo(num), MOD - 2, MOD);
}
bool isprime(long long num) {
  if (num == 1) return 0;
  for (long long i = 2; i * i <= num; i++) {
    if (num % i == 0) return 0;
  }
  return 1;
}
long long smallest(long long num) {
  for (long long i = 2;; i++) {
    if (num % i == 0) return i;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long num;
  cin >> num;
  long long cnt = 0;
  while (!isprime(num)) {
    if (num % 2 == 0) {
      cnt = cnt + (num / 2) - 1;
      break;
    }
    cnt++;
    num = num - smallest(num);
  }
  cout << cnt + 1;
  return 0;
}
