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 prime[100005];
bool isprime[100005];
long long pn = 0;
void calPrime() {
  memset(isprime, 1, sizeof(isprime));
  isprime[1] = false;
  isprime[2] = true;
  prime[pn++] = 2;
  for (int j = 2; j <= 100000; j++) {
    if (isprime[j]) {
      prime[pn++] = j;
      for (int i = 2;; i++) {
        int t = i * j;
        if (t > 100000)
          break;
        else
          isprime[t] = false;
      }
    }
  }
}
bool isPrime(long long a) {
  long long t = sqrt(a);
  for (long long i = 2; i <= t; i++) {
    if (a % i == 0) return false;
  }
  return true;
}
int main() {
  long long n;
  while (cin >> n) {
    long long ans = 0;
    calPrime();
    while (n != 0) {
      if (isPrime(n)) {
        ans++;
        break;
      } else {
        long long sub = 2;
        for (long long i = 0; i < pn; i++) {
          if (n % prime[i] == 0) {
            ans = ans + (n - n / prime[i]) / prime[i];
            n /= prime[i];
            break;
          }
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}
