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;
const int maxn = 2e5 + 100;
const int mod = 1e9 + 7;
int x[maxn];
long long pri[maxn];
int cnt;
void Ini() {
  cnt = 0;
  for (long long i = 2; i < maxn; i++) {
    if (x[i] == 0) {
      pri[cnt++] = i;
      for (long long j = i * 2; j < maxn; j += i) {
        x[j] = 1;
      }
    }
  }
}
void solve() {
  Ini();
  long long n;
  cin >> n;
  int q = sqrt(n);
  int flag = 0;
  for (int i = 2; i <= q + 1; i++) {
    if (q % i == 0) {
      flag = 1;
      break;
    }
  }
  if (!flag) {
    printf("1\n");
    return;
  } else {
    long long res = 0;
    for (int i = 0; i < cnt; i++) {
      if (n % pri[i] == 0) {
        res = n / pri[i];
        break;
      }
    }
    printf("%lld\n", res);
  }
}
int main() {
  solve();
  return 0;
}
