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 = 100006;
bool notprime[MAXN];
int prime[10000];
int cnt = 0;
void init() {
  memset(notprime, false, sizeof(notprime));
  notprime[0] = notprime[1] = true;
  for (int i = 2; i < MAXN; i++)
    if (!notprime[i]) {
      prime[cnt++] = i;
      if (i > MAXN / i) continue;
      for (int j = i * i; j < MAXN; j += i) notprime[j] = true;
    }
}
int main() {
  init();
  long long n;
  long long ans = 0;
  scanf("%lld", &n);
  if (n == 0)
    printf("0\n");
  else {
    for (int i = 0; i < cnt; i++) {
      while (n % prime[i] == 0) {
        n = n / prime[i];
        ans++;
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}
