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>
int prost(unsigned long long &n);
int main() {
  long long unsigned int n, count = 0, p = 1;
  scanf("%W", n);
  if (n == 0) return 0;
  if ((n % 2 != 0) && (n % 3 != 0) && (n % 5 != 0) && (n % 7 != 0) ||
      (n == 2) || (n == 3) || (n == 5) || (n == 7)) {
    printf("1");
    return 0;
  } else
    while (n != 0) {
      prost(n);
      count++;
    }
  printf("%d", count);
  return 0;
}
int prost(unsigned long long &n) {
  if (n % 2 == 0) {
    n -= 2;
    return n;
  }
  if (n % 3 == 0) {
    n -= 3;
    return n;
  }
  if (n % 5 == 0) {
    n -= 5;
    return n;
  }
  if (n % 7 == 0) {
    n -= 7;
    return n;
  }
}
