Problem:
Score : 100 points

Problem Statement

How many ways are there to choose two distinct positive integers totaling N, disregarding the order?

Constraints

1 \leq N \leq 10^6

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

4

Sample Output 1

1

There is only one way to choose two distinct integers totaling 4: to choose 1 and 3. (Choosing 3 and 1 is not considered different from this.)

Sample Input 2

999999

Sample Output 2

499999

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
int acs(const void *a, const void *b) { return *(int *)a - *(int *)b; }                            /* 1,2,3,4.. */
int des(const void *a, const void *b) { return *(int *)b - *(int *)a; }                            /* 8,7,6,5.. */
int cmp_char(const void *a, const void *b) { return *(char *)a - *(char *)b; }                     /* a,b,c,d.. */
int cmp_str(const void *a, const void *b) { return strcmp(*(const char **)a, *(const char **)b); } /* aaa,aab.. */
#define min(a, b) (a < b ? a : b)
#define max(a, b) (a > b ? a : b)
#define rep(i, l, r) for (int i = l; i < r; i++)
#define MAX 100001
#define MOD 1000000007

typedef long long int lli;

int main(void)
{
  int n;
  scanf("%d", &n);

  int ans = (n - 1) / 2;
  printf("%d\n", ans);
  return 0;
}