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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define ll long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define pi 3.141592653589793
using namespace std;


pair<int, int> a[200000];
bool f(int i, int j) {
    return a[i + 1].first - a[i].first == a[j + 1].first - a[j].first ? i < j : a[i + 1].first - a[i].first > a[j + 1].first - a[j].first;
}

int main() {
//    cout << fixed << setprecision(10);
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);



    ll n;
    cin >> n;
    cout << (n - 1) / 2;


}

//20
//668 684 759 389 948 900 70 733 713 903 940 207 297 687 512 891 498 2 564 465