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 <iostream>
#include <fstream>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <unordered_set>
#include <bitset>
#include <limits>
#include <algorithm>

#define MOD_BIG 1000000007
#define MOD_BIG2 998244353
using std::string;
using std::cin;
using std::cout;
using std::cerr;
using std::endl;
using ll = long long;
using std::vector;

template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) {
    vec.resize(len);
    for (auto &v : vec) ndarray(v, args...);
}

//////////////////////////////////////

//////////////////////////////////////

int main() {
    
    int N;
    cin >> N;
    
    ll ans = 0;
    for (int i = 1; i <= N/2; i++) {
        if (N - i > 1 && N-i != i) {
            ans += 1;
        } else {
            break;
        }
    }
    cout << ans << endl;
    return 0;
}