Problem:
Score : 300 points

Problem Statement

There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).

We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:

Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.

Find the minimum total cost required to collect all the balls when we optimally choose p and q.

Constraints

1 \leq N \leq 50

|x_i|, |y_i| \leq 10^9

If i \neq j, x_i \neq x_j or y_i \neq y_j.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
:
x_N y_N

Output

Print the minimum total cost required to collect all the balls.

Sample Input 1

2
1 1
2 2

Sample Output 1

1

If we choose p = 1, q = 1, we can collect all the balls at a cost of 1 by collecting them in the order (1, 1), (2, 2).

Sample Input 2

3
1 4
4 6
7 8

Sample Output 2

1

If we choose p = -3, q = -2, we can collect all the balls at a cost of 1 by collecting them in the order (7, 8), (4, 6), (1, 4).

Sample Input 3

4
1 1
1 2
2 1
2 2

Sample Output 3

2

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

    #include <bits/stdc++.h>
    using namespace std;

    //long long
    using ll = long long;

    // pair<int, int>
    using PII = pair<int, int>;

    //最大値、mod
    const int MOD = 1000000007;
    const int INF = 1000000000;
    const long long LINF = 1e18;

    //出力系
    #define print(x) cout << x << endl
    #define prints(x) cout << fixed << setprecision(10) << x << endl
    #define printc(x) cout << setw(2) << setfill('0') << x << endl;
    #define yes cout << "Yes" << endl
    #define YES cout << "YES" << endl
    #define no cout << "No" << endl
    #define NO cout << "NO" << endl
    
    // begin() end()
    #define all(x) (x).begin(),(x).end()

    //for
    #define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)

    //最大公約数 
    unsigned gcd(unsigned a, unsigned b) {
      if(a < b) return gcd(b, a);
      unsigned r;
      while ((r=a%b)) {
        a = b;
        b = r;
      }
      return b;
    }

    // 最小公倍数
    unsigned lcm(unsigned a, unsigned b){
        return a / gcd(a, b) * b; 
    }

    // a = max(a, b), a = min(a, b)
    template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
    template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

    //int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};









int main(){
  int N;
  cin >> N;
  int cnt = 0;
  vector<ll> x(N), y(N);
  REP(i, N) cin >> x.at(i) >> y.at(i);
  for(int i = 0; i < N; i++){
    for(int j = i + 1; j < N; j++){
      int tmp = 0;
      int a = x.at(i) - x.at(j);
      int b = y.at(i) - y.at(j);
      for(int i1 = 0; i1 < N; i1++)
        for(int j1 = i1 + 1; j1 < N; j1++){
          if(a == x.at(i1) - x.at(j1) && b == y.at(i1) - y.at(j1) && i1 != j1) tmp++;
        }
        chmax(cnt, tmp);
    }
  }
  int ans = N - cnt;
  print(ans);
  return 0;
}