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++ (GCC 9.2.1) (judge verdict: Accepted)

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

// エイリアス
using  ll = long signed long;
using ull = long unsigned long;
using  ld = long double;
using   P = pair<int, int>;
using llP = pair<ll, ll>;
using DoP = pair<double, double>;

// 汎用マクロ
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define rep(i, n) for (int i = 0; i < n; ++i)
#define FOREACH(i,q) for (auto &i : q)
#define UNIQUE(v) do { sort((v).begin(), (v).end()); (v).erase(unique((v).begin(), (v).end()), (v).end()); } while (false)
#define SZ(x) ((int)(x).size())
template <class T>inline bool chmin(T &a, const T b) {if (a > b) { a = b; return true;} return false;}
template <class T>inline bool chmax(T &a, const T b) {if (a < b) { a = b; return true;} return false;}
template <class T>inline void line_out(const vector<T> vec) {int n = SZ(vec); rep(i, n) { cout << vec[i]; if(i < n-1) cout << " ";}cout << endl;}
const int di[] = {0, 1, 0, -1, 1, 1, -1, -1};
const int dj[] = {1, 0, -1, 0, 1, -1, 1, -1};
const int INF = 1 << 28;
const ll INF64 = 1ll << 55;
const int mod = 1000000007;
//const int mod = 998244353;
struct mint
{
  ll x; // typedef long long ll;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint operator-() const { return mint(-x); }
  mint &operator+=(const mint a)
  {
    if ((x += a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator-=(const mint a)
  {
    if ((x += mod - a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator*=(const mint a)
  {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const
  {
    mint res(*this);
    return res += a;
  }
  mint operator-(const mint a) const
  {
    mint res(*this);
    return res -= a;
  }
  mint &operator++()
  {
    if ((x += 1) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator--()
  {
    if ((x += mod - 1) >= mod)
      x -= mod;
    return *this;
  }
  mint operator*(const mint a) const
  {
    mint res(*this);
    return res *= a;
  }
  mint pow(ll t) const
  {
    if (!t)
      return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1)
      a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const
  {
    return pow(mod - 2);
  }
  mint &operator/=(const mint a)
  {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const
  {
    mint res(*this);
    return res /= a;
  }
};

int main()
{
  int n; cin >> n;
  vector<llP> point(n);
  rep(i, n) cin >> point[i].first >> point[i].second;
  if(n == 1) {
    cout << 1 << endl;
    return 0;
  }
  sort(ALL(point));

  vector<ll> p, q;

  rep(i, n) rep(j, n) {
    if(i == j) continue;
    p.push_back(point[i].first - point[j].first);
    q.push_back(point[i].second - point[j].second);
  }

  int m = p.size();
  int ans = INF;
  rep(k, m) {
    int tmp = n;
    rep(i, n) {
      if(binary_search(ALL(point), make_pair(point[i].first-p[k], point[i].second-q[k]))) --tmp;
    }
    chmin(ans, tmp);
  }
  cout << ans << endl;
}