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: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i, n) for (ll i = 0; i < (ll)(n); ++i)
#define rep2(i, s, n) for (ll i = s; i < (ll)(n); i++)
#define repr(i, n) for (ll i = n; i >= 0; i--)
#define pb push_back
#define COUT(x) cout << (x) << "\n"
#define COUTF(x) cout << setprecision(15) << (x) << "\n"
#define ENDL cout << "\n"
#define DF(x) x.erase(x.begin())
#define ALL(x) x.begin(), x.end()
#define SZ(x) (ll) x.size()
#define SORT(x) sort(ALL(x))
#define REVERSE(x) reverse(ALL(x))
#define ANS cout << ans << "\n"
#define init() \
  cin.tie(0);  \
  ios::sync_with_stdio(false)
#define LINE cerr << "[debug] line: " << __LINE__ << "\n";
#define debug(x) cerr << "[debug] " << #x << ": " << x << "\n";
#define debugV(v)                       \
  cerr << "[debugV] " << #v << ":";     \
  rep(i, v.size()) cerr << " " << v[i]; \
  cerr << "\n";
using namespace std;
using ll = long long;
using ld = long double;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using mll = map<ll, ll>;
using qll = queue<ll>;
using P = pair<ll, ll>;

constexpr ll INF = 0x3f3f3f3f3f3f3f3f;
constexpr ld PI = 3.141592653589793238462643383279;
ll get_digit(ll x) {
  return to_string(x).size();
}

ll gcd(ll x, ll y) {
  return y ? gcd(y, x % y) : x;
}

ll lcm(ll a, ll b) {
  return a / gcd(a, b) * b;
}

vector<P> factorize(ll n) {
  vector<P> result;
  for (ll i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      result.pb({i, 0});
      while (n % i == 0) {
        n /= i;
        result.back().second++;
      }
    }
  }
  if (n != 1) {
    result.pb({n, 1});
  }
  return result;
}

vll divisor(ll n) {
  vll ret;
  for (ll i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i * i != n) ret.push_back(n / i);
    }
  }
  SORT(ret);
  return (ret);
}

const int mod = 1000000007;
struct mint {
  ll x;
  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 {
    return mint(*this) += a;
  }
  mint operator-(const mint a) const {
    return mint(*this) -= a;
  }
  mint operator*(const mint a) const {
    return mint(*this) *= 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 {
    return mint(*this) /= a;
  }
};
istream& operator>>(istream& is, const mint& a) {
  return is >> a.x;
}
ostream& operator<<(ostream& os, const mint& a) {
  return os << a.x;
}

struct combination {
  vector<mint> fact, ifact;
  combination(ll n) : fact(n + 1), ifact(n + 1) {
    assert(n < mod);
    fact[0] = 1;
    for (ll i = 1; i <= n; ++i) {
      fact[i] = fact[i - 1] * i;
    }
    ifact[n] = fact[n].inv();
    for (ll i = n; i >= 1; --i) {
      ifact[i - 1] = ifact[i] * i;
    }
  }
  mint operator()(ll n, ll k) {
    if (k < 0 || k > n) return 0;
    return fact[n] * ifact[k] * ifact[n - k];
  }
};

signed main() {
  init();
  ll N;
  cin >> N;
  vector<pair<ld, ld>> C(N);
  rep(i, N) cin >> C[i].first >> C[i].second;
  ll ans = INF;
  rep(ii, N) {
    rep2(i, ii + 1, N) {
      ld gradient = (C[ii].first - C[i].first) / (C[ii].second - C[i].second);
      // debug(gradient);
      vector<bool> done(N, false);
      ll count = 0;
      rep(k, N) {
        if (!done[k]) {
          done[k] = true;
          count++;
          rep(j, N) {
            ld g = (C[k].first - C[j].first) / (C[k].second - C[j].second);
            if (gradient == g) done[j] = true;
          }
        }
      }
      // debug(count);
      ans = min(ans, count);
    }
  }
  ANS;
  return 0;
}