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: Accepted)

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

#define rep(i,n) for(int i = 0; i < n; i++)
#define REP(i,n) for(int i = 1; i < n; i++)
#define per(i,n) for(int i = n-1; i >= 0; i--)
#define PER(i,n) for(int i = n; i >= 1; i--)
#define sz(x) int(x.size())
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define YesorNo(a) printf(a ? "Yes\n" : "No\n")
#define endl '\n'
#define fi first
#define se second
using ll = long long;
using P = pair<int,int>;
using Pl = pair<ll,ll>;
using vi = vector<int>;
using vii = vector<vi>;
using vl = vector<ll>;
using vll = vector<vl>;
using vs = vector<string>;
using vb = vector<bool>;
using vc = vector<char>;
using vcc = vector<vc>;
using vP = vector<P>;
using vPl = vector<Pl>;
const int dx[] = {0,1,0,-1,1,1,-1,-1};
const int dy[] = {1,0,-1,0,1,-1,-1,1};
const int inf = (1<<30)-1;
const ll infll = (1LL<<62)-1;
ll ceil(const ll &a, const ll &b){return ((a)+(b)-1)/b;}
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }


struct UnionFind { // UnionFindTree
  vector<int> d;
  UnionFind(int n): d(n,-1) {}
  int find(int x) {
    if(d[x] < 0) return x;
    return d[x] = find(d[x]);
  }
  bool unite(int x, int y) {
    x = find(x);
    y = find(y);
    if(x == y) return false;
    if(-d[x] < -d[y]) swap(x,y);
    d[x] += d[y];
    d[y] = x;
    return true;
  }
  bool same(int x, int y) { return find(x) == find(y); }
  int size(int x) { return -d[find(x)]; }
};


int main(){
  ios::sync_with_stdio(false);
  cin.tie(nullptr);

  int n;
  cin >> n;
  vi x(n), y(n);
  rep(i,n) cin >> x[i] >> y[i];
  int ans = inf;
  rep(i,n) {
    rep(j,n) {
      int dix = x[i] - x[j];
      int diy = y[i] - y[j];
      UnionFind u(n);
      rep(k,n) {
        rep(l,n) {
          if(x[k] - x[l] == dix && y[k] - y[l] == diy) {
            u.unite(k,l);
          }
        }
      }
      set<int> st;
      rep(k,n) {
        st.insert(u.find(k));
      }
      ans = min(ans, sz(st));
    }
  }
  cout << ans << endl;
}