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;
struct Fast {
    Fast() {
        std::cin.tie(0);
        ios::sync_with_stdio(false);
        cout.precision(20);
    }
} fast;

/* define */
#define FOR(I, X, Y) for (long long(I) = (X); (I) < (Y); (I)++)
#define REP(I, X, Y) for (long long(I) = (Y)-1; (I) >= (X); (I)--)
#define ALL(X) (X).begin(), (X).end()
#define pb push_back
#define COUNT(V, X)                           \
    (upper_bound((V).begin(), (V).end(), X) - \
     lower_bound((V).begin(), (V).end(), X))
#define debug(x) cerr << #x << ':' << x << endl;
#define DEBUG(v)                             \
    {                                        \
        cerr << #v << ':';                   \
        for (auto xv : v) cerr << xv << ' '; \
        cerr << endl;                        \
    }
#define Yes(X) cout << (X ? "Yes" : "No") << endl;
#define YES(X) cout << (X ? "YES" : "NO") << endl;
#define ctoi(C) (C - '0')
#define pow2(x) ((long long)((long long)1 << x))

/* alias */
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vii = vector<vector<int>>;
using vl = vector<long long>;
using vll = vector<vector<long long>>;
using pi = pair<int, int>;
using pl = pair<long long, long long>;
template <typename T>
using PQ = priority_queue<T>;
template <typename T>
using minPQ = priority_queue<T, vector<T>, greater<T>>;

/* const */
const long long dx[] = {1, 0, -1, 0};
const long long dy[] = {0, 1, 0, -1};
const long long dx8[] = {1, 1, 0, -1, -1, -1, 0, 1};
const long long dy8[] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long dx9[] = {1, 1, 0, -1, -1, -1, 0, 1, 0};
const long long dy9[] = {0, 1, 1, 1, 0, -1, -1, -1, 0};
const int INF = 1000000007;
const long long LINF = 1000000000000000007;

/* func */

/* main */

signed main() {
    int N;
    cin >> N;
    ll x[N], y[N];
    FOR(i, 0, N) {
        cin >> x[i] >> y[i];
    }
    if (N == 1) {
        cout << 1 << endl;
        return 0;
    }
    ll ans = 0;
    FOR(i, 0, N) {
        FOR(j, i + 1, N) {
            ll p = x[j] - x[i];
            ll q = y[j] - y[i];
            ll counter = 0;
            FOR(k, 0, N) {
                FOR(l, 0, N) {
                    if (x[l] - x[k] == p && y[l] - y[k] == q) {
                        counter += 1;
                    }
                }
            }
            ans = max(ans, counter);
        }
    }
    cout << N - ans << endl;
}