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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
	// public static void printArray(int[] a) {
	// 	for (int i = 0; i < a.length; i++) {
	// 		System.out.print(a[i] + ", ");
	// 	}
	// 	System.out.println();
	// }

	public static<T> void printArrayT(T[] a) {
		for (int i = 0; i < a.length; i++) {
			System.out.print(a[i] + ", ");
		}
		System.out.println();
	}

	public static class Tuple implements Comparable<Tuple> {
		public int X, Y;
		Tuple(int X, int Y) {
			this.X = X;
			this.Y = Y;
		}

		@Override
		public String toString() {
			return String.format("(%d, %d)", X, Y);
		}

		@Override
		public int compareTo(Tuple o) {
			int a = this.X - o.X;
			if (a != 0) return a;
			int b = this.Y - o.Y;
			return b;
		}
	}

	static int count(int N, Tuple[] xy, int rest, long flag, HashMap<Long, Integer> cache) {
		if (flag == (1 << N) - 1) {
			return 0;
		}
		if (cache.containsKey(flag)) {
			return cache.get(flag);
		}

		int min = (rest + 1) / 2;
		for (int i = 0; i < N; i++) {
			long flagI = 1 << i;
			if ((flag & flagI) != 0) continue;
			for (int j = i + 1; j < N; j++) {
				long flagJ = 1 << j;
				if ((flag & flagJ) != 0) continue;

				long p = ((long)xy[j].X) - ((long)xy[i].X);
				long q = ((long)xy[j].Y) - ((long)xy[i].Y);
				if (p == 0 && q == 0) continue;

				int cnt = 0;
				long copy = flag | flagI | flagJ;
				long nextX = ((long)xy[j].X) + p;
				long nextY = ((long)xy[j].Y) + q;

				for (int k = j + 1; k < N; k++) {
					if (xy[k].X < nextX) continue;
					if (xy[k].X > nextX) break;
					long flagK = 1 << k;
					if ((flag & flagK) != 0) continue;
					if (xy[k].Y == nextY) {
						copy |= flagK;
						cnt++;
						nextX += p;
						nextY += q;
					}
				}
				if (cnt > 0) {
					int nextRest = rest - cnt - 2;
					min = Math.min(min, count(N, xy, nextRest, copy, cache)) + 1;
				}
			}
		}
		cache.put(flag, min);
		return min;
	}

	// static int solve(int N, Tuple[] xy) {
	// 	int result = 0;
	// 	int rest = N;

	// 	long end = (1 << N) - 1;
	// 	long flag = 0;
	// 	while (rest > 0) {
	// 		int max = 0;
	// 		long maxFlag = 0;

	// 		for (int i = 0; i < N; i++) {
	// 			long flagI = 1 << i;
	// 			if ((flag & flagI) != 0) continue;
	// 			for (int j = i + 1; j < N; j++) {
	// 				long flagJ = 1 << j;
	// 				if ((flag & flagJ) != 0) continue;

	// 				long p = ((long)xy[j].X) - ((long)xy[i].X);
	// 				long q = ((long)xy[j].Y) - ((long)xy[i].Y);
	// 				if (p == 0 && q == 0) continue;

	// 				int cnt = 0;
	// 				long copy = flagI | flagJ;
	// 				long nextX = ((long)xy[j].X) + p;
	// 				long nextY = ((long)xy[j].Y) + q;

	// 				for (int k = j + 1; k < N; k++) {
	// 					if (xy[k].X < nextX) continue;
	// 					if (xy[k].X > nextX) break;
	// 					long flagK = 1 << k;
	// 					if ((flag & flagK) != 0) continue;
	// 					if (xy[k].Y == nextY) {
	// 						copy |= flagK;
	// 						cnt++;
	// 						nextX += p;
	// 						nextY += q;
	// 					}
	// 				}
	// 				if (max < cnt) {
	// 					max = cnt;
	// 					maxFlag = copy;
	// 				}
	// 			}
	// 		}
	// 		if (max >= 0) {
	// 			flag |= maxFlag;
	// 			result += 1;
	// 			rest -= (max + 2);
	// 		} else {
	// 			return result + (rest + 1) / 2;
	// 		}
	// 	}
	// 	return result;
	// }


	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		Tuple[] xy = new Tuple[N];
		for (int i = 0; i < N; i++) {
			int x = sc.nextInt();
			int y = sc.nextInt();
			xy[i] = new Tuple(x, y);
		}
		Arrays.sort(xy);
		// printArrayT(xy);
		int result = count(N, xy, N, 0, new HashMap<>());
		// int result = solve(N, xy);
		System.out.println(result);
	}
}