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# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;

namespace AtCoder20190615
{
	public static class Util
	{
		// https://codereview.stackexchange.com/questions/91808/permutations-in-c
		public static IEnumerable<T[]> GetPermutations<T>(T[] values)
		{
			if (values.Length == 1)
				return new[] { values };

			return values.SelectMany(v => GetPermutations(values.Except(new[] { v }).ToArray()),
				(v, p) => new[] { v }.Concat(p).ToArray());
		}
	}

	class ProgramB
	{
		class Ball
		{
			public Ball(int x, int y) { X = x; Y = y; }
			public int X;
			public int Y;
		}
		struct PQ
		{
			public PQ(int p, int q) { P = p; Q = q; }
			public int P;
			public int Q;
		}
		static void Main(string[] args)
		{
			if (true)
			{
				var N = int.Parse(Console.ReadLine());
				var balls = new List<Ball>();
				for (int i = 0; i < N; i++)
				{
					var array = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();
					var ball = new Ball(array[0], array[1]);
					balls.Add(ball);
				}
				B(N, balls.ToArray());
			}
			else
			{
				B(2, new Ball[] { new Ball(1, 1), new Ball(2, 2) });
				B(3, new Ball[] { new Ball(1, 4), new Ball(4, 6), new Ball(7, 8) });
				B(4, new Ball[] { new Ball(1, 1), new Ball(1, 2), new Ball(2, 1), new Ball(2, 2) });

				var r = new Random();
				List<Ball> balls = new List<Ball>();
				for (int i = 0; i < 50; i++)
				{
					balls.Add(new Ball(-(r.Next() % 1000000000), -(r.Next() % 1000000000)));
				}
				B(50, balls.ToArray());
			}
		}

		// 巡回順序の全パターン
		static IEnumerable<int[]> visitPatterns(int N)
		{
			return Util.GetPermutations(Enumerable.Range(0, N).ToArray());
		}

		// コスト計算
		static int calcCost(int N, Ball[] balls, int p, int q)
		{
			int bestCost = int.MaxValue;
			foreach (var visitPattern in visitPatterns(N))
			{
				int totalCost = 0;
				Ball prev = null;
				foreach (var i in visitPattern)
				{
					var ball = balls[i];
					var cost = 1;
					if (prev != null)
					{
						if (ball.X - p == prev.X && ball.Y - q == prev.Y) cost = 0;
					}
					totalCost += cost;
					prev = ball;
				}
				if (totalCost < bestCost) bestCost = totalCost;
			}
			return bestCost;
		}

		// 最も効率的な p, q を探す
		static PQ findGoodPq(int N, Ball[] balls)
		{
			// あり得る p, q を抽出
			HashSet<int> ps = new HashSet<int>();
			HashSet<int> qs = new HashSet<int>();
			foreach (var visitPattern in visitPatterns(N))
			{
				Ball prev = null;
				foreach (var i in visitPattern)
				{
					var ball = balls[i];
					if (prev != null)
					{
						var goodP = ball.X - prev.X;
						var goodQ = ball.Y - prev.Y;
						ps.Add(goodP);
						qs.Add(goodQ);
					}
					prev = ball;
				}
			}

			// 全ての p, q の組み合わせでコスト算出
			int bestCost = int.MaxValue;
			int bestP = 0;
			int bestQ = 0;
			foreach (int p in ps)
			{
				foreach (int q in qs)
				{
					// p == 0 && q == 0 は対象から外す
					if (p == 0 && q == 0) continue;

					var cost = calcCost(N, balls, p, q);
					if (cost < bestCost)
					{
						bestCost = cost;
						bestP = p;
						bestQ = q;
					}
				}
			}
			return new PQ(bestP, bestQ);
		}

		static void B(int N, Ball[] balls)
		{
			// 実際の計算
			PQ pq = findGoodPq(N, balls);
			var result = calcCost(N, balls, pq.P, pq.Q);
			Console.WriteLine(result);
		}
	}
}