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: Wrong Answer)

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

namespace AtCoder20190615
{
	class ProgramB
	{
		struct Ball
		{
			public Ball(int x, int y) { X = x; Y = y; }
			public int X;
			public int Y;
		}
		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());
			}
		}

		// コスト計算
		static int calcCost(int N, Ball[] balls, int p, int q)
		{
			var totalCost = 0;
			for (int i = 0; i < N; i++)
			{
				var ball = balls[i];
				var cost = 1;
				if (i > 0)
				{
					var prev = balls[i - 1];
					if (ball.X - p == prev.X && ball.Y - q == prev.Y) cost = 0;
				}
				totalCost += cost;
			}
			return totalCost;
		}

		// 最も効率的な p, q を探す
		static Tuple<int, int> 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>();
			for (int i = 0; i < N; i++)
			{
				var ball = balls[i];
				if (i > 0)
				{
					var prev = balls[i - 1];
					var goodP = ball.X - prev.X;
					var goodQ = ball.Y - prev.Y;
					ps.Add(goodP);
					qs.Add(goodQ);
				}
			}

			// 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 Tuple.Create(bestP, bestQ);
		}

		static void B(int N, Ball[] balls)
		{
			// 実際の計算
			Tuple<int, int> PQ = findGoodPq(N, balls);
			int P = PQ.Item1;
			int Q = PQ.Item2;
			var result = calcCost(N, balls, P, Q);
			Console.WriteLine(result);
		}
	}
}