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: Accepted)

using System;
using System.Collections.Generic;

namespace diverta2019
{
	class Program
	{
		static void Main(string[] args)
		{
			// 入力
			int N = int.Parse(Console.ReadLine());
			int[] x = new int[N];
			int[] y = new int[N];
			for (int i = 0; i < N; i++)
			{
				int[] inp = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
				x[i] = inp[0];
				y[i] = inp[1];
			}
			// 処理(N<=50)
			// 1:(p,q)の候補は、(x[i]-x[j],y[i]-y[j])(0<=i,j<N)の全ての組み合わせ(O(N^2))
			// 2:(p,q)が決まったとして、そのコストを求めるには(x[i]-x[j],y[i]-y[j])(0<=j<N,j!=i)となるjが存在するiの個数を求めれば良い(O(N^2))
			// 上手くデータを保存しておけば、同時並行でデータを作っていけるのでは？
			// 1
			List<List<int>> subXList=new List<List<int>>(N);
			List<List<int>> subYList=new List<List<int>>(N);
			for (int i = 0; i < N; i++)
			{
				List<int> listX = new List<int>(N);
				List<int> listY = new List<int>(N);
				for (int j = 0; j < N; j++)
				{
					listX.Add(x[i] - x[j]);
					listY.Add(y[i] - y[j]);
				}
				subXList.Add(listX);
				subYList.Add(listY);
			}
			// 2
			int minCost = N;
			for (int a = 0; a < N; a++)
			{
				for (int b = 0; b < N; b++)
				{
					if (a==b)
					{
						// i=jとなる場合は計算しない
						continue;
					}
					int subX = subXList[a][b];
					int subY = subYList[a][b];
					int count = 0;
					for (int c = 0; c < N; c++)
					{
						for (int d = 0; d < N; d++)
						{
							// 同一
							if (subXList[c][d]==subX && subYList[c][d]==subY)
							{
								count++;
								break;
							}
						}
					}
					// 最小コスト更新
					int cost = N - count;
					if (minCost>cost)
					{
						minCost = cost;
					}
				}
			}
			if (minCost==0)
			{
				minCost = 1;
			}
			Console.WriteLine(minCost);
		}
	}
}