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;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

class Program
{
    const int MOD = 1000000007;
    static void Main(string[] args)
    {
        new Program().Solve();
    }

    void Solve()
    {
        Scanner cin = new Scanner();
        int n = cin.NextInt();
        long[] x = new long[n];
        long[] y = new long[n];
        long[] diffx = new long[n];
        long[] diffy = new long[n];

        for (int i = 0; i < n; i++)
        {
            if (i == 0)
            {
                x[i] = cin.NextLong();
                y[i] = cin.NextLong();
                continue;
            }
            x[i] = cin.NextLong();
            diffx[i - 1] = x[i] - x[i - 1];
            y[i] = cin.NextLong();
            diffy[i - 1] = y[i] - y[i - 1];
        }

        var diffxUnique = diffx.Distinct();
        var diffyUnique = diffy.Distinct();

        Dictionary<long, long> countx = new Dictionary<long, long>();
        Dictionary<long, long> county = new Dictionary<long, long>();
        foreach (var num in diffxUnique)
        {
            long c = diffx.Where(e => e == num).Count();
            countx.Add(num, c);
        }

        foreach (var num in diffyUnique) 
        {
            long c = diffy.Where(e => e == num).Count();
            county.Add(num, c);
        }

        long maxCountX = countx.Select(e => e.Value).Max();
        long maxKeyX = countx.Where(e => e.Value == maxCountX).First().Key;
        long maxCountY = county.Select(e => e.Value).Max();
        long maxKeyY = county.Where(e => e.Value == maxCountY).First().Key;

        long commonDiffNum = Math.Min(maxCountX, maxCountY);
        Console.WriteLine(n - commonDiffNum);
    }
}


class Scanner
{
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner()
    {
        s = new string[0];
        i = 0;
    }

    public string Next()
    {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        if (s.Length == 0) return Next();
        i = 0;
        return s[i++];
    }

    public int NextInt()
    {
        return int.Parse(Next());
    }

    public char[] ArrayChar()
    {
        return Next().ToCharArray();
    }
    public int[] ArrayInt(int N, int add = 0)
    {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = NextInt() + add;
        }
        return Array;
    }

    public long NextLong()
    {
        return long.Parse(Next());
    }

    public long[] ArrayLong(int N, long add = 0)
    {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = NextLong() + add;
        }
        return Array;
    }

    public double NextDouble()
    {
        return double.Parse(Next());
    }


    public double[] ArrayDouble(int N, double add = 0)
    {
        double[] Array = new double[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = NextDouble() + add;
        }
        return Array;
    }
}