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

import java.io.*;
import java.io.ObjectInputStream.*;
import java.util.*;
import java.util.regex.*;


public class Main {

	final static long MOD = 1000000007;
	//final static long MOD = 998244353;

	public static String[] Ws = null;
	public static int wsIndx = 0;
	public static BufferedReader in = null;
	public static long cnt=0;
	public static PrintWriter pw = new PrintWriter(System.out);
	public static void main(String[] args) throws Exception {
		slove(args);
		if(in!=null){
			in.close();
		}
		pw.flush();
		pw.close();
	}

	public static void slove(String[] args) throws Exception{

		int N = nextInt();
		int[][] p = new int[N][];
		for(int i=0;i<N;i++) {
			p[i]=nextInts();
		}


		Arrays.sort(p,(s1,s2)->s1[0]-s2[0]);


		int ans = 1;
		HashMap<String,Integer> map = new HashMap<String, Integer>();
		for(int i=0;i<N;i++) {
			for(int k=0;k<N;k++) {
				if(i==k)
					continue;
				String key = String.valueOf(p[k][0]-p[i][0])+","+String.valueOf(p[k][1]-p[i][1]);
				map.put(key,map.getOrDefault(key, 0)+1);
			}
		}
		for(int val:map.values()) {
			ans=Math.max(ans,val);
		}
		pw.println(N-ans);

	}


	public static void check() throws Exception{
		if(in == null){
			in = new BufferedReader(new InputStreamReader(System.in));
		}
		if(Ws==null || Ws.length<=wsIndx){
			Ws = in.readLine().split(" ");
			wsIndx=0;
		}
	}
	public static int nextInt()throws Exception{
		check();
		return Integer.parseInt(Ws[wsIndx++]);
	}

	public static long nextLong()throws Exception{
		check();
		return Long.parseLong(Ws[wsIndx++]);
	}

	public static String nextString()throws Exception{
		check();
		return Ws[wsIndx++];
	}

	public static int[] nextInts()throws Exception{
		check();
		int[] tmp = new int[Ws.length];
		for(int i=0;i<tmp.length;i++){
			tmp[i]=Integer.parseInt(Ws[i]);
		}
		wsIndx=Ws.length;
		return tmp;
	}

	public static long[] nextLongs()throws Exception{
		check();
		long[] tmp = new long[Ws.length];
		for(int i=0;i<tmp.length;i++){
			tmp[i]=Long.parseLong(Ws[i]);
		}
		wsIndx=Ws.length;
		return tmp;
	}

	public static String[] nextStrings()throws Exception{
		check();
		wsIndx=Ws.length;
		return Ws;
	}


}