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.

Examples

Input

2
1 1
2 2


Output

1


Input

3
1 4
4 6
7 8


Output

1


Input

4
1 1
1 2
2 1
2 2


Output

2
import java.util.*;
import java.io.*;

public class Main {
	static int mod = (int) 1e9 + 7;
	static int DX[] = { -1, 0, 1, 0 }, DY[] = { 0, -1, 0, 1 };
	static final int INF = Integer.MAX_VALUE / 3;
	static final long LINF = Long.MAX_VALUE / 3;
	static int p[][];
	static boolean visited[];
	static int n;

	public static void main(String[] args) {
		FastScanner fs = new FastScanner(System.in);
		n = fs.nextInt();
		p = new int[n][2];
		for(int i=0;i<n;i++) {
			int x = fs.nextInt(), y = fs.nextInt();
			p[i][0] = x; p[i][1] = y;
		}
		
		if(n==1) {
			System.out.println(1);
			return;
		}
		
		Arrays.sort(p,(p1,p2)->{
			if(p1[0]==p2[0]) return p1[1] - p2[1];
			return p1[0] - p2[0];
		});
		
		List<int[]> ls = new ArrayList<>();
		for(int i=0;i<n;i++) {
			for(int j=0;j<n;j++) {
				if(i==j)continue;
				int t[] = {p[i][0] - p[j][0], p[i][1] - p[j][1]};
				ls.add(t);
			}
		}
		
		int ans = INF;
		for(int[] t:ls) {
			int v = t[0], w = t[1];
			visited = new boolean[n];
			int res = 0;
			
			for(int i=0;i<n;i++) {
				if(visited[i])continue;
				res++;
				dfs(i,v,w);
			}
			ans = Math.min(ans, res);
		}
		System.out.println(ans);
	}

	static void dfs(int x, int v, int w) {
		visited[x]=true;
		int nx = p[x][0] + v, ny = p[x][1] + w;
		for(int i=0;i<n;i++) {
			if(p[i][0]==nx&&p[i][1]==ny) {
				dfs(i, v, w);
				break;
			}
		}
	}
	
	
	//MOD culculations
	static long plus(long x, long y) {
		long res = (x + y) % mod;
		return res < 0 ? res + mod : res;
	}

	static long sub(long x, long y) {
		long res = (x - y) % mod;
		return res < 0 ? res + mod : res;
	}

	static long mul(long x, long y) {
		long res = (x * y) % mod;
		return res < 0 ? res + mod : res;
	}

	static long div(long x, long y) {
		long res = x * pow(y, mod - 2) % mod;
		return res < 0 ? res + mod : res;
	}

	static long pow(long x, long y) {
		if (y < 0)
			return 0;
		if (y == 0)
			return 1;
		if (y % 2 == 1)
			return (x * pow(x, y - 1)) % mod;
		long root = pow(x, y / 2);
		return root * root % mod;
	}
}

//高速なScanner
class FastScanner {
	private BufferedReader reader = null;
	private StringTokenizer tokenizer = null;

	public FastScanner(InputStream in) {
		reader = new BufferedReader(new InputStreamReader(in));
		tokenizer = null;
	}

	public String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public String nextLine() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken("\n");
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}