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 PyPy3 (2.4.0) (judge verdict: Accepted)

#!usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
import itertools
sys.setrecursionlimit(10**5)
stdin = sys.stdin
bisect_left = bisect.bisect_left
bisect_right = bisect.bisect_right
def LI(): return list(map(int, stdin.readline().split()))
def LF(): return list(map(float, stdin.readline().split()))
def LI_(): return list(map(lambda x: int(x)-1, stdin.readline().split()))
def II(): return int(stdin.readline())
def IF(): return float(stdin.readline())
def LS(): return list(map(list, stdin.readline().split()))
def S(): return list(stdin.readline().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = float('INF')

#A
def A():
    n, k = LI()
    a = n - k
    if k == 1:
        print(0)
        return
    print(a)
    return

#B
def B():
    n = II()
    xy = LIR(n)
    if n == 1:
        print(1)
        return
    ans = inf
    for i in range(n):
        for k in range(n):
            if i != k:
                b = 1
                check = []
                check.append((i,k))
                p = xy[i][0] - xy[k][0]
                q = xy[i][1] - xy[k][1]
                for l in range(n):
                    for m in range(n):
                        if not((l,m) in check) and p == xy[l][0] - xy[m][0] and q == xy[l][1] - xy[m][1]:
                            b += 1
                            check.append((l, m))
                ans = min(ans, n-b)
    print(ans)

    return

#C
def C():
    m = II()
    a = LI()
    a.sort()
    check = [0] * m
    anslis = []
    ans = 0
    if m % 2:
        ans += a[m // 2]
        for i in range(m//2):
            anslis.append((a[i], a[-1 - i]))
            anslis.append((ans,-a[-1 - i]+a[i]))
            ans += a[-1 - i] - a[i]
    else:
        ans += a[m//2] - a[m//2-1]
        for i in range(m//2-1):
            anslis.append((a[i], a[-1 - i]))
            anslis.append((ans,-a[-1 - i]+a[i]))
            ans += a[-1 - i] - a[i]
    print(ans)
    for a, b in anslis:
        print(a,b)
    return

#D
def D():
    return

#E
def E():
    return

#F
def F():
    return

#G
def G():
    return

#H
def H():
    return

#Solve
if __name__ == '__main__':
    B()