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 Python (3.4.3) (judge verdict: Accepted)

import sys
sys.setrecursionlimit(10**7)
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x) - 1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def II(): return int(sys.stdin.readline())
def SI(): return sys.stdin.readline().strip()
INF = 10 ** 18
MOD = 10 ** 9 + 7


import collections
import itertools
import operator


class UnionFind:
    def __init__(self, elems=None):
        class KeyDict(dict):
            def __missing__(self, key):
                self[key] = key
                return key

        self.parent = KeyDict()
        self.rank = collections.defaultdict(int)

        if elems is not None:
            for elem in elems:
                _, _ = self.parent[elem], self.rank[elem]

    def find(self, x):
        if self.parent[x] == x:
            return x
        else:
            self.parent[x] = self.find(self.parent[x])
            return self.parent[x]

    def unite(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if self.rank[x] < self.rank[y]:
            self.parent[x] = y
        else:
            self.parent[y] = x
        if self.rank[x] == self.rank[y]:
            self.rank[x] += 1

    def are_same(self, x, y):
        return self.find(x) == self.find(y)

    def grouper(self):
        roots = [(x, self.find(x_par)) for x, x_par in self.parent.items()]
        root = operator.itemgetter(1)
        for _, group in itertools.groupby(sorted(roots, key=root), root):
            yield [x for x, _ in group]


def main(): 
    N = II()
    position = []
    for _ in range(N):
        position.append(LI())

    min_ = INF
    import itertools
    for A, B in itertools.combinations(range(N), 2):
        uf = UnionFind(range(N))
        p = position[A][0] - position[B][0]
        q = position[A][1] - position[B][1]
        # print(p,q)
        uf.unite(A, B)
        for X, Y in itertools.combinations(range(N), 2):
            if A == X and B == Y:
                continue
            if (p == position[X][0] - position[Y][0] and q == position[X][1] - position[Y][1]) or (p == position[Y][0] - position[X][0] and q == position[Y][1] - position[X][1]):
                uf.unite(X, Y)
        
        min_ = min(min_, len(list(uf.grouper())))
    print(1 if N == 1 else min_)

main()