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: Time Limit Exceeded)

#!/usr/bin/env python3
import sys
import itertools

def solve(N: int, x: "List[int]", y: "List[int]"):
    if N == 1:
        print(1)
        return

    PL = list(zip(x,y))

    mincost = float('inf')
    for seq in itertools.permutations(PL):
        cost = 1
        p = seq[1][0] - seq[0][0]
        q = seq[1][1] - seq[0][1]
        #print(seq)
        for i in range(1,len(seq)):
            p2 = seq[i][0] - seq[i-1][0] 
            q2 = seq[i][1] - seq[i-1][1] 
            #print(p,p2,q,q2)
            if p != p2 or  q != q2:
                cost += 1
            if cost >= mincost:
                break
        mincost = min(mincost,cost)

    print(mincost)
            


# Generated by 1.1.4 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    N = int(next(tokens))  # type: int
    x = [int()] * (N)  # type: "List[int]" 
    y = [int()] * (N)  # type: "List[int]" 
    for i in range(N):
        x[i] = int(next(tokens))
        y[i] = int(next(tokens))
    solve(N, x, y)

if __name__ == '__main__':
    main()