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 Ruby (2.3.3) (judge verdict: Runtime Error)

def s()
  gets().chomp
end
def i()
  gets.to_i
end
def li()
  gets.split.map(&:to_i)
end
def ili()
  n = gets.to_i
  as = gets.split.map(&:to_i)
  [n, as]
end
def inli()
  n = gets.to_i
  as = []
  n.times {
    as << gets.split.map(&:to_i)
  }
  [n, as]
end
class Array
  def sum
    total = 0
    self.each do |item|
      total += item
    end
    total
  end
end

def eratosthenes(size)
  prime_list = []
  is_prime = Array.new(size, true)
  is_prime[0] = false
  is_prime[1] = false

  2.upto(size-1){|i|
    if is_prime[i]
      prime_list << i
      (2).upto((size.to_f/i).ceil-1){|j|
        is_prime[j*i] = false
      }
    end
  }
  [prime_list, is_prime]
end

def dijkstra(s, e, cost)
  size = cost.length
  if s==e
    return 0
  end
  dp = {}
  dp[s] = 0
  undefined_node = {}
  size.times{|i|
    if i != s
      undefined_node[i] = cost[s][i]
    end
  }
  while undefined_node.length != 0
    min = undefined_node.min_by{|_, v| v}
    stock = {}
    temp = undefined_node.clone
    temp.each{|k, v|
      if min[1] == v
        undefined_node.delete(k)
        if k == e
          return v
        end
        dp[k] = v
        stock[k] = v
      end
    }

    stock.each{|k, v|
      undefined_node.each{|k2, v2|
        undefined_node[k2] = v + cost[k][k2] < v2 ?  v + cost[k][k2] : v2
      }
    }
  end
end

def get_bits(len)
  res =  []
  (2**len).times{|i|
    res << i.to_s(2).rjust(len, '0')
  }
  res
end

def get_GCD(m, n)
  tmp = [n, m].max
  n = [n, m].min
  m = tmp
  while n != 0
    tmp = m % n
    m = n
    n = tmp
  end
  m
end


n, as = inli()
map = {}
0.upto(n-2) { |i|
  (i+1).upto(n-1) {|j|
    key = (as[i][0]-as[j][0]).to_s + " " + (as[i][1] - as[j][1]).to_s
    if map.has_key?(key)
      map[key] += 1
    else
      map[key] = 1
    end
  }
}

puts n-map.max{ |x, y| x[1] <=> y[1] }[1]