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 LuaJIT (2.0.4) (judge verdict: Wrong Answer)

local mce, mfl, msq, mmi, mma = math.ceil, math.floor, math.sqrt, math.min, math.max
local n = io.read("*n")
local t = {}
for i = 1, n do
  t[i] = {}
  t[i].x, t[i].y = io.read("*n", "*n")
end
table.sort(t, function(v, w)
  if v.x == w.x then
    return v.y < w.y
  else
    return v.x < w.x
  end
end
)

local function getcost(p, q)
  local cost = 0
  if p == 0 then
    local done = 0
    local donelist = {}
    for i = 1, n do donelist[i] = false end
    while done < n do
      cost = cost + 1
      local idx = 1
      for i = 1, n do
        if donelist[i] then idx = i break end
      end
      donelist[idx] = true
      done = done + 1
      local nx, ny = t[idx].x, t[idx].y + q
      local rep = true
      while rep do
        rep = false
        for i = 1, n do
          if nx == t[i].x and ny == t[i].y and not donelist[i] then
            ny = ny + q
            donelist[i] = true
            done = done + 1
            rep = true
          end
        end
      end
    end
  else
    local done = 0
    local donelist = {}
    for i = 1, n do donelist[i] = false end
    while done < n do
      cost = cost + 1
      local idx = 1
      for i = 1, n do
        if donelist[i] then idx = i break end
      end
      donelist[idx] = true
      done = done + 1
      local nx, ny = t[idx].x + p, t[idx].y + q
      for i = idx + 1, n do
        if nx == t[i].x and ny == t[i].y then
          nx, ny = nx + p, ny + q
          donelist[i] = true
          done = done + 1
        end
      end
    end
  end
  return cost
end

local ret = 50
for i = 1, n - 1 do
  for j = i + 1, n do
    local c = getcost(t[j].x - t[i].x, t[j].y - t[i].y)
    ret = mmi(ret, c)
  end
end
print(ret)