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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

import kotlin.math.ceil

// https://atcoder.jp/contests/diverta2019-2/tasks/diverta2019_2_b
// TODO: try to AC
// N=50なんで全組み合わせで探索とかでいけそうな気もするが、N!だと収まらない

fun main() {
    divelta2019_2b()
}

fun divelta2019_2b() {
    val n = readLine()!!.toInt()
    val plots = mutableListOf<Pair<Long, Long>>()
    repeat(n) {
        val (x, y) = readLine()!!.split(" ").map { it.toLong() }
        plots.add(x to y)
    }
    val costMap = mutableMapOf<Pair<Long, Long>, MutableList<Pair<Int, Int>>>()
    for (i in 0 until n) {
        val base = plots[i]
        for (j in 0 until n) {
            val t = plots[j]
            val p = t.first - base.first
            val q = t.second - base.second
            if (p == 0L && q == 0L) continue
            if (p to q in costMap) {
                costMap[p to q]!!.add(i to j)
            } else {
                costMap[p to q] = mutableListOf(i to j)
            }
        }
    }
    var cost = 0
    val visited = mutableSetOf<Int>()
    val sortedCosts = costMap.values.filter { it.size > 1 }.sortedByDescending { it.size }
    for (s in sortedCosts) {
        var maxCount = 0
        var maxLines = mutableSetOf<Int>()
        for ((i, j) in s) {
            if (i in visited || j in visited) continue
            val lines = mutableSetOf<Int>(i,j)
            var left = i
            var right = j
            var count = 2
            for ((i, j) in s) {
                if (i in visited || j in visited) continue
                if (i == right) {
                    right = j
                    lines.add(j)
                    count++
                } else if (j == left) {
                    left = i
                    count++
                    lines.add(i)
                }
            }
            if (maxCount < count) {
                maxCount = count
                maxLines = lines
            }
        }
        if (maxLines.isNotEmpty()) {
            visited.addAll(maxLines)
            cost++
        }
        if (visited.size == n) break
    }
    if (visited.size < n) {
        // 最後に1つだけ余ったりする、もしくは基点がすでに他のもので使われてしまっている場合の対策場
        // あぶれたのが複数ある場合、その2点を直線で繋げばコスト1で2点回収できる
        cost += ceil((n - visited.size) / 2.0).toInt()
    }
    println(cost)
}