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 (1.0.0) (judge verdict: Wrong Answer)

fun main(args : Array<String>) {
    val N = readLine()!!.toInt()
    val points: Array<Pair<Long, Long>?> = Array(N, { null })
    val map = hashMapOf<Pair<Long, Long>, Int>()
    (1..N).forEach {
        val (x, y) = readLine()!!.split(" ").map { it.toLong() }
        val pair = Pair(x, y)
        points[it - 1] = pair
        map[pair] = it - 1
    }

    var answer = Int.MAX_VALUE
    val possiblePairs = mutableListOf<Pair<Long, Long>>()
    (0..(N - 1)).forEach { a ->
        (0..(N - 1)).forEach { b ->
            if (a != b) {
                val p = points[b]!!.first - points[a]!!.first
                val q = points[b]!!.second - points[a]!!.second
                possiblePairs.add(Pair(p, q))
            }
        }
    }

    var cost = Int.MAX_VALUE
    possiblePairs.forEach {
        val p = it.first
        val q = it.second

        val unionFind = UnionFind(N)
        (0..(N-1)).forEach { i ->
            val origin = points[i]
            val index = map[Pair(origin!!.first + p, origin!!.second + q)]
            if (index != null) {
                unionFind.merge(i, index)
            }
        }

        cost = Math.min(cost, unionFind.parent.distinct().size)
    }

    println(cost)
}
class UnionFind(size: Int) {

    var parent: IntArray
    var height: IntArray
    var size: IntArray

    init {
        this.parent = IntArray(size)
        this.height = IntArray(size)
        this.size = IntArray(size)

        for (i in 0 until size) {
            makeSet(i)
        }
    }

    fun merge(x: Int, y: Int) {

        val rootX = root(x)
        val rootY = root(y)

        if (rootX == rootY) {
            return
        }

        if (height[rootX] > height[rootY]) {
            parent[rootY] = rootX
            size[rootX] += size[rootY]

        } else if (height[rootX] < height[rootY]) {
            parent[rootX] = rootY
            size[rootY] += size[rootX]

        } else {
            parent[rootY] = rootX
            size[rootX] += size[rootY]
            height[rootX]++
        }
    }

    fun isSame(x: Int, y: Int) : Boolean {
        return root(x) == root(y)
    }

    fun size(x: Int): Int {
        val rootX = root(x)
        return size[rootX]
    }

    private fun root(x: Int): Int {
        if (x != parent[x]) {
            parent[x] = root(parent[x])
        }
        return parent[x]
    }


    private fun makeSet(i: Int) {
        parent[i] = i
        height[i] = 0
        size[i] = 1
    }
}