Problem:
Score : 600 points

Problem Statement

There are N dots in a two-dimensional plane. The coordinates of the i-th dot are (x_i, y_i).

We will repeat the following operation as long as possible:

Choose four integers a, b, c, d (a \neq c, b \neq d) such that there are dots at exactly three of the positions (a, b), (a, d), (c, b) and (c, d), and add a dot at the remaining position.

We can prove that we can only do this operation a finite number of times. Find the maximum number of times we can do the operation.

Constraints

1 \leq N \leq 10^5

1 \leq x_i, y_i \leq 10^5

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 maximum number of times we can do the operation.

Sample Input 1

3
1 1
5 1
5 5

Sample Output 1

1

By choosing a = 1, b = 1, c = 5, d = 5, we can add a dot at (1, 5). We cannot do the operation any more, so the maximum number of operations is 1.

Sample Input 2

2
10 10
20 20

Sample Output 2

0

There are only two dots, so we cannot do the operation at all.

Sample Input 3

9
1 1
2 1
3 1
4 1
5 1
1 2
1 3
1 4
1 5

Sample Output 3

16

We can do the operation for all choices of the form a = 1, b = 1, c = i, d = j (2 \leq i,j \leq 5), and no more. Thus, the maximum number of operations is 16.

Solution in Kotlin (1.0.0) (judge verdict: Time Limit Exceeded)

fun main(args: Array<String>) {
    val n = readLine()!!.toInt()
    val p = (1..n).map { val a = (readLine()!!.split(" ").map { it.toInt() }); Point(it-1, a[0], a[1]) }.sorted()

    val xy = Array(100001) { mutableListOf<Int>() }
    val yx = Array(100001) { mutableListOf<Int>() }
    val uf = UnionFind(n)
    p.forEach {
        xy[it.x].add(it.y)
        yx[it.y].add(it.x)
        uf.xs[it.id].add(it.x)
        uf.ys[it.id].add(it.y)
    }

    (1..100000).forEach { pos ->
        if(xy[pos].size>=2) xy[pos].drop(1).forEach {
            val x = pos
            val y = it
            val i = p.binarySearch(Point(-1,x,xy[x][0]))
            val j = p.binarySearch(Point(-1,x,y))
            uf.unite(p[i].id, p[j].id)
        }
        if(yx[pos].size>=2) yx[pos].drop(1).forEach {
            val y = pos
            val x = it
            val i = p.binarySearch(Point(-1,yx[y][0],y))
            val j = p.binarySearch(Point(-1,x,y))
            uf.unite(p[i].id, p[j].id)
        }

    }

    val sum = uf.roots.map { 1L * uf.xs[it].size * uf.ys[it].size }.sum()
    println(sum-n)
}

data class Point(val id: Int, val x: Int, val y: Int) : Comparable<Point>{
    override fun compareTo(other: Point): Int =
        if (this.x != other.x) this.x.compareTo(other.x)
        else this.y.compareTo(other.y)
}

class UnionFind(N: Int) {
    private val parent = IntArray(N) { it }
    val roots = (0..N-1).toList().toMutableSet()
    private val rank = IntArray(N) { 0 }
    private val size = IntArray(N) { 1 }
    private var numTree = N

    val xs = Array(N){ mutableSetOf<Int>() }
    val ys = Array(N){ mutableSetOf<Int>() }

    fun size() = numTree
    fun find(x: Int): Int = if (parent[x] == x) x else { parent[x] = find(parent[x]); parent[x] }

    fun same(x: Int, y: Int) = find(x) == find(y)
    fun sizeOf(x: Int) = size[find(x)]
    fun unite(x: Int, y: Int) {
        val rx = find(x)
        val ry = find(y)
        if (rx == ry) return else numTree--

        if (rank[rx] < rank[ry]) {
            parent[rx] = ry
            roots.remove(rx)
            size[ry] += size[rx]
            xs[ry].addAll(xs[rx])
            ys[ry].addAll(ys[rx])
        } else {
            parent[ry] = rx
            roots.remove(ry)
            size[rx] += size[ry]
            xs[rx].addAll(xs[ry])
            ys[rx].addAll(ys[ry])
            if (rank[rx] == rank[ry]) rank[rx]++
        }
    }
}