Problem:
Score : 400 points

Problem Statement

There are N pieces of sushi. Each piece has two parameters: "kind of topping" t_i and "deliciousness" d_i.
You are choosing K among these N pieces to eat.
Your "satisfaction" here will be calculated as follows:

The satisfaction is the sum of the "base total deliciousness" and the "variety bonus".

The base total deliciousness is the sum of the deliciousness of the pieces you eat.

The variety bonus is x*x, where x is the number of different kinds of toppings of the pieces you eat.

You want to have as much satisfaction as possible.
Find this maximum satisfaction.

Constraints

1 \leq K \leq N \leq 10^5

1 \leq t_i \leq N

1 \leq d_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
t_1 d_1
t_2 d_2
.
.
.
t_N d_N

Output

Print the maximum satisfaction that you can obtain.

Sample Input 1

5 3
1 9
1 7
2 6
2 5
3 1

Sample Output 1

26

If you eat Sushi 1,2 and 3:

The base total deliciousness is 9+7+6=22.

The variety bonus is 2*2=4.

Thus, your satisfaction will be 26, which is optimal.

Sample Input 2

7 4
1 1
2 1
3 1
4 6
4 5
4 5
4 5

Sample Output 2

25

It is optimal to eat Sushi 1,2,3 and 4.

Sample Input 3

6 5
5 1000000000
2 990000000
3 980000000
6 970000000
6 960000000
4 950000000

Sample Output 3

4900000016

Note that the output may not fit into a 32-bit integer type.

Solution in Kotlin (1.0.0) (judge verdict: Runtime Error)

fun main(args: Array<String>) {
    val (n, k) = readLine()!!.split(' ').map(String::toInt)
    val tdList = (1..n).map { readLine()!!.split(' ').map(String::toLong).let { Pair(it[0], it[1]) } }.sortedByDescending { it.second }

    val selected = tdList.take(k).toMutableList()//美味しさだけで選ばれたk個の寿司
    val notSelected = tdList.takeLast(n - k).toMutableList()//選ばれなかった寿司
    val ds = selected.map { it.first }.toSet()//選択されたネタの一覧
    var dc = ds.count()//ネタの種類数
    var maxScore = selected.map { it.second }.sum() + dc * dc//暫定満足ポイント

    //選ばれた寿司の中で、ネタが重複しているもののうち、スコアが低い方のリストを作成する
    val changeableSelected = mutableListOf<Pair<Long, Long>>()
    run {
        val tempD = mutableSetOf<Long>()
        var i = 0
        while (true) {
            if (selected.size == i) break
            if (tempD.contains(selected[i].first)) {
                changeableSelected.add(selected[i])
            } else {
                tempD.add(selected[i].first)
            }
            i++
        }
    }

    //選ばれなかった寿司の中で、選んだすしに含まれないもののうち、スコアが高い方のリストを作成する
    val changeableNotSelected = mutableListOf<Pair<Long, Long>>()
    run {
        val tempD = mutableSetOf<Long>()
        var i = 0
        while (true) {
            if (notSelected.size == i) break
            val d = selected[i].first
            if(tempD.contains(d).not() && ds.contains(d).not()){
                changeableNotSelected.add(notSelected[i])
                tempD.add(notSelected[i].first)
            }
            i++
        }
    }

    //選んだ寿司の中でネタが重複しているもののうち、スコアが小さいものから1つ消し、選ばなかったものから1つ追加する、という処理を実行できなくなるまで繰り返し、その中で最大の満足度を探す
    var tempMax = maxScore
    while (changeableSelected.size > 0 && changeableNotSelected.size > 0) {
        tempMax = tempMax - changeableSelected.last().second + changeableNotSelected.first().second - dc * dc + (dc + 1) * (dc + 1)
        dc++
        changeableSelected.remove(changeableSelected.last())
        changeableNotSelected.remove(changeableNotSelected.first())
        maxScore = Math.max(maxScore, tempMax)
    }
    println(maxScore)
}