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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;

public class Main {
	private static void solve(){
		//Implement solution here.
		int n = ni();
		int k = ni();
		Map<Integer,Queue<Integer>> map = new HashMap<>();
		Queue<Integer> nq = new PriorityQueue<>(Collections.reverseOrder());
		Queue<Integer> eq = new PriorityQueue<>(Collections.reverseOrder());
		for(int i = 0; i < n; i++) {
			int t = ni();
			int d = ni();
			if(!map.containsKey(t)) {
				map.put(t, new PriorityQueue<Integer>(Collections.reverseOrder()));
			} 
			map.get(t).add(d);
		}
		
		for(int key:map.keySet()) {
			nq.add(map.get(key).poll());
			while(!map.get(key).isEmpty()) {
				eq.add(map.get(key).poll());
			}
		}

		long dsum = 0;
		long var = 0;
		Queue<Integer> rq = new PriorityQueue<>();
		while(rq.size() < k && !nq.isEmpty()) {
			int dd = nq.poll();			
			rq.add(dd);
			dsum += dd;
			var++;
		}		
		long result = dsum + var * var;
		
		for(int i = 0; i < k; i++) {
			if(eq.isEmpty())break;
			int dd = eq.poll();
			dsum += dd;
			if(rq.size() >= k) {				
				dsum -= rq.poll();			
				var--;
			}
			rq.add(dd);
			result = Math.max(result, dsum + var * var);
		}
		
		System.out.println(result);
	}

	//Switch input source (stdin/file)
	private static BufferedReader br  = new BufferedReader(new InputStreamReader(System.in));
	public static void main(String[] args){
		String debugDataPath = System.getenv("DD_PATH");        
		if(debugDataPath != null){
			try{
				br = new BufferedReader(new InputStreamReader(new FileInputStream(new File(debugDataPath))));
			}catch(Exception e){
				throw new RuntimeException(e);
			}
		}
		solve();
	}

	//Input read utility
	private static StringTokenizer tokenizer = null;
	public static String next() {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}
	//Get next single int
	private static int ni() {
		return Integer.parseInt(next());
	}
	//Get next single long
	private static long nl() {
		return Long.parseLong(next());
	}
	//Get next single double
	private static double nd() {
		return Double.parseDouble(next());
	}
	//Get next int array from one line
	private static int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}
	//Get next char array from one line
	private static char[] ns() {
		return next().toCharArray();
	}
	//Get next long array from one line
	private static long[] nal(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nl();
		return a;
	}
}