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 Rust (1.15.1) (judge verdict: Accepted)

// -----------------------------------------------
// Framework <https://github.com/vain0x/procon>
//
// See the bottom of file for solution.
// -----------------------------------------------

#![allow(unused_imports)]
#![allow(non_snake_case)]

use std::cell::RefCell;
use std::cmp::{max, min, Ordering};
use std::collections::*;
use std::fmt::{Debug, Display, Formatter, Write as FmtWrite};
use std::io::{stderr, stdin, BufRead, Write};
use std::mem::{replace, swap};
use std::ops::*;
use std::rc::Rc;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), stderr());
            writeln!(err, "\x1B[33m{}\x1B[0m = {:?}", e, $e).unwrap()
        })*
    };
}

#[allow(unused_macros)]
macro_rules! read {
    ([$t:ty] ; $n:expr) =>
        ((0..$n).map(|_| read!([$t])).collect::<Vec<_>>());
    ($($t:ty),+ ; $n:expr) =>
        ((0..$n).map(|_| read!($($t),+)).collect::<Vec<_>>());
    ([$t:ty]) =>
        (rl().split_whitespace().map(|w| w.parse().unwrap()).collect::<Vec<$t>>());
    ($($t:ty),*) => {{
        let buf = rl();
        let mut w = buf.split_whitespace();
        ($(w.next().unwrap().parse::<$t>().unwrap()),*)
    }};
}

#[allow(dead_code)]
fn rl() -> String {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    buf.trim_right().to_owned()
}

// -----------------------------------------------
// Solution
// -----------------------------------------------

// 問題概要: 品物が N 個ある。i 番目の品物の種類は t で、価値は d である。品物を K 個選んだ。選ばれた品物の種類数を x とし、価値の総和を s とするとき、スコア (s + x^2) の値が (選びかたの中で) 最大だった。この最大スコアを求めよ。

fn main() {
    let (N, K) = read!(usize, usize);
    let T = read![usize, i64; N];

    let mut T = T.into_iter().map(|(t, d)| (d, t - 1)).collect::<Vec<_>>();
    T.sort();
    T.reverse();

    debug!(T);

    let mut s = 0_i64;
    let mut x = 0_i64;
    let mut Q = vec![];

    // done[t] = true : 種類 t の品物を選んでいる
    let mut done = vec![false; N + 1];

    for i in 0..K {
        let (d, t) = T[i];

        if done[t] {
            // 同じ種類の2個目以降の品物は選ばないほうがいい可能性がある
            Q.push(d);
        } else {
            // この種類で価値が最大の品物は必ず選ぶべきなので Q には入れない
            done[t] = true;
            x += 1;
        }

        s += d;
    }

    let mut max_score = s + x * x;
    debug!(s, x, max_score);

    for i in K..N {
        let (d, t) = T[i];

        if done[t] {
            continue;
        }

        // この品物はまだ選んでいない。

        // 同じ種類の品物で2つ以上選んでいるもののうち、価値が最も低いものを選ばないことにする。
        if let Some(d) = Q.pop() {
            s -= d;
        } else {
            // 除去できる品物がないなら終了。
            break;
        }

        // この品物を選ぶ。価値の合計は減るが、種類数が増えるので、スコアが大きくなる可能性がある。
        x += 1;
        s += d;
        done[t] = true;
        let score = s + x * x;

        max_score = max(max_score, score);
        debug!(d, x, s, x * x, score, max_score);
    }

    println!("{}", max_score);
}