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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define per(i, n) for (int i = (int)(n-1); i >= 0; i--)
#define sz(x) int(x.size())
#define all(x) (x).begin(), (x).end()
#define INF 2e9
#define LINF 1e18
template<class T> inline bool chmax(T &a, const T &b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T &a, const T &b) { if (a > b) { a = b; return 1; } return 0; }
typedef long long ll;
typedef pair<int, int> P;

int main() {
    int n, k; cin >> n >> k;
    vector<vector<int> > tmp(n);
    rep(i, n) {
        int t0, d0;
        cin >> t0 >> d0;
        t0--;
        tmp[t0].push_back(d0);
    }

    rep(i, n) sort(all(tmp[i]), greater<int>()); //各ネタごとに美味しさ降順でソート

    vector<int> y1, y0;
    rep(i, n) {
        if (sz(tmp[i]) == 0) continue;
        y1.push_back(tmp[i][0]); //各寿司の一番美味しいやつの味を格納
        for (int j=1; j<sz(tmp[i]); j++) {
            y0.push_back(tmp[i][j]); //一番美味しいやつ以外の味を格納
        }
    }
    sort(all(y1), greater<int>());
    sort(all(y0), greater<int>());
    //以降、各寿司を食う場合は一番美味しいやつを食えばよく
    //その一番美味しいやつを何個食うかで調べていけば良い
    ll ans = 0;
    for (int i=1; i<=min(sz(y1), k); i++) {
        ll tmp = (ll)i*i; //何種類の寿司を食うか
        rep(j, i) tmp += y1[j];
        int m = k-i; //あと何個一番美味しい寿司以外食うか
        if (sz(y0) < m) continue;
        rep(j, m) {
            tmp += y0[j];
        }
        chmax(ans, tmp);
    }

    cout << ans << endl;
}