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: Runtime Error)

#include <bits/stdc++.h>  // clang-format off
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define each(x,v) for(auto& x : v)
#define all(v) (v).begin(),(v).end()
#define sz(v) ((int)(v).size())
#define ini(...) int __VA_ARGS__; in(__VA_ARGS__)
#define inl(...) long long __VA_ARGS__; in(__VA_ARGS__)
#define ins(...) string __VA_ARGS__; in(__VA_ARGS__)
#define inc(...) char __VA_ARGS__; in(__VA_ARGS__)
#define in2(s,t)     rep(i,sz(s)){in(s[i] , t[i]);}
#define in3(s,t,u)   rep(i,sz(s)){in(s[i] , t[i] , u[i]);}
#define in4(s,t,u,v) rep(i,sz(s)){in(s[i] , t[i] , u[i] , v[i]);}
#ifdef ONLINE_JUDGE
  #define rep(i,N) for(int i = 0; i < (int)(N); i++)
  #define repr(i,N) for(int i = (int)(N) - 1; i >= 0; i--)
  #define rep1(i,N) for(int i = 1; i <= (int)(N) ; i++)
  #define repr1(i,N) for(int i = (N) ; (int)(i) > 0 ; i--)
#else
  #define rep(i,N) for(long long i = 0; i < (long long)(N); i++)
  #define repr(i,N) for(long long i = (long long)(N) - 1; i >= 0; i--)
  #define rep1(i,N) for(long long i = 1; i <= (long long)(N) ; i++)
  #define repr1(i,N) for(long long i = (N) ; (long long)(i) > 0 ; i--)
#endif
using namespace std; void solve();
using ll = long long; template<class T = ll> using V = vector<T>;
using vi = V<int>; using vl = V<>; using vvi = V< V<int> >;
using vd = V<double>; using vs = V<string>; using vvl = V< V<> >;
constexpr int inf = 1001001001; constexpr ll infLL = (1LL << 61) - 1;
template<typename T, typename U> inline bool amin(T &x, U y) { return (y < x) ? (x = y, true) : false; }
template<typename T, typename U> inline bool amax(T &x, U y) { return (x < y) ? (x = y, true) : false; }
template<typename T,typename U>ll ceil(T a,U b){return (a + b - 1) / b;}
template<typename T, typename U> ostream& operator <<(ostream& os, const pair<T, U> &p) { os << p.first << " " << p.second; return os; }
template<typename T, typename U> istream& operator >>(istream& is, pair<T, U> &p) { is >> p.first >> p.second; return is; }
template<typename T> ostream& operator <<(ostream& os, const vector<T> &v) { int s = (int)v.size(); for(int i=0;i<s;i++) os << (i ? " " : "") << v[i]; return os; }
template<typename T> istream& operator >>(istream& is, vector<T> &v) { for(auto &x : v) is >> x; return is; }
void in(){} template <typename T,class... U> void in(T &t,U &...u){ cin >> t; in(u...);}
void out(){cout << "\n";} template <typename T,class... U> void out(const T &t,const U &...u){ cout << t; if(sizeof...(u)) cout << " "; out(u...);}
template<typename T>void die(T x){out(x); exit(0);}

#ifdef NyaanDebug
  #include "NyaanDebug.h"
  #define trc(...) do { cerr << #__VA_ARGS__ << " = "; dbg_out(__VA_ARGS__);} while(0)
  #define trca(v,N) do { cerr << #v << " = "; array_out(v , N);} while(0)
  #define trcc(v) do { cerr << "name : " << #v << "\n"; int cnt = 0; each(x , v){cerr << (cnt++) << " : "; trc(x); } } while(0)
#else
  #define trc(...)
  #define trca(...)
  #define trcc(...)
  int main(){solve();}
#endif

constexpr ll TEN(int n){ll ret=1,x=10;while(n){if(n&1)ret*=x;x*=x;n>>=1;}return ret;}

struct IoSetupNya {IoSetupNya() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); cerr << fixed << setprecision(7);} } iosetupnya;
using P = pair<ll,ll>; using vp = V<P>;
constexpr int MOD = /**/ 1000000007; //*/ 998244353;

// clang-format on
/////////////

void solve(){
  ini(N , K);
  vp a(N); in(a);
  sort(all(a),[](P x,P y){
    return x.se > y.se;
  });

  // 大きい順
  priority_queue<ll,vl,greater<>> Q;
  ll val = 0 , kind = 0;
  vi cnt(N + 1);
  vp b;
  rep(i , N){
    if(sz(Q) < K && cnt[a[i].fi] == 0){
      cnt[a[i].fi]++;
      Q.emplace(a[i].se);
      kind++;
      val += a[i].se;
    }
    else b.eb(a[i]);
  }

  int nya = 0;
  // 愚直でよい
  if(sz(Q) < K){
    exit(1);
    rep(i,sz(b)){
      if(sz(Q)==K) break;
      //Q.emplace(a[i].se);
      val += b[i].se;
      nya++;
    }
  }
  
  //trc(val , kind , b);
  // 愚直ダメ
  for(int i = nya ; i < sz(b) ; i++){
    if(Q.empty()) break;
    auto tp = Q.top();
    if(tp + kind * kind >= b[i].se + (kind - 1) * (kind - 1)) break;
    val += b[i].se - tp;
    kind--;
    Q.pop();
  }

  out(val + kind * kind);

}