#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define pb push_back
#define mp make_pair
#define rep(i,n) for(int i=0;i<(n);++i)

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n,k;cin >> n >> k;
	vector<pii> a(n);
	rep(i,n) cin >> a.at(i).second >> a.at(i).first;
	sort(a.rbegin(),a.rend());
	map<int,bool> m;
	stack<int> s;
	ll sum=0,type=0;
	vector<ll> ans(n+1);
	rep(i,n){
		if(i<k){
			if(m[a.at(i).second]){
				sum+=a.at(i).first;
				s.push(a.at(i).first);
				ans[type]=sum;
			}
			else{
				sum+=a.at(i).first-type*type;
				type++;
				sum+=type*type;
				m[a.at(i).second]=true;
				ans[type]=sum;
			}
		}
		else{
			if(!m[a.at(i).second] && s.size()!=0){
				sum-=s.top()+type*type;
				type++;
				sum+=a.at(i).first+type*type;
				ans[type]=sum;
				s.pop();
				m[a.at(i).second]=true;
			}
			else ans[type]=sum;
		}
	}
	ll res=0;
	rep(i,n+1) res=max(res,ans[i]);
	cout << res << endl;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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.

Examples

Input

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


Output

26


Input

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


Output

25


Input

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


Output

4900000016

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define pb push_back
#define mp make_pair
#define rep(i,n) for(int i=0;i<(n);++i)

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n,k;cin >> n >> k;
	vector<pii> a(n);
	rep(i,n) cin >> a.at(i).second >> a.at(i).first;
	sort(a.rbegin(),a.rend());
	map<int,bool> m;
	stack<int> s;
	ll sum=0,type=0;
	vector<ll> ans(n+1);
	rep(i,n){
		if(i<k){
			if(m[a.at(i).second]){
				sum+=a.at(i).first;
				s.push(a.at(i).first);
				ans[type]=sum;
			}
			else{
				sum+=a.at(i).first-type*type;
				type++;
				sum+=type*type;
				m[a.at(i).second]=true;
				ans[type]=sum;
			}
		}
		else{
			if(!m[a.at(i).second] && s.size()!=0){
				sum-=s.top()+type*type;
				type++;
				sum+=a.at(i).first+type*type;
				ans[type]=sum;
				s.pop();
				m[a.at(i).second]=true;
			}
			else ans[type]=sum;
		}
	}
	ll res=0;
	rep(i,n+1) res=max(res,ans[i]);
	cout << res << endl;
}
```