#include <bits/stdc++.h>
#define sqr(x) ((LL)(x)*(x))
using namespace std;
typedef long long LL;
vector <pair<LL,int> > a; 
int n,k;
stack <LL> stc;
set <int> types;
int main()
{
	cin>>n>>k;
	for(int i=0;i<n;i++)
	{
		int t;
		LL d;
		cin>>t>>d;
		a.push_back(make_pair(d,t));
	}
	sort(a.begin(),a.end());
	reverse(a.begin(),a.end());
	LL ans=0;
	for(int i=0;i<k;i++)
	{
		ans+=a[i].first;
		if(types.count(a[i].second))stc.push(a[i].first);
		types.insert(a[i].second);
	}
	ans+=sqr(types.size());
	LL res=ans;
	for(int i=k;i<n;i++)
	{
		if(stc.empty())break;
		if(!types.count(a[i].second))
		{
			res=res-stc.top()+a[i].first-sqr(types.size())+sqr(types.size()+1);
			stc.pop();
			types.insert(a[i].second);
			ans=max(ans,res);
		}
	}
	cout<<ans<<endl;
	return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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>
#define sqr(x) ((LL)(x)*(x))
using namespace std;
typedef long long LL;
vector <pair<LL,int> > a; 
int n,k;
stack <LL> stc;
set <int> types;
int main()
{
	cin>>n>>k;
	for(int i=0;i<n;i++)
	{
		int t;
		LL d;
		cin>>t>>d;
		a.push_back(make_pair(d,t));
	}
	sort(a.begin(),a.end());
	reverse(a.begin(),a.end());
	LL ans=0;
	for(int i=0;i<k;i++)
	{
		ans+=a[i].first;
		if(types.count(a[i].second))stc.push(a[i].first);
		types.insert(a[i].second);
	}
	ans+=sqr(types.size());
	LL res=ans;
	for(int i=k;i<n;i++)
	{
		if(stc.empty())break;
		if(!types.count(a[i].second))
		{
			res=res-stc.top()+a[i].first-sqr(types.size())+sqr(types.size()+1);
			stc.pop();
			types.insert(a[i].second);
			ans=max(ans,res);
		}
	}
	cout<<ans<<endl;
	return 0;
}

```