#include<bits/stdc++.h>
#define ll long long
#define pll pair<ll,ll>
#define F first
#define S second
#define mp make_pair
using namespace std;
const int N=1e5+5;
stack<ll>st;
ll n,cnt,k,num;
pll p[N];
ll ans,sum;
bool use[N];
int main()
{
	cin>>n>>k;
	for(int i=0;i<n;i++)cin>>p[i].S>>p[i].F;
	sort(p,p+n);
	reverse(p,p+n);
	for(int i=0;i<n;i++)swap(p[i].F,p[i].S);
	for(int i=0;i<n;i++)
	{
		if(cnt<k)
		{
			if(use[p[i].F])
				st.push(p[i].S);
			else
				num++,use[p[i].F]=true;
			sum+=p[i].S;
			ans=max(ans,sum+num*num);
			cnt++;
		}
		else
		{
			if(st.empty())break;
			if(use[p[i].F])continue;
			use[p[i].F]=true;
			num++;
			sum-=st.top();
			sum+=p[i].S;
			st.pop();
			ans=max(ans,sum+num*num);
		}
	}
	cout<<ans<<endl;
	return 0;
}
### 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>
#define ll long long
#define pll pair<ll,ll>
#define F first
#define S second
#define mp make_pair
using namespace std;
const int N=1e5+5;
stack<ll>st;
ll n,cnt,k,num;
pll p[N];
ll ans,sum;
bool use[N];
int main()
{
	cin>>n>>k;
	for(int i=0;i<n;i++)cin>>p[i].S>>p[i].F;
	sort(p,p+n);
	reverse(p,p+n);
	for(int i=0;i<n;i++)swap(p[i].F,p[i].S);
	for(int i=0;i<n;i++)
	{
		if(cnt<k)
		{
			if(use[p[i].F])
				st.push(p[i].S);
			else
				num++,use[p[i].F]=true;
			sum+=p[i].S;
			ans=max(ans,sum+num*num);
			cnt++;
		}
		else
		{
			if(st.empty())break;
			if(use[p[i].F])continue;
			use[p[i].F]=true;
			num++;
			sum-=st.top();
			sum+=p[i].S;
			st.pop();
			ans=max(ans,sum+num*num);
		}
	}
	cout<<ans<<endl;
	return 0;
}
```