#include<cstdio>
#include<vector>
#include<algorithm>

using namespace std;

using ll=long long;
using vi=vector<int>;
using pii=pair<int,int>;
using vpii=vector<pii>;


int max_u(ll&m, ll v)
{
	if(m<v)
	{
		m=v;
		return 1;
	}
	return 0;
}


int main(void)
{
	int n, k;

	while(scanf("%d%d", &n, &k)==2)
	{
		vpii dt(n);
		vi c(n+1);
		vi f(n);
		ll sum=0, tn=0, ans=0;

		for(int i=0;i<n;i++)
		{
			scanf("%d%d", &dt[i].second, &dt[i].first);
		}
		sort(dt.rbegin(), dt.rend());

		for(int i=0;i<k;i++)
		{
			int&rc=c[dt[i].second];

			sum+=dt[i].first;
			if(!rc) tn++;
			rc++;
		}

		max_u(ans, sum+tn*tn);
		int t=k;
		for(int i=k-1;i>=0;i--)
		{
			int&rc=c[dt[i].second];
			if(rc<=1) continue;
			rc--;
			for(;t<n;t++)
			{
				if(!c[dt[t].second]) break;
			}
			if(t>=n) break;

			sum+=dt[t].first-dt[i].first;
			c[dt[t].second]++;
			tn++;
			max_u(ans, sum+tn*tn);
		}

		printf("%lld\n", ans);
	}

	return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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<cstdio>
#include<vector>
#include<algorithm>

using namespace std;

using ll=long long;
using vi=vector<int>;
using pii=pair<int,int>;
using vpii=vector<pii>;


int max_u(ll&m, ll v)
{
	if(m<v)
	{
		m=v;
		return 1;
	}
	return 0;
}


int main(void)
{
	int n, k;

	while(scanf("%d%d", &n, &k)==2)
	{
		vpii dt(n);
		vi c(n+1);
		vi f(n);
		ll sum=0, tn=0, ans=0;

		for(int i=0;i<n;i++)
		{
			scanf("%d%d", &dt[i].second, &dt[i].first);
		}
		sort(dt.rbegin(), dt.rend());

		for(int i=0;i<k;i++)
		{
			int&rc=c[dt[i].second];

			sum+=dt[i].first;
			if(!rc) tn++;
			rc++;
		}

		max_u(ans, sum+tn*tn);
		int t=k;
		for(int i=k-1;i>=0;i--)
		{
			int&rc=c[dt[i].second];
			if(rc<=1) continue;
			rc--;
			for(;t<n;t++)
			{
				if(!c[dt[t].second]) break;
			}
			if(t>=n) break;

			sum+=dt[t].first-dt[i].first;
			c[dt[t].second]++;
			tn++;
			max_u(ans, sum+tn*tn);
		}

		printf("%lld\n", ans);
	}

	return 0;
}

```