#include <bits/stdc++.h>

using pii = std::pair<int, int>;

int main()
{
	int n, k;
	scanf("%d%d", &n, &k);
	std::vector<pii> sushi(n);
	for (auto& e: sushi)
	{
		scanf("%d%d", &e.second, &e.first);
		e.second--;
	}
	std::sort(sushi.rbegin(), sushi.rend());
	std::vector<bool> ate(n);
	int x{};
	long long max{};
	std::stack<int> contain;
	for (int i{}; i < k; i++)
	{
		if (!ate[sushi[i].second])
		{
			ate[sushi[i].second] = true;
			max += 2 * x + 1;
			x++;
		}
		else
			contain.push(sushi[i].first);
		max += sushi[i].first;
	}

	long long tmp_max{max};
	for (int i{k}; i < n; i++)
	{
		if (contain.empty()) break;
		if (ate[sushi[i].second]) continue;
		tmp_max += sushi[i].first + 2 * x + 1 - contain.top();
		x++;
		max = std::max(max, tmp_max);
		contain.pop();
		ate[sushi[i].second] = true;
	}

	printf("%lld\n", max);

	return 0;
}
### Prompt

Please formulate 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>

using pii = std::pair<int, int>;

int main()
{
	int n, k;
	scanf("%d%d", &n, &k);
	std::vector<pii> sushi(n);
	for (auto& e: sushi)
	{
		scanf("%d%d", &e.second, &e.first);
		e.second--;
	}
	std::sort(sushi.rbegin(), sushi.rend());
	std::vector<bool> ate(n);
	int x{};
	long long max{};
	std::stack<int> contain;
	for (int i{}; i < k; i++)
	{
		if (!ate[sushi[i].second])
		{
			ate[sushi[i].second] = true;
			max += 2 * x + 1;
			x++;
		}
		else
			contain.push(sushi[i].first);
		max += sushi[i].first;
	}

	long long tmp_max{max};
	for (int i{k}; i < n; i++)
	{
		if (contain.empty()) break;
		if (ate[sushi[i].second]) continue;
		tmp_max += sushi[i].first + 2 * x + 1 - contain.top();
		x++;
		max = std::max(max, tmp_max);
		contain.pop();
		ate[sushi[i].second] = true;
	}

	printf("%lld\n", max);

	return 0;
}
```