#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

int main(){
	int n, k;
	cin>>n>>k;
	vector<pair<int, int> > arr(n);
	for(auto &x : arr){
		cin>>x.second>>x.first;
		x.second -= 1;
	}
	sort(arr.rbegin(), arr.rend());
	vector<int> mask(n, 0);
	map<int, int> mp;
	ll sum = 0, cnt = 0;
	for(int i = 0; i < k; i++){
		if(mask[arr[i].second]){
			mp[arr[i].first]++;
		}else{
			mask[arr[i].second] = 1;
			cnt++;
		}
		sum += arr[i].first;
	}
	ll ans = sum + cnt*cnt;
	for(int i = k; i < n; i++){
		if(cnt == k)break;
		if(mask[arr[i].second])continue;
		mask[arr[i].second] = 1;
		int val = (*mp.begin()).first;
		mp[val]--; if(mp[val] == 0)mp.erase(val);
		sum += arr[i].first - val; cnt++;
		ans = max(ans, sum + cnt*cnt);
	}
	cout<<ans<<endl;
}
### Prompt

Create a solution in CPP for 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 namespace std;

typedef long long ll;

int main(){
	int n, k;
	cin>>n>>k;
	vector<pair<int, int> > arr(n);
	for(auto &x : arr){
		cin>>x.second>>x.first;
		x.second -= 1;
	}
	sort(arr.rbegin(), arr.rend());
	vector<int> mask(n, 0);
	map<int, int> mp;
	ll sum = 0, cnt = 0;
	for(int i = 0; i < k; i++){
		if(mask[arr[i].second]){
			mp[arr[i].first]++;
		}else{
			mask[arr[i].second] = 1;
			cnt++;
		}
		sum += arr[i].first;
	}
	ll ans = sum + cnt*cnt;
	for(int i = k; i < n; i++){
		if(cnt == k)break;
		if(mask[arr[i].second])continue;
		mask[arr[i].second] = 1;
		int val = (*mp.begin()).first;
		mp[val]--; if(mp[val] == 0)mp.erase(val);
		sum += arr[i].first - val; cnt++;
		ans = max(ans, sum + cnt*cnt);
	}
	cout<<ans<<endl;
}
```