#include <iostream>
#include <cstdio>
#include <algorithm>
typedef long long ll;
#define N 100005
#define PII pair<ll, ll>
#define T second
#define D first
using namespace std;

ll n, k, sum, mx, v[N], x[N], y[N], xi, yi;
PII a[N];
int main() {
	ll i;
	scanf("%lld%lld", &n, &k);
	for(i=0; i<n; i++) scanf("%lld%lld", &a[i].T, &a[i].D);
	sort(a, a+n, greater<PII>());
	for(i=0; i<n; i++) {
		if(!v[a[i].T]) {
			v[a[i].T] = 1;
			x[xi] = a[i].D + xi*2+1;
			xi++;
			if(xi<=k) sum += x[xi-1];
		} else y[yi++] = a[i].D;
	}
	mx = sum;
	for(i=0; i<k; i++) {
		sum += y[i] - x[k-1-i];
		mx = max(mx, sum);
	}
	cout<<mx;
	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 <iostream>
#include <cstdio>
#include <algorithm>
typedef long long ll;
#define N 100005
#define PII pair<ll, ll>
#define T second
#define D first
using namespace std;

ll n, k, sum, mx, v[N], x[N], y[N], xi, yi;
PII a[N];
int main() {
	ll i;
	scanf("%lld%lld", &n, &k);
	for(i=0; i<n; i++) scanf("%lld%lld", &a[i].T, &a[i].D);
	sort(a, a+n, greater<PII>());
	for(i=0; i<n; i++) {
		if(!v[a[i].T]) {
			v[a[i].T] = 1;
			x[xi] = a[i].D + xi*2+1;
			xi++;
			if(xi<=k) sum += x[xi-1];
		} else y[yi++] = a[i].D;
	}
	mx = sum;
	for(i=0; i<k; i++) {
		sum += y[i] - x[k-1-i];
		mx = max(mx, sum);
	}
	cout<<mx;
	return 0;
}
```