#include <bits/stdc++.h>
#define N 100005
using namespace std;
int n, m, i, j, u[N], v[N], c[N];
long long k, s, ans;
struct AI{
	int a, i;
	bool operator < (const AI &A) const{
		return a > A.a;
	}
} a[N], b[N];
//int byi(AI a, AI b){
//	if(a.i == b.i) return a.a > b.a;
//	return a.i < b.i;
//}
//int bya(AI a, AI b){
//	if(a.a == b.a) return a.i < b.i;
//	return a.a > b.a;
//}
int main(){
	scanf("%d%d", &n, &m);
	for(i=1; i<=n; i++){
		scanf("%d%d", &a[i].i, &a[i].a);
	}
//	memcpy(b, a, sizeof(a));
//	sort(a+1, a+n+1, byi);
//	sort(b+1, b+n+1, bya);
	sort(a+1, a+n+1);
	for(i=1; i<=n; i++){
		if(!u[a[i].i]){
			v[i] = u[a[i].i] = 1;
			c[++k] = i;
			s += a[i].a;
			if(k == m) break;
		}
	}
	for(i=1; i<=n; i++){
		if(v[i]) continue;
		if(k < m) m--, s += a[i].a, v[i] = 1;
	}
	ans = s + k * k;
	for(i=1; i<=n; i++){
		if(v[i]) continue;
		if(a[i].a <= a[c[k]].a) break;
		s += a[i].a - a[c[k--]].a;
		if(s+k*k > ans) ans = s + k*k;
	}
	printf("%lld\n", ans);
	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>
#define N 100005
using namespace std;
int n, m, i, j, u[N], v[N], c[N];
long long k, s, ans;
struct AI{
	int a, i;
	bool operator < (const AI &A) const{
		return a > A.a;
	}
} a[N], b[N];
//int byi(AI a, AI b){
//	if(a.i == b.i) return a.a > b.a;
//	return a.i < b.i;
//}
//int bya(AI a, AI b){
//	if(a.a == b.a) return a.i < b.i;
//	return a.a > b.a;
//}
int main(){
	scanf("%d%d", &n, &m);
	for(i=1; i<=n; i++){
		scanf("%d%d", &a[i].i, &a[i].a);
	}
//	memcpy(b, a, sizeof(a));
//	sort(a+1, a+n+1, byi);
//	sort(b+1, b+n+1, bya);
	sort(a+1, a+n+1);
	for(i=1; i<=n; i++){
		if(!u[a[i].i]){
			v[i] = u[a[i].i] = 1;
			c[++k] = i;
			s += a[i].a;
			if(k == m) break;
		}
	}
	for(i=1; i<=n; i++){
		if(v[i]) continue;
		if(k < m) m--, s += a[i].a, v[i] = 1;
	}
	ans = s + k * k;
	for(i=1; i<=n; i++){
		if(v[i]) continue;
		if(a[i].a <= a[c[k]].a) break;
		s += a[i].a - a[c[k--]].a;
		if(s+k*k > ans) ans = s + k*k;
	}
	printf("%lld\n", ans);
	return 0;
}
```