#include<bits/stdc++.h>
#define rep(i,a,b) for (int i=(a); i<=(b); i++)
#define per(i,a,b) for (int i=(a); i>=(b); i--)
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1; char ch=getchar();
    while (!(ch>='0'&&ch<='9')) {if (ch=='-')f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9') {x=x*10+(ch-'0'); ch=getchar();}
    return x*f;
}

const int N = 100005;

int n,k,m1,m2;
int t[N],d[N],a[N],b[N],c[N];
bool used[N];
LL sumb[N],sumc[N],ans = 0;

inline bool cmp(int x,int y) {if (d[x]==d[y]) return t[x]>t[y]; return d[x]>d[y];}

int main() {

	n=read(),k=read(); rep(i,1,n) t[i]=read(),d[i]=read(),a[i]=i;
	sort(a+1,a+n+1,cmp);
	rep(i,1,n) {
		if (used[t[a[i]]]) {
			b[++m1] = a[i];
		} else {
			used[t[a[i]]]=1;
			c[++m2] = a[i];
		}
	}
	rep(i,1,m1) sumb[i]=sumb[i-1]+d[b[i]];
	rep(i,1,m2) sumc[i]=sumc[i-1]+d[c[i]];
	rep(i,1,min(k,m2)) ans=max(ans,sumc[i]+sumb[k-i]+(LL)i*i);
	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 rep(i,a,b) for (int i=(a); i<=(b); i++)
#define per(i,a,b) for (int i=(a); i>=(b); i--)
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1; char ch=getchar();
    while (!(ch>='0'&&ch<='9')) {if (ch=='-')f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9') {x=x*10+(ch-'0'); ch=getchar();}
    return x*f;
}

const int N = 100005;

int n,k,m1,m2;
int t[N],d[N],a[N],b[N],c[N];
bool used[N];
LL sumb[N],sumc[N],ans = 0;

inline bool cmp(int x,int y) {if (d[x]==d[y]) return t[x]>t[y]; return d[x]>d[y];}

int main() {

	n=read(),k=read(); rep(i,1,n) t[i]=read(),d[i]=read(),a[i]=i;
	sort(a+1,a+n+1,cmp);
	rep(i,1,n) {
		if (used[t[a[i]]]) {
			b[++m1] = a[i];
		} else {
			used[t[a[i]]]=1;
			c[++m2] = a[i];
		}
	}
	rep(i,1,m1) sumb[i]=sumb[i-1]+d[b[i]];
	rep(i,1,m2) sumc[i]=sumc[i-1]+d[c[i]];
	rep(i,1,min(k,m2)) ans=max(ans,sumc[i]+sumb[k-i]+(LL)i*i);
	printf("%lld\n",ans);	

	return 0;
	
}

```