#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
typedef long long ll;
int main(){
    ll n,k,t,d;
    cin>>n>>k;
    vector<pair<ll,ll>> A,B,C;
    int var[200001]={0};
    rep(i,n) cin>>t>>d,A.push_back(make_pair(d,t));
    sort(A.begin(),A.end(),greater<pair<ll,ll>>());
    rep(i,k) B.push_back(A[i]),var[A[i].second]++;
    rep(i,n-k) C.push_back(A[i+k]);
    reverse(B.begin(),B.end());
    ll c=0,ans=0,sum=0,x=0;
    rep(i,k) sum+=B[i].first;
    rep(i,n) if(var[i+1]) x++;
    ans=sum+x*x;
    rep(i,k){
        if(var[B[i].second]>1){
            while(c<n-k&&var[C[c].second]!=0) c++;
            if(c==n-k) break;
            var[B[i].second]--;
            sum-=B[i].first;
            B[i].first=C[c].first;
            B[i].second=C[c].second;
            var[B[i].second]++;
            sum+=B[i].first;
            x++;
        }
        ans=max(ans,sum+x*x);
    }
    cout<<ans;
}
### 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;
#define rep(i,n) for(int i=0;i<n;i++)
typedef long long ll;
int main(){
    ll n,k,t,d;
    cin>>n>>k;
    vector<pair<ll,ll>> A,B,C;
    int var[200001]={0};
    rep(i,n) cin>>t>>d,A.push_back(make_pair(d,t));
    sort(A.begin(),A.end(),greater<pair<ll,ll>>());
    rep(i,k) B.push_back(A[i]),var[A[i].second]++;
    rep(i,n-k) C.push_back(A[i+k]);
    reverse(B.begin(),B.end());
    ll c=0,ans=0,sum=0,x=0;
    rep(i,k) sum+=B[i].first;
    rep(i,n) if(var[i+1]) x++;
    ans=sum+x*x;
    rep(i,k){
        if(var[B[i].second]>1){
            while(c<n-k&&var[C[c].second]!=0) c++;
            if(c==n-k) break;
            var[B[i].second]--;
            sum-=B[i].first;
            B[i].first=C[c].first;
            B[i].second=C[c].second;
            var[B[i].second]++;
            sum+=B[i].first;
            x++;
        }
        ans=max(ans,sum+x*x);
    }
    cout<<ans;
}
```