#include <bits/stdc++.h>
#define fi first
#define se second
using namespace std;

long long n, k, cnt = 0;
vector<pair<long long, long long>> s;
long long num[100005] = {0};

long long solve();

int main() {
  cin >> n >> k;
  for(long long i = 0; i < n; ++i) {
    long long t, d;
    cin >> t >> d;
    --t;
    s.push_back(make_pair(d, t));
  }
  sort(s.begin(), s.end(),
       greater<pair<long long, long long>>());
  cout << solve() << endl;
  return 0;
}

long long solve() {
  long long now = 0, ans = 0, r = k - 1;
  for(long long i = 0; i < k; ++i) {
    now += s[i].fi;
    if(++num[s[i].se] == 1) ++cnt;
  }
  ans = now + cnt * cnt;
  for(long long i = k; i < n; ++i)
    if(num[s[i].se] == 0) {
      while(r >= 0 && num[s[r].se] == 1) --r;
      if(r < 0) break;
      now += s[i].fi - s[r].fi;
      ++cnt;
      ++num[s[i].se];
      --num[s[r].se];
      --r;
      ans = max(ans, now + cnt * cnt);
    }
  return ans;
}
### Prompt

Your task is to create 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 fi first
#define se second
using namespace std;

long long n, k, cnt = 0;
vector<pair<long long, long long>> s;
long long num[100005] = {0};

long long solve();

int main() {
  cin >> n >> k;
  for(long long i = 0; i < n; ++i) {
    long long t, d;
    cin >> t >> d;
    --t;
    s.push_back(make_pair(d, t));
  }
  sort(s.begin(), s.end(),
       greater<pair<long long, long long>>());
  cout << solve() << endl;
  return 0;
}

long long solve() {
  long long now = 0, ans = 0, r = k - 1;
  for(long long i = 0; i < k; ++i) {
    now += s[i].fi;
    if(++num[s[i].se] == 1) ++cnt;
  }
  ans = now + cnt * cnt;
  for(long long i = k; i < n; ++i)
    if(num[s[i].se] == 0) {
      while(r >= 0 && num[s[r].se] == 1) --r;
      if(r < 0) break;
      now += s[i].fi - s[r].fi;
      ++cnt;
      ++num[s[i].se];
      --num[s[r].se];
      --r;
      ans = max(ans, now + cnt * cnt);
    }
  return ans;
}
```