#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define MOD 1000000007
#define INF (1<<29)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;

vector<P> jewels;
vector<Int> jewel[216000];

Int val2(Int a){
  return jewel[a][0] + jewel[a][1];
}

Int val3(Int a){
  return jewel[a][0] + jewel[a][1] + jewel[a][2];
}

Int minval(multiset<Int> &S){
  return *S.begin();
}


Int maxval(multiset<Int> &S){
  auto it = S.end();it--;
  return *it;
}
Int minval2(multiset<P> &S){
  return val2(S.begin()->second);
}

Int maxval2(multiset<P> &S){
  auto it = S.end();it--;
  return val2(it->second);
}

Int maxval3(multiset<P> &S){
  auto it = S.end();it--;
  return val3(it->second);
}


Int cnt[216000];
multiset<Int> PP, Q;
multiset<P> R, S, T;
Int ans[216000];

Int n, k, c, v;

int main(){
  fill(ans, ans + 216000, -2);
  cin >> n >> k;
  for(Int i = 0;i < n;i++){
    cin >> c >> v;
    jewel[c].push_back(v*2);
  }
  for(Int i = 0;i < n;i++){
    if(jewel[i].empty())continue;
    sort(jewel[i].rbegin(), jewel[i].rend());
    Int sum = jewel[i][0] + jewel[i][1];
    jewel[i][0] = jewel[i][1] = sum / 2;
    for(Int j = 0;j < jewel[i].size();j++){
      v = jewel[i][j];
      jewels.push_back(P(v, i));
    }
    S.insert(P(val2(i), i));
    if(jewel[i].size() >= 3){
      T.insert(P(val3(i), i));
    }
  }
  sort(jewels.rbegin(), jewels.rend());
  Int sum = 0;
  for(Int i = 0;i < n;i++){
    v = jewels[i].first;
    c = jewels[i].second;
    cnt[c]++;
    if(cnt[c] >= 2){
      sum += v;
      ans[i+1] = sum;
      if(cnt[c] == 2){
	R.insert(P(val2(c),c));
	S.erase(P(val2(c),c));
	if(jewel[c].size() >= 3)
	  T.erase(P(val3(c), c));
	for(Int j = 2;j < jewel[c].size();j++){
	  Q.insert(jewel[c][j]);
	}
      }
      else{
	R.erase(P(val2(c),c));
	Q.erase(v);
	PP.insert(v);
      }
      continue;
    }
    if(!Q.empty())ans[i+1] = max(ans[i+1], sum + maxval(Q));
    if(!PP.empty() && !S.empty())
      ans[i+1]  = max(ans[i+1], sum - minval(PP) + maxval2(S));
    if(!R.empty() && !T.empty())
      ans[i+1]  = max(ans[i+1], sum - minval2(R) + maxval3(T));
    sum += v;    
  }

    

  for(Int i = 0;i < n;i++)cout << ans[i+1] / 2 << endl;
  
  return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

There are N jewels, numbered 1 to N. The color of these jewels are represented by integers between 1 and K (inclusive), and the color of Jewel i is C_i. Also, these jewels have specified values, and the value of Jewel i is V_i.

Snuke would like to choose some of these jewels to exhibit. Here, the set of the chosen jewels must satisfy the following condition:

* For each chosen jewel, there is at least one more jewel of the same color that is chosen.



For each integer x such that 1 \leq x \leq N, determine if it is possible to choose exactly x jewels, and if it is possible, find the maximum possible sum of the values of chosen jewels in that case.

Constraints

* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq \lfloor N/2 \rfloor
* 1 \leq C_i \leq K
* 1 \leq V_i \leq 10^9
* For each of the colors, there are at least two jewels of that color.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N K
C_1 V_1
C_2 V_2
:
C_N V_N


Output

Print N lines. In the i-th line, if it is possible to choose exactly i jewels, print the maximum possible sum of the values of chosen jewels in that case, and print -1 otherwise.

Examples

Input

5 2
1 1
1 2
1 3
2 4
2 5


Output

-1
9
6
14
15


Input

5 2
1 1
1 2
2 3
2 4
2 5


Output

-1
9
12
12
15


Input

8 4
3 2
2 3
4 5
1 7
3 11
4 13
1 17
2 19


Output

-1
24
-1
46
-1
64
-1
77


Input

15 5
3 87
1 25
1 27
3 58
2 85
5 19
5 39
1 58
3 12
4 13
5 54
4 100
2 33
5 13
2 55


Output

-1
145
173
285
318
398
431
491
524
576
609
634
653
666
678

### Response

```cpp
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define MOD 1000000007
#define INF (1<<29)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;

vector<P> jewels;
vector<Int> jewel[216000];

Int val2(Int a){
  return jewel[a][0] + jewel[a][1];
}

Int val3(Int a){
  return jewel[a][0] + jewel[a][1] + jewel[a][2];
}

Int minval(multiset<Int> &S){
  return *S.begin();
}


Int maxval(multiset<Int> &S){
  auto it = S.end();it--;
  return *it;
}
Int minval2(multiset<P> &S){
  return val2(S.begin()->second);
}

Int maxval2(multiset<P> &S){
  auto it = S.end();it--;
  return val2(it->second);
}

Int maxval3(multiset<P> &S){
  auto it = S.end();it--;
  return val3(it->second);
}


Int cnt[216000];
multiset<Int> PP, Q;
multiset<P> R, S, T;
Int ans[216000];

Int n, k, c, v;

int main(){
  fill(ans, ans + 216000, -2);
  cin >> n >> k;
  for(Int i = 0;i < n;i++){
    cin >> c >> v;
    jewel[c].push_back(v*2);
  }
  for(Int i = 0;i < n;i++){
    if(jewel[i].empty())continue;
    sort(jewel[i].rbegin(), jewel[i].rend());
    Int sum = jewel[i][0] + jewel[i][1];
    jewel[i][0] = jewel[i][1] = sum / 2;
    for(Int j = 0;j < jewel[i].size();j++){
      v = jewel[i][j];
      jewels.push_back(P(v, i));
    }
    S.insert(P(val2(i), i));
    if(jewel[i].size() >= 3){
      T.insert(P(val3(i), i));
    }
  }
  sort(jewels.rbegin(), jewels.rend());
  Int sum = 0;
  for(Int i = 0;i < n;i++){
    v = jewels[i].first;
    c = jewels[i].second;
    cnt[c]++;
    if(cnt[c] >= 2){
      sum += v;
      ans[i+1] = sum;
      if(cnt[c] == 2){
	R.insert(P(val2(c),c));
	S.erase(P(val2(c),c));
	if(jewel[c].size() >= 3)
	  T.erase(P(val3(c), c));
	for(Int j = 2;j < jewel[c].size();j++){
	  Q.insert(jewel[c][j]);
	}
      }
      else{
	R.erase(P(val2(c),c));
	Q.erase(v);
	PP.insert(v);
      }
      continue;
    }
    if(!Q.empty())ans[i+1] = max(ans[i+1], sum + maxval(Q));
    if(!PP.empty() && !S.empty())
      ans[i+1]  = max(ans[i+1], sum - minval(PP) + maxval2(S));
    if(!R.empty() && !T.empty())
      ans[i+1]  = max(ans[i+1], sum - minval2(R) + maxval3(T));
    sum += v;    
  }

    

  for(Int i = 0;i < n;i++)cout << ans[i+1] / 2 << endl;
  
  return 0;
}
```