#include <bits/stdc++.h>
using namespace std;

const long long INF=1000000000000000ll;

int main()
{
  int n,K;
  scanf("%d%d",&n,&K);
  vector<int> C(n);
  vector<long long> V(n);
  for(int i=0;i<n;i++){
    scanf("%d%lld",&C[i],&V[i]);
    C[i]--;
  }
  vector<vector<long long> > CV(K);
  for(int i=0;i<n;i++){
    CV[C[i]].push_back(V[i]);
  }
  vector<pair<long long,int> > V2(K+1);
  multiset<long long> V3;
  for(int j=0;j<K;j++){
    sort(CV[j].begin(),CV[j].end(),greater<long long>());
    V2[j]=make_pair(CV[j][0]+CV[j][1],j);
    if(CV[j].size()>=3){
      V3.insert(CV[j][0]+CV[j][1]+CV[j][2]);
    }
  }
  V2[K]=make_pair(-INF,-1);
  V3.insert(-INF);
  sort(V2.begin(),V2.end(),greater<pair<long long,int> >());
  multiset<long long> S;
  S.insert(-INF);
  long long S1=INF,S2=INF;
  vector<long long> ans(n+1);
  ans[0]=0;
  for(int i=0,j=0;i<n;){
    int k=V2[j].second;
    long long s1=*S.rbegin(),s2=V2[j].first;
    //printf("%d %lld %lld\n",k,s1,s2);
    if(2*s1>=s2){
      ans[i+1]=ans[i]+s1;
      S1=s1;
      S.erase(S.find(s1));
      i++;
    }
    else{
      ans[i+1]=max(-1ll,max(ans[i]+s1,max(ans[i]-S1+s2,ans[i]-S2+*V3.rbegin())));
      ans[i+2]=ans[i]+s2;
      S2=s2;
      for(int t=2;t<CV[k].size();t++){
	S.insert(CV[k][t]);
      }
      if(CV[k].size()>=3){
	V3.erase(V3.find(CV[k][0]+CV[k][1]+CV[k][2]));
      }
      i+=2;
      j++;
    }
  }
  for(int i=1;i<=n;i++){
    printf("%lld\n",ans[i]);
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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 <bits/stdc++.h>
using namespace std;

const long long INF=1000000000000000ll;

int main()
{
  int n,K;
  scanf("%d%d",&n,&K);
  vector<int> C(n);
  vector<long long> V(n);
  for(int i=0;i<n;i++){
    scanf("%d%lld",&C[i],&V[i]);
    C[i]--;
  }
  vector<vector<long long> > CV(K);
  for(int i=0;i<n;i++){
    CV[C[i]].push_back(V[i]);
  }
  vector<pair<long long,int> > V2(K+1);
  multiset<long long> V3;
  for(int j=0;j<K;j++){
    sort(CV[j].begin(),CV[j].end(),greater<long long>());
    V2[j]=make_pair(CV[j][0]+CV[j][1],j);
    if(CV[j].size()>=3){
      V3.insert(CV[j][0]+CV[j][1]+CV[j][2]);
    }
  }
  V2[K]=make_pair(-INF,-1);
  V3.insert(-INF);
  sort(V2.begin(),V2.end(),greater<pair<long long,int> >());
  multiset<long long> S;
  S.insert(-INF);
  long long S1=INF,S2=INF;
  vector<long long> ans(n+1);
  ans[0]=0;
  for(int i=0,j=0;i<n;){
    int k=V2[j].second;
    long long s1=*S.rbegin(),s2=V2[j].first;
    //printf("%d %lld %lld\n",k,s1,s2);
    if(2*s1>=s2){
      ans[i+1]=ans[i]+s1;
      S1=s1;
      S.erase(S.find(s1));
      i++;
    }
    else{
      ans[i+1]=max(-1ll,max(ans[i]+s1,max(ans[i]-S1+s2,ans[i]-S2+*V3.rbegin())));
      ans[i+2]=ans[i]+s2;
      S2=s2;
      for(int t=2;t<CV[k].size();t++){
	S.insert(CV[k][t]);
      }
      if(CV[k].size()>=3){
	V3.erase(V3.find(CV[k][0]+CV[k][1]+CV[k][2]));
      }
      i+=2;
      j++;
    }
  }
  for(int i=1;i<=n;i++){
    printf("%lld\n",ans[i]);
  }
  return 0;
}

```