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

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

const int INF=1001001001001001001ll;

int N,K;
int C[222222],V[222222];
vint lis[222222];

int ans[222222];

signed main(){
	cin>>N>>K;
	rep(i,N){
		cin>>C[i]>>V[i];
		C[i]--;
		lis[C[i]].pb(V[i]);
	}
	rep(i,K){
		sort(all(lis[i]));
		reverse(all(lis[i]));
	}
	vector<pair<pint,int>>ord;
	rep(i,K){
		ord.pb({{lis[i][0]+lis[i][1],(lis[i].size()>2?lis[i][2]:-INF)},i});
	}
	sort(all(ord));reverse(all(ord));

	vint uku(ord.size()+1);
	uku.back()=-INF;
	for(int i=ord.size()-1;i>=0;i--){
		uku[i]=uku[i+1];
		int k=ord[i].se;
		if(lis[k].size()<3)continue;
		chmax(uku[i],lis[k][0]+lis[k][1]+lis[k][2]);
	}



	fill_n(ans,N+1,-INF);

	int n=0;
	int cur=0;
	vector<int>vs;
	priority_queue<int>nex;
	int sum=0;
	while(true){
		chmax(ans[n],sum);
		if(uku[cur]!=-INF)chmax(ans[n+3],sum+uku[cur]);


		if(n==N)break;
		bool ok=true;
		if(nex.size()==0){
			if(vs.size()){
				
				sum-=vs.back();
				nex.push(vs.back());
				vs.pop_back();
				n--;
			}
			ok=false;
		}
		else if(vs.size()==0){}
		else if(cur<K&&ord[cur].fi.fi>vs.back()+nex.top()){
			sum-=vs.back();
			nex.push(vs.back());
			vs.pop_back();
			n--;
			ok=false;
		}

		if(ok){
			sum+=nex.top();
			vs.pb(nex.top());
			nex.pop();
			n++;
		}
		else{
			int k=ord[cur].se;
			sum+=ord[cur].fi.fi;
			for(int i=2;i<lis[k].size();i++)nex.push(lis[k][i]);
			cur++;
			n+=2;
		}
	}
	for(int i=1;i<=N;i++){
		if(ans[i]==-INF)cout<<-1<<endl;
		else cout<<ans[i]<<endl;
	}
	return 0;
}
### Prompt

In cpp, your task is to solve 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;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

const int INF=1001001001001001001ll;

int N,K;
int C[222222],V[222222];
vint lis[222222];

int ans[222222];

signed main(){
	cin>>N>>K;
	rep(i,N){
		cin>>C[i]>>V[i];
		C[i]--;
		lis[C[i]].pb(V[i]);
	}
	rep(i,K){
		sort(all(lis[i]));
		reverse(all(lis[i]));
	}
	vector<pair<pint,int>>ord;
	rep(i,K){
		ord.pb({{lis[i][0]+lis[i][1],(lis[i].size()>2?lis[i][2]:-INF)},i});
	}
	sort(all(ord));reverse(all(ord));

	vint uku(ord.size()+1);
	uku.back()=-INF;
	for(int i=ord.size()-1;i>=0;i--){
		uku[i]=uku[i+1];
		int k=ord[i].se;
		if(lis[k].size()<3)continue;
		chmax(uku[i],lis[k][0]+lis[k][1]+lis[k][2]);
	}



	fill_n(ans,N+1,-INF);

	int n=0;
	int cur=0;
	vector<int>vs;
	priority_queue<int>nex;
	int sum=0;
	while(true){
		chmax(ans[n],sum);
		if(uku[cur]!=-INF)chmax(ans[n+3],sum+uku[cur]);


		if(n==N)break;
		bool ok=true;
		if(nex.size()==0){
			if(vs.size()){
				
				sum-=vs.back();
				nex.push(vs.back());
				vs.pop_back();
				n--;
			}
			ok=false;
		}
		else if(vs.size()==0){}
		else if(cur<K&&ord[cur].fi.fi>vs.back()+nex.top()){
			sum-=vs.back();
			nex.push(vs.back());
			vs.pop_back();
			n--;
			ok=false;
		}

		if(ok){
			sum+=nex.top();
			vs.pb(nex.top());
			nex.pop();
			n++;
		}
		else{
			int k=ord[cur].se;
			sum+=ord[cur].fi.fi;
			for(int i=2;i<lis[k].size();i++)nex.push(lis[k][i]);
			cur++;
			n+=2;
		}
	}
	for(int i=1;i<=N;i++){
		if(ans[i]==-INF)cout<<-1<<endl;
		else cout<<ans[i]<<endl;
	}
	return 0;
}
```