#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<LL,LL> pii;
#define N 300000
#define P(x,y) make_pair(x,y)

LL n,m,x,y,now,f[N],g[N],ans;
bool vis[N];
vector<LL> a[N];
priority_queue<LL> pq1;
priority_queue<pii> pq2,pq3;
priority_queue<LL,vector<LL>,greater<LL> > fin1,fin2;

int main(){
	scanf("%lld%lld",&n,&m);
	for (LL i=1;i<=n;++i){
		scanf("%lld%lld",&x,&y);
		a[x].push_back(y);
	}
	for (LL i=1;i<=m;++i){
		sort(a[i].begin(),a[i].end(),greater<LL>());
		f[i]=a[i][0]+a[i][1];
		pq2.push(P(f[i],i));
		if (a[i].size()>2){
			g[i]=f[i]+a[i][2];
			pq3.push(P(g[i],i));
		}
	}
	for (LL i=1;i<=n;++i){
		LL t1=pq1.empty()?-1:pq1.top(),t2=pq2.empty()?-1:pq2.top().second;
		if (t1==-1&&t2==-1){puts("-1"); continue;}
		if (t2==-1||t1*2>f[t2]){
			++now; pq1.pop();
			fin1.push(t1);
			printf("%lld\n",ans+=t1);
		}
		else{
			if (now+1==i){
				LL tmp=0;
				if (t1>=0) tmp=ans+t1;
				if (!fin1.empty()) tmp=max(tmp,ans+f[t2]-fin1.top());
				if (!fin2.empty()&&!pq3.empty()) tmp=max(tmp,ans+pq3.top().first-fin2.top());
				if (!tmp) puts("-1");
				else printf("%lld\n",tmp);
			}
			else{
				now+=2; pq2.pop(); vis[t2]=1;
				for (;!pq3.empty()&&vis[pq3.top().second];pq3.pop());
				for (LL i=2;i<(LL)a[t2].size();++i) pq1.push(a[t2][i]);
				fin2.push(f[t2]);
				printf("%lld\n",ans+=f[t2]);
			}
		}
	}
	
	return 0;
}
			
### Prompt

Please create a solution in Cpp 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;

typedef long long LL;
typedef pair<LL,LL> pii;
#define N 300000
#define P(x,y) make_pair(x,y)

LL n,m,x,y,now,f[N],g[N],ans;
bool vis[N];
vector<LL> a[N];
priority_queue<LL> pq1;
priority_queue<pii> pq2,pq3;
priority_queue<LL,vector<LL>,greater<LL> > fin1,fin2;

int main(){
	scanf("%lld%lld",&n,&m);
	for (LL i=1;i<=n;++i){
		scanf("%lld%lld",&x,&y);
		a[x].push_back(y);
	}
	for (LL i=1;i<=m;++i){
		sort(a[i].begin(),a[i].end(),greater<LL>());
		f[i]=a[i][0]+a[i][1];
		pq2.push(P(f[i],i));
		if (a[i].size()>2){
			g[i]=f[i]+a[i][2];
			pq3.push(P(g[i],i));
		}
	}
	for (LL i=1;i<=n;++i){
		LL t1=pq1.empty()?-1:pq1.top(),t2=pq2.empty()?-1:pq2.top().second;
		if (t1==-1&&t2==-1){puts("-1"); continue;}
		if (t2==-1||t1*2>f[t2]){
			++now; pq1.pop();
			fin1.push(t1);
			printf("%lld\n",ans+=t1);
		}
		else{
			if (now+1==i){
				LL tmp=0;
				if (t1>=0) tmp=ans+t1;
				if (!fin1.empty()) tmp=max(tmp,ans+f[t2]-fin1.top());
				if (!fin2.empty()&&!pq3.empty()) tmp=max(tmp,ans+pq3.top().first-fin2.top());
				if (!tmp) puts("-1");
				else printf("%lld\n",tmp);
			}
			else{
				now+=2; pq2.pop(); vis[t2]=1;
				for (;!pq3.empty()&&vis[pq3.top().second];pq3.pop());
				for (LL i=2;i<(LL)a[t2].size();++i) pq1.push(a[t2][i]);
				fin2.push(f[t2]);
				printf("%lld\n",ans+=f[t2]);
			}
		}
	}
	
	return 0;
}
			
```