#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pll pair<ll,ll>

vector<ll> v[100010];
vector<pll> v1;

multiset<ll> st1,st2,st3,st4;

int cnt[100010],n,m;

inline int rd()
{
	int x=0;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar());
	for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x;
}

inline void print(ll x)
{
	static char s[233];
	if (x<0) putchar('-'),x=-x;
	if (!x) { putchar('0');putchar('\n');return; }
	int tot=0;
	for (;x;x/=10) s[++tot]=x%10+'0';
	for (;tot;tot--) putchar(s[tot]);
	putchar('\n');
}

int main()
{
	n=rd();m=rd();
	for (int i=1;i<=n;i++)
	{
		int x=rd(),y=rd();
		v[x].push_back(y);
	}
	for (int i=1;i<=m;i++)
	{
		sort(v[i].begin(),v[i].end());reverse(v[i].begin(),v[i].end());
		ll hh=v[i][0]+v[i][1];
		v[i][0]=v[i][1]=hh;
		for (int j=2;j<v[i].size();j++) v[i][j]*=2;
		for (int j=0;j<v[i].size();j++) v1.push_back(pll(v[i][j],i));
	}
	sort(v1.begin(),v1.end());reverse(v1.begin(),v1.end());
	for (int i=1;i<=m;i++) if (v[i].size()>=3) st4.insert(v[i][0]+v[i][1]+v[i][2]);
	ll sum=0;
	for (int i=1;i<=n;i++)
	{
		pll hh=v1[i-1];
		ll x=hh.second,y=hh.first,ans=-2;
		if (!cnt[x])
		{
			if (st1.size()) ans=max(ans,sum+*st1.rbegin());
			if (st2.size()) ans=max(ans,sum-*st2.begin()+y*2);
			if (st3.size()&&st4.size()) ans=max(ans,sum-*st3.begin()+*st4.rbegin());
			if (v[x].size()>=3) st4.erase(st4.find(v[x][0]+v[x][1]+v[x][2]));
			st3.insert(y*2);
			if (v[x].size()>2) st1.insert(v[x][2]);
			cnt[x]=1;sum+=y*2;
			print(ans/2);print(sum/2);
		}
		else if (cnt[x]>1)
		{
			sum+=y;ans=sum;
			if (cnt[x]==2) st3.erase(st3.find(v[x][0]*2));
			if (cnt[x]>2) st2.erase(st2.find(v[x][cnt[x]-1]));
			st2.insert(v[x][cnt[x]]);st1.erase(st1.find(v[x][cnt[x]]));
			if (cnt[x]+1<v[x].size()) st1.insert(v[x][cnt[x]+1]);
			cnt[x]++;
			print(ans/2);
		}
		else cnt[x]++;
	}
	return 0;
}
### Prompt

Your task is to create 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;
#define ll long long
#define pll pair<ll,ll>

vector<ll> v[100010];
vector<pll> v1;

multiset<ll> st1,st2,st3,st4;

int cnt[100010],n,m;

inline int rd()
{
	int x=0;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar());
	for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x;
}

inline void print(ll x)
{
	static char s[233];
	if (x<0) putchar('-'),x=-x;
	if (!x) { putchar('0');putchar('\n');return; }
	int tot=0;
	for (;x;x/=10) s[++tot]=x%10+'0';
	for (;tot;tot--) putchar(s[tot]);
	putchar('\n');
}

int main()
{
	n=rd();m=rd();
	for (int i=1;i<=n;i++)
	{
		int x=rd(),y=rd();
		v[x].push_back(y);
	}
	for (int i=1;i<=m;i++)
	{
		sort(v[i].begin(),v[i].end());reverse(v[i].begin(),v[i].end());
		ll hh=v[i][0]+v[i][1];
		v[i][0]=v[i][1]=hh;
		for (int j=2;j<v[i].size();j++) v[i][j]*=2;
		for (int j=0;j<v[i].size();j++) v1.push_back(pll(v[i][j],i));
	}
	sort(v1.begin(),v1.end());reverse(v1.begin(),v1.end());
	for (int i=1;i<=m;i++) if (v[i].size()>=3) st4.insert(v[i][0]+v[i][1]+v[i][2]);
	ll sum=0;
	for (int i=1;i<=n;i++)
	{
		pll hh=v1[i-1];
		ll x=hh.second,y=hh.first,ans=-2;
		if (!cnt[x])
		{
			if (st1.size()) ans=max(ans,sum+*st1.rbegin());
			if (st2.size()) ans=max(ans,sum-*st2.begin()+y*2);
			if (st3.size()&&st4.size()) ans=max(ans,sum-*st3.begin()+*st4.rbegin());
			if (v[x].size()>=3) st4.erase(st4.find(v[x][0]+v[x][1]+v[x][2]));
			st3.insert(y*2);
			if (v[x].size()>2) st1.insert(v[x][2]);
			cnt[x]=1;sum+=y*2;
			print(ans/2);print(sum/2);
		}
		else if (cnt[x]>1)
		{
			sum+=y;ans=sum;
			if (cnt[x]==2) st3.erase(st3.find(v[x][0]*2));
			if (cnt[x]>2) st2.erase(st2.find(v[x][cnt[x]-1]));
			st2.insert(v[x][cnt[x]]);st1.erase(st1.find(v[x][cnt[x]]));
			if (cnt[x]+1<v[x].size()) st1.insert(v[x][cnt[x]+1]);
			cnt[x]++;
			print(ans/2);
		}
		else cnt[x]++;
	}
	return 0;
}
```