#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <utility>
#include <set>
#define llint long long

using namespace std;
typedef pair<llint, pair<llint, llint> > J;

llint n, k;
vector<llint> vec[200005];
vector<J> vec2, vec2r;
priority_queue<J> Q;
set<pair<llint, llint> > S;
llint preS, preP;

int main(void)
{
	cin >> n >> k;
	llint c, v;
	for(int i = 1; i <= n; i++){
		cin >> c >> v;
		vec[c].push_back(2*v);
	}
	for(int i = 1; i <= k; i++){
		sort(vec[i].rbegin(), vec[i].rend());
		llint m = vec[i][0] + vec[i][1];
		vec[i][0] = vec[i][1] = m/2;
		for(int j = 0; j < vec[i].size(); j++) vec2.push_back(make_pair(vec[i][j], make_pair(-i, -j)));
	}
	sort(vec2.rbegin(), vec2.rend());
	
	preS = preP = -1;
	for(int i = 1; i <= k; i++){
		if(vec[i].size() >= 3) S.insert(make_pair(vec[i][0] + vec[i][1] + vec[i][2], i));
	}
	
	llint sum = 0;
	for(int i = 1; i <= n; i++){
		sum += vec2[i-1].first;
		
		llint c = -vec2[i-1].second.first, ans = sum;
		if(Q.size() && Q.top() == vec2[i-1]){
			Q.pop();
			preS = vec2[i-1].first;
		}
		if(-vec2[i-1].second.second == 1){
			for(int j = 2; j < vec[c].size(); j++) Q.push(make_pair(vec[c][j], make_pair(-c, -j)));
			preP = vec[c][0] + vec[c][1];
			S.erase(make_pair(vec[c][0] + vec[c][1] + vec[c][2], c));
		}
		
		if(i < n && vec2[i-1].second.first == vec2[i].second.first && vec2[i-1].second.second == 0){
			ans = -2;
			if(Q.size()) ans = max(ans, sum - vec2[i-1].first + Q.top().first);
			if(S.size() && preP != -1){
				auto p = S.end();
				p--;
				ans = max(ans, sum - (preP + vec2[i-1].first) + p->first);
			}
			if(preS != -1) ans = max(ans, sum - preS + vec2[i].first);
		}
		cout << ans/2 << endl;
	}
	
	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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <utility>
#include <set>
#define llint long long

using namespace std;
typedef pair<llint, pair<llint, llint> > J;

llint n, k;
vector<llint> vec[200005];
vector<J> vec2, vec2r;
priority_queue<J> Q;
set<pair<llint, llint> > S;
llint preS, preP;

int main(void)
{
	cin >> n >> k;
	llint c, v;
	for(int i = 1; i <= n; i++){
		cin >> c >> v;
		vec[c].push_back(2*v);
	}
	for(int i = 1; i <= k; i++){
		sort(vec[i].rbegin(), vec[i].rend());
		llint m = vec[i][0] + vec[i][1];
		vec[i][0] = vec[i][1] = m/2;
		for(int j = 0; j < vec[i].size(); j++) vec2.push_back(make_pair(vec[i][j], make_pair(-i, -j)));
	}
	sort(vec2.rbegin(), vec2.rend());
	
	preS = preP = -1;
	for(int i = 1; i <= k; i++){
		if(vec[i].size() >= 3) S.insert(make_pair(vec[i][0] + vec[i][1] + vec[i][2], i));
	}
	
	llint sum = 0;
	for(int i = 1; i <= n; i++){
		sum += vec2[i-1].first;
		
		llint c = -vec2[i-1].second.first, ans = sum;
		if(Q.size() && Q.top() == vec2[i-1]){
			Q.pop();
			preS = vec2[i-1].first;
		}
		if(-vec2[i-1].second.second == 1){
			for(int j = 2; j < vec[c].size(); j++) Q.push(make_pair(vec[c][j], make_pair(-c, -j)));
			preP = vec[c][0] + vec[c][1];
			S.erase(make_pair(vec[c][0] + vec[c][1] + vec[c][2], c));
		}
		
		if(i < n && vec2[i-1].second.first == vec2[i].second.first && vec2[i-1].second.second == 0){
			ans = -2;
			if(Q.size()) ans = max(ans, sum - vec2[i-1].first + Q.top().first);
			if(S.size() && preP != -1){
				auto p = S.end();
				p--;
				ans = max(ans, sum - (preP + vec2[i-1].first) + p->first);
			}
			if(preS != -1) ans = max(ans, sum - preS + vec2[i].first);
		}
		cout << ans/2 << endl;
	}
	
	return 0;
}
```