#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

int main(){
	int N,K; cin >> N >> K;
	vector<vector<LL>> A(K);
	repp(i,0,N){
		int C; LL V; cin >> C >> V;
		A[C-1].push_back(V);
	}
	vector<int> t(K),w(K);
	repp(i,0,K){
		sort(A[i].begin(),A[i].end(),greater<LL>());
		A[i].push_back(-1e18);
		w[i] = t[i] = i;
	}
	sort(w.begin(),w.end(),[&](const int &i, const int &j){return A[i][0]+A[i][1] > A[j][0]+A[j][1];});
	sort(t.begin(),t.end(),[&](const int &i, const int &j){return A[i][0]+A[i][1]+A[i][2] > A[j][0]+A[j][1]+A[j][2];});
	vector<LL> ans(N+3,-1);
	priority_queue<LL> Q;
	vector<int> b(K,0);
	ans[1] = A[w[0]][0]+A[w[0]][1];
	repp(i,2,A[w[0]].size()-1) Q.push(A[w[0]][i]);
	int	wp = 1, tp = 0, m = 0;
	b[w[0]] = 1;
	while(tp < K && A[t[tp]][2] < 0) ++tp;
	if(A[t[0]][2] >= 0){
		ans[2] = A[t[0]][0]+A[t[0]][1]+A[t[0]][2];
	}
	while(tp < K && (b[t[tp]] == 1 || A[t[tp]][2] < 0)) ++tp;
	repp(i,3,N){
		if(m == 0){
			int z = 0;
			if(Q.size() >= 2){
				LL r = Q.top(); Q.pop();
				if(ans[i] < ans[i-2]+r+Q.top()){
					ans[i] = ans[i-2]+r+Q.top();
					z = 1;
				}
				Q.push(r);
			}
			if(wp < K && ans[i] < ans[i-2]+A[w[wp]][0]+A[w[wp]][1]){
				ans[i] = ans[i-2]+A[w[wp]][0]+A[w[wp]][1];
				z = 2;
			}
			if(tp < K) ans[i+1] = ans[i-2]+A[t[tp]][0]+A[t[tp]][1]+A[t[tp]][2];
			if(z == 2){
				if(!Q.empty() && tp < K) ans[i+2] = ans[i-2]+Q.top()+A[t[tp]][0]+A[t[tp]][1]+A[t[tp]][2];
				b[w[wp]] = 1;
				repp(i,2,A[w[wp]].size()-1) Q.push(A[w[wp]][i]);
				++wp;
				while(tp < K && (b[t[tp]] == 1 || A[t[tp]][2] < 0)) ++tp;
				m = 1;
			} else {
				if(!Q.empty()) Q.pop();
				m = 0;
			}
		} else {
			if(!Q.empty()) ans[i] = max(ans[i],ans[i-1]+Q.top());
			m = 0;
		}
	}
	repp(i,0,N) cout << ans[i] << 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 <bits/stdc++.h>

using namespace std;

typedef long long LL;
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

int main(){
	int N,K; cin >> N >> K;
	vector<vector<LL>> A(K);
	repp(i,0,N){
		int C; LL V; cin >> C >> V;
		A[C-1].push_back(V);
	}
	vector<int> t(K),w(K);
	repp(i,0,K){
		sort(A[i].begin(),A[i].end(),greater<LL>());
		A[i].push_back(-1e18);
		w[i] = t[i] = i;
	}
	sort(w.begin(),w.end(),[&](const int &i, const int &j){return A[i][0]+A[i][1] > A[j][0]+A[j][1];});
	sort(t.begin(),t.end(),[&](const int &i, const int &j){return A[i][0]+A[i][1]+A[i][2] > A[j][0]+A[j][1]+A[j][2];});
	vector<LL> ans(N+3,-1);
	priority_queue<LL> Q;
	vector<int> b(K,0);
	ans[1] = A[w[0]][0]+A[w[0]][1];
	repp(i,2,A[w[0]].size()-1) Q.push(A[w[0]][i]);
	int	wp = 1, tp = 0, m = 0;
	b[w[0]] = 1;
	while(tp < K && A[t[tp]][2] < 0) ++tp;
	if(A[t[0]][2] >= 0){
		ans[2] = A[t[0]][0]+A[t[0]][1]+A[t[0]][2];
	}
	while(tp < K && (b[t[tp]] == 1 || A[t[tp]][2] < 0)) ++tp;
	repp(i,3,N){
		if(m == 0){
			int z = 0;
			if(Q.size() >= 2){
				LL r = Q.top(); Q.pop();
				if(ans[i] < ans[i-2]+r+Q.top()){
					ans[i] = ans[i-2]+r+Q.top();
					z = 1;
				}
				Q.push(r);
			}
			if(wp < K && ans[i] < ans[i-2]+A[w[wp]][0]+A[w[wp]][1]){
				ans[i] = ans[i-2]+A[w[wp]][0]+A[w[wp]][1];
				z = 2;
			}
			if(tp < K) ans[i+1] = ans[i-2]+A[t[tp]][0]+A[t[tp]][1]+A[t[tp]][2];
			if(z == 2){
				if(!Q.empty() && tp < K) ans[i+2] = ans[i-2]+Q.top()+A[t[tp]][0]+A[t[tp]][1]+A[t[tp]][2];
				b[w[wp]] = 1;
				repp(i,2,A[w[wp]].size()-1) Q.push(A[w[wp]][i]);
				++wp;
				while(tp < K && (b[t[tp]] == 1 || A[t[tp]][2] < 0)) ++tp;
				m = 1;
			} else {
				if(!Q.empty()) Q.pop();
				m = 0;
			}
		} else {
			if(!Q.empty()) ans[i] = max(ans[i],ans[i-1]+Q.top());
			m = 0;
		}
	}
	repp(i,0,N) cout << ans[i] << endl;
	return 0;
}

```