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

int N, K;
vector<int64_t> V[200000];

priority_queue<int64_t> Q;
priority_queue<int64_t, vector<int64_t>, greater<int64_t>> P, R;
priority_queue<pair<int64_t, int>> T;

vector<int> c_order;
bitset<200000> c_used;
int co = 0;

int64_t irregular = 0;

int64_t pick(){
    if(irregular){
        int64_t ret = irregular;
        R.push(2*ret);
        for(int i=2; i<V[c_order[co]].size(); i++) Q.push(V[c_order[co]][i]);
        c_used[c_order[co]] = 1;
        while(T.size() && c_used[T.top().second]) T.pop();
        co++;
        irregular = 0;
        return ret;
    }

    int64_t a = (Q.size() ? Q.top() : 0);
    int64_t b = (co < K ? V[c_order[co]][0] : 0);
    if(a > b){
        Q.pop();
        P.push(a);
        return a;
    }else{
        irregular = b;
        return b;
    }
}

int main(){
    cin >> N >> K;
    for(int i=0; i<N; i++){
        int c, v;
        cin >> c >> v;
        V[c-1].push_back(2*v);
    }

    for(int c=0; c<K; c++){
        sort(V[c].rbegin(), V[c].rend());
        V[c][0] = V[c][1] = (V[c][0] + V[c][1]) / 2;
    }

    for(int i=0; i<K; i++) c_order.push_back(i);
    sort(c_order.begin(), c_order.end(), [&](int a, int b){return V[a][0] > V[b][0];});
    c_used[c_order[0]] = 1;
    for(int c=0; c<K; c++) if(V[c].size() >= 3) T.push({V[c][0] + V[c][1] + V[c][2], c});

    int64_t sum = 0;
    for(int i=0; i<N; i++){
        int64_t add = pick();
        int64_t ans;
        if(irregular){
            ans = -2;
            if(Q.size()) ans = max(ans, sum + Q.top());
            if(P.size()) ans = max(ans, sum - P.top() + 2*add);
            if(R.size() && T.size()) ans = max(ans, sum - R.top() + T.top().first);
        }else{
            ans = sum + add;
        }
        cout << ans/2 << endl;
        sum += add;
    }
    return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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;

int N, K;
vector<int64_t> V[200000];

priority_queue<int64_t> Q;
priority_queue<int64_t, vector<int64_t>, greater<int64_t>> P, R;
priority_queue<pair<int64_t, int>> T;

vector<int> c_order;
bitset<200000> c_used;
int co = 0;

int64_t irregular = 0;

int64_t pick(){
    if(irregular){
        int64_t ret = irregular;
        R.push(2*ret);
        for(int i=2; i<V[c_order[co]].size(); i++) Q.push(V[c_order[co]][i]);
        c_used[c_order[co]] = 1;
        while(T.size() && c_used[T.top().second]) T.pop();
        co++;
        irregular = 0;
        return ret;
    }

    int64_t a = (Q.size() ? Q.top() : 0);
    int64_t b = (co < K ? V[c_order[co]][0] : 0);
    if(a > b){
        Q.pop();
        P.push(a);
        return a;
    }else{
        irregular = b;
        return b;
    }
}

int main(){
    cin >> N >> K;
    for(int i=0; i<N; i++){
        int c, v;
        cin >> c >> v;
        V[c-1].push_back(2*v);
    }

    for(int c=0; c<K; c++){
        sort(V[c].rbegin(), V[c].rend());
        V[c][0] = V[c][1] = (V[c][0] + V[c][1]) / 2;
    }

    for(int i=0; i<K; i++) c_order.push_back(i);
    sort(c_order.begin(), c_order.end(), [&](int a, int b){return V[a][0] > V[b][0];});
    c_used[c_order[0]] = 1;
    for(int c=0; c<K; c++) if(V[c].size() >= 3) T.push({V[c][0] + V[c][1] + V[c][2], c});

    int64_t sum = 0;
    for(int i=0; i<N; i++){
        int64_t add = pick();
        int64_t ans;
        if(irregular){
            ans = -2;
            if(Q.size()) ans = max(ans, sum + Q.top());
            if(P.size()) ans = max(ans, sum - P.top() + 2*add);
            if(R.size() && T.size()) ans = max(ans, sum - R.top() + T.top().first);
        }else{
            ans = sum + add;
        }
        cout << ans/2 << endl;
        sum += add;
    }
    return 0;
}
```