#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, x, n) for(int i = x; i <= n; i++)
#define rep3(i, x, n) for(int i = x; i >= n; i--)
#define elif else if
#define sp(x) fixed << setprecision(x)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const ll MOD = 1e9+7;
//const ll MOD = 998244353;
const int inf = (1<<30)-1;
const ll INF = (1LL<<60)-1;
const ld EPS = 1e-10;
template<typename T> bool chmax(T &x, const T &y) {return (x < y)? (x = y, true) : false;};
template<typename T> bool chmin(T &x, const T &y) {return (x > y)? (x = y, true) : false;};

int main(){
    int N, K;
    cin >> N >> K;
    vector<ll> v[K];
    rep(i, N){
        int C; ll V;
        cin >> C >> V; C--;
        v[C].pb(V);
    }

    vector<pli> p;
    priority_queue<pli> que1, que4; 
    priority_queue<ll, vector<ll>, greater<ll> > que2;
    priority_queue<pli, vector<pli>, greater<pli> > que3;
    rep(i, K){
        sort(all(v[i]), greater<ll>());
        rep(j, 2) p.pb(pli(v[i][0]+v[i][1], i));
        rep2(j, 2, sz(v[i])-1) p.pb(pli(2*v[i][j], i));
        if(sz(v[i]) >= 3){
            que4.push(pli((v[i][0]+v[i][1]+v[i][2])*2, i));
        }
    }
    sort(all(p), greater<pli>());

    vector<int> cnt(K, 0);
    ll sum = 0;
    rep(i, N){
        int j = p[i].second;
        cnt[j]++;
        if(cnt[j] == 1){
            ll tmp = -INF;
            if(!que1.empty()) chmax(tmp, que1.top().first);
            if(!que2.empty()) chmax(tmp, p[i].first*2-que2.top());
            if(!que3.empty() && !que4.empty()){
                chmax(tmp, que4.top().first-que3.top().first);
            }
            if(tmp == -INF) cout << -1 << endl;
            else cout << (sum+tmp)/2 << endl;
        }
        else{
            cout << (sum+p[i].first)/2 << endl;
            while(!que1.empty()){
                int k = que1.top().second;
                if(cnt[k] == sz(v[k]) || que1.top().first > v[k][cnt[k]]*2) que1.pop();
                else break;
            }
            if(cnt[j] < sz(v[j])){
                que1.push(pli(v[j][cnt[j]]*2, j));
            }
            if(cnt[j] >= 3){
                que2.push((v[j][cnt[j]-1])*2);
            }
            while(!que3.empty()){
                int k = que3.top().second;
                if(cnt[k] > 2) que3.pop();
                else break;
            }
            if(cnt[j] == 2){
                que3.push(pli(p[i].first*2, j));
            }
            while(!que4.empty()){
                int k = que4.top().second;
                if(cnt[k] > 0) que4.pop();
                else break;
            }
        }
        sum += p[i].first;
    }
}
### 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;
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, x, n) for(int i = x; i <= n; i++)
#define rep3(i, x, n) for(int i = x; i >= n; i--)
#define elif else if
#define sp(x) fixed << setprecision(x)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const ll MOD = 1e9+7;
//const ll MOD = 998244353;
const int inf = (1<<30)-1;
const ll INF = (1LL<<60)-1;
const ld EPS = 1e-10;
template<typename T> bool chmax(T &x, const T &y) {return (x < y)? (x = y, true) : false;};
template<typename T> bool chmin(T &x, const T &y) {return (x > y)? (x = y, true) : false;};

int main(){
    int N, K;
    cin >> N >> K;
    vector<ll> v[K];
    rep(i, N){
        int C; ll V;
        cin >> C >> V; C--;
        v[C].pb(V);
    }

    vector<pli> p;
    priority_queue<pli> que1, que4; 
    priority_queue<ll, vector<ll>, greater<ll> > que2;
    priority_queue<pli, vector<pli>, greater<pli> > que3;
    rep(i, K){
        sort(all(v[i]), greater<ll>());
        rep(j, 2) p.pb(pli(v[i][0]+v[i][1], i));
        rep2(j, 2, sz(v[i])-1) p.pb(pli(2*v[i][j], i));
        if(sz(v[i]) >= 3){
            que4.push(pli((v[i][0]+v[i][1]+v[i][2])*2, i));
        }
    }
    sort(all(p), greater<pli>());

    vector<int> cnt(K, 0);
    ll sum = 0;
    rep(i, N){
        int j = p[i].second;
        cnt[j]++;
        if(cnt[j] == 1){
            ll tmp = -INF;
            if(!que1.empty()) chmax(tmp, que1.top().first);
            if(!que2.empty()) chmax(tmp, p[i].first*2-que2.top());
            if(!que3.empty() && !que4.empty()){
                chmax(tmp, que4.top().first-que3.top().first);
            }
            if(tmp == -INF) cout << -1 << endl;
            else cout << (sum+tmp)/2 << endl;
        }
        else{
            cout << (sum+p[i].first)/2 << endl;
            while(!que1.empty()){
                int k = que1.top().second;
                if(cnt[k] == sz(v[k]) || que1.top().first > v[k][cnt[k]]*2) que1.pop();
                else break;
            }
            if(cnt[j] < sz(v[j])){
                que1.push(pli(v[j][cnt[j]]*2, j));
            }
            if(cnt[j] >= 3){
                que2.push((v[j][cnt[j]-1])*2);
            }
            while(!que3.empty()){
                int k = que3.top().second;
                if(cnt[k] > 2) que3.pop();
                else break;
            }
            if(cnt[j] == 2){
                que3.push(pli(p[i].first*2, j));
            }
            while(!que4.empty()){
                int k = que4.top().second;
                if(cnt[k] > 0) que4.pop();
                else break;
            }
        }
        sum += p[i].first;
    }
}
```