#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MAXN = 200005, INF = 0x3f3f3f3f3f3f3f3f;
struct ii {
    ll a, b;
    bool operator < (ii o) const {
        return tie(a, b) < tie(o.a, o.b);
    }
};
struct iii {
    ll a, b, c;
    bool operator > (iii o) const {
        return tie(a, b, c) > tie(o.a, o.b, o.c);
    }
};

ll N, K, cnt[MAXN];
vector<ll> col[MAXN], doneCol[MAXN];
multiset<ll> masuk1, masuk2, masuk3, keluar1, keluar2;

ll lastTwo(const vector<ll> &v) {
    return v[v.size()-2] + v.back();
}

int main() {
    ios::sync_with_stdio(0); cin.tie(0);

    cin >> N >> K;
    for (ll i = 1; i <= N; i++) {
        ll c, v;
        cin >> c >> v;
        col[c].push_back(2*v);
    }
    vector<iii> proses;
    proses.push_back({INF, INF, INF}); //dummy 1-index
    for (ll i = 1; i <= K; i++) {
        sort(col[i].begin(), col[i].end(), greater<ll>());
        proses.push_back({(col[i][0]+col[i][1])/2, i});
        proses.push_back({(col[i][0]+col[i][1])/2, i});
        for (ll j = 2; j < (ll)col[i].size(); j++) {
            proses.push_back({col[i][j], i});
        }
        masuk2.insert(col[i][0]+col[i][1]);
        if (col[i].size() > 2) masuk3.insert(col[i][0] + col[i][1] + col[i][2]);
    }
    sort(proses.begin(), proses.end(), greater<iii>());
    
    ll sum = 0;
    for (ll i = 1; i <= N; i++) {
        //~ cerr << i << " i awal\n";
        if (i < N && proses[i].b == proses[i+1].b && cnt[proses[i].b] == 0) {
            //~ if (!masuk3.empty() && !keluar2.empty()) cerr << i << " " << *masuk3.rbegin() << " " << *keluar2.begin() << endl;
            ll add = -INF;
            if (!masuk1.empty()) add = max(add, *masuk1.rbegin());
            if (!masuk2.empty() && !keluar1.empty()) add = max(add, *masuk2.rbegin() - *keluar1.begin());
            if (!masuk3.empty() && !keluar2.empty()) add = max(add, *masuk3.rbegin() - *keluar2.begin());
            if (add == -INF) {
                cout << "-1\n";
            } else {
                cout << (sum + add)/2 << "\n";
            }
        } else {
            //~ cerr << i << " masuk\n";
            cout << (sum+proses[i].a)/2 << "\n";
        }
        sum += proses[i].a;
        ll curCol = proses[i].b;
        //~ cerr << i << " i akhir\n";
        if (cnt[curCol] >= 2) keluar1.insert(proses[i].a);
        if (cnt[curCol] == 1 || cnt[curCol] >= 3) keluar2.insert(doneCol[curCol].back() + proses[i].a);
        if (cnt[curCol] == 2) keluar2.erase(keluar2.find(lastTwo(doneCol[curCol])));
        if (cnt[curCol] == 1) {
            for (ll j = 2; j < (ll)col[curCol].size(); j++) {
                masuk1.insert(col[curCol][j]);
            }
        }
        //~ cerr << i<< " " << cnt[curCol] << " i\n";
        if (cnt[curCol] == 1) {
            //~ cerr << *masuk2.find(col[curCol][0] + col[curCol][1]) << "benar" << endl;
            masuk2.erase(masuk2.find(col[curCol][0] + col[curCol][1]));
            if (col[curCol].size() > 2) masuk3.erase(masuk3.find(col[curCol][0] + col[curCol][1] + col[curCol][2]));
        }
        if (cnt[curCol] >= 2) {
            masuk1.erase(masuk1.find(proses[i].a));
        }
        
        cnt[curCol]++;
        doneCol[curCol].push_back(proses[i].a);
    }

}

### 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;
const ll MAXN = 200005, INF = 0x3f3f3f3f3f3f3f3f;
struct ii {
    ll a, b;
    bool operator < (ii o) const {
        return tie(a, b) < tie(o.a, o.b);
    }
};
struct iii {
    ll a, b, c;
    bool operator > (iii o) const {
        return tie(a, b, c) > tie(o.a, o.b, o.c);
    }
};

ll N, K, cnt[MAXN];
vector<ll> col[MAXN], doneCol[MAXN];
multiset<ll> masuk1, masuk2, masuk3, keluar1, keluar2;

ll lastTwo(const vector<ll> &v) {
    return v[v.size()-2] + v.back();
}

int main() {
    ios::sync_with_stdio(0); cin.tie(0);

    cin >> N >> K;
    for (ll i = 1; i <= N; i++) {
        ll c, v;
        cin >> c >> v;
        col[c].push_back(2*v);
    }
    vector<iii> proses;
    proses.push_back({INF, INF, INF}); //dummy 1-index
    for (ll i = 1; i <= K; i++) {
        sort(col[i].begin(), col[i].end(), greater<ll>());
        proses.push_back({(col[i][0]+col[i][1])/2, i});
        proses.push_back({(col[i][0]+col[i][1])/2, i});
        for (ll j = 2; j < (ll)col[i].size(); j++) {
            proses.push_back({col[i][j], i});
        }
        masuk2.insert(col[i][0]+col[i][1]);
        if (col[i].size() > 2) masuk3.insert(col[i][0] + col[i][1] + col[i][2]);
    }
    sort(proses.begin(), proses.end(), greater<iii>());
    
    ll sum = 0;
    for (ll i = 1; i <= N; i++) {
        //~ cerr << i << " i awal\n";
        if (i < N && proses[i].b == proses[i+1].b && cnt[proses[i].b] == 0) {
            //~ if (!masuk3.empty() && !keluar2.empty()) cerr << i << " " << *masuk3.rbegin() << " " << *keluar2.begin() << endl;
            ll add = -INF;
            if (!masuk1.empty()) add = max(add, *masuk1.rbegin());
            if (!masuk2.empty() && !keluar1.empty()) add = max(add, *masuk2.rbegin() - *keluar1.begin());
            if (!masuk3.empty() && !keluar2.empty()) add = max(add, *masuk3.rbegin() - *keluar2.begin());
            if (add == -INF) {
                cout << "-1\n";
            } else {
                cout << (sum + add)/2 << "\n";
            }
        } else {
            //~ cerr << i << " masuk\n";
            cout << (sum+proses[i].a)/2 << "\n";
        }
        sum += proses[i].a;
        ll curCol = proses[i].b;
        //~ cerr << i << " i akhir\n";
        if (cnt[curCol] >= 2) keluar1.insert(proses[i].a);
        if (cnt[curCol] == 1 || cnt[curCol] >= 3) keluar2.insert(doneCol[curCol].back() + proses[i].a);
        if (cnt[curCol] == 2) keluar2.erase(keluar2.find(lastTwo(doneCol[curCol])));
        if (cnt[curCol] == 1) {
            for (ll j = 2; j < (ll)col[curCol].size(); j++) {
                masuk1.insert(col[curCol][j]);
            }
        }
        //~ cerr << i<< " " << cnt[curCol] << " i\n";
        if (cnt[curCol] == 1) {
            //~ cerr << *masuk2.find(col[curCol][0] + col[curCol][1]) << "benar" << endl;
            masuk2.erase(masuk2.find(col[curCol][0] + col[curCol][1]));
            if (col[curCol].size() > 2) masuk3.erase(masuk3.find(col[curCol][0] + col[curCol][1] + col[curCol][2]));
        }
        if (cnt[curCol] >= 2) {
            masuk1.erase(masuk1.find(proses[i].a));
        }
        
        cnt[curCol]++;
        doneCol[curCol].push_back(proses[i].a);
    }

}

```