#include <bits/stdc++.h>
using namespace std;
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template<typename T> void Max(T &m, const T q) { if (m < q) m = q; }
template<typename T> void Min(T &m, const T q) { if (m > q) m = q; }
template<typename T1, typename T2> pair<T1, T2> operator+(pair<T1, T2> &l, pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(pair<T1, T2> &l, pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;

int N, K;
map<int, vector<lint>> ma, ma2;

int main()
{
    cin >> N >> K;
    REP(_, N)
    {
        int C;
        lint V;
        cin >> C >> V;
        ma[C].push_back(V * 2);
    }
    priority_queue<plint> doublet, triplet, pq1, pq2;

    FOR(c, 1, K + 1)
    {
        if (ma[c].size() > 1)
        {
            vector<lint> vec = ma[c];
            sort(vec.rbegin(), vec.rend());
            doublet.push(plint(vec[0] + vec[1], c));
            pq1.push(plint((vec[0] + vec[1]) / 2, c));
            pq1.push(plint((vec[0] + vec[1]) / 2, c));
            if (vec.size() > 2) triplet.push(plint(vec[0] + vec[1] + vec[2], c));
            FOR(i, 2, vec.size()) ma2[c].push_back(vec[i]);
        }
    }

    vector<lint> res(N + 3, -2);
    res[0] = 0;
    map<int, int> c_count;

    lint tmpsum = 0;
    bool flg = true;

    set<lint> se;
    REP(i, N)
    {
        while (!doublet.empty() and c_count[doublet.top().second] >= 2) doublet.pop();
        while (!triplet.empty() and c_count[triplet.top().second] >= 2) triplet.pop();

        if (flg)
        {
            if (!triplet.empty()) Max(res[i + 3], tmpsum + triplet.top().first);
            if (!doublet.empty()) Max(res[i + 2], tmpsum + doublet.top().first);
            if (!pq2.empty()) Max(res[i + 1], tmpsum + pq2.top().first);
        }
        else
        {
            Max(res[i + 1], tmpsum + pq1.top().first);
            if (!se.empty()) Max(res[i], tmpsum + pq1.top().first - *se.begin());
        }

        plint pa;
        if (pq2.empty() or (!pq1.empty() and pq1.top().first >= pq2.top().first))
        {
            pa = pq1.top();
            pq1.pop();
        }
        else
        {
            pa = pq2.top();
            pq2.pop();
            se.insert(pa.first);
        }

        tmpsum += pa.first;
        c_count[pa.second]++;
        flg = (c_count[pa.second] >= 2);

        if (c_count[pa.second] == 2)
        {
            for (auto v : ma2[pa.second]) pq2.push(plint(v, pa.second));
        }
    }
    FOR(i, 1, N + 1) printf("%lld\n", res[i] / 2);
}

### Prompt

Please create a solution in cpp 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;
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template<typename T> void Max(T &m, const T q) { if (m < q) m = q; }
template<typename T> void Min(T &m, const T q) { if (m > q) m = q; }
template<typename T1, typename T2> pair<T1, T2> operator+(pair<T1, T2> &l, pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(pair<T1, T2> &l, pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;

int N, K;
map<int, vector<lint>> ma, ma2;

int main()
{
    cin >> N >> K;
    REP(_, N)
    {
        int C;
        lint V;
        cin >> C >> V;
        ma[C].push_back(V * 2);
    }
    priority_queue<plint> doublet, triplet, pq1, pq2;

    FOR(c, 1, K + 1)
    {
        if (ma[c].size() > 1)
        {
            vector<lint> vec = ma[c];
            sort(vec.rbegin(), vec.rend());
            doublet.push(plint(vec[0] + vec[1], c));
            pq1.push(plint((vec[0] + vec[1]) / 2, c));
            pq1.push(plint((vec[0] + vec[1]) / 2, c));
            if (vec.size() > 2) triplet.push(plint(vec[0] + vec[1] + vec[2], c));
            FOR(i, 2, vec.size()) ma2[c].push_back(vec[i]);
        }
    }

    vector<lint> res(N + 3, -2);
    res[0] = 0;
    map<int, int> c_count;

    lint tmpsum = 0;
    bool flg = true;

    set<lint> se;
    REP(i, N)
    {
        while (!doublet.empty() and c_count[doublet.top().second] >= 2) doublet.pop();
        while (!triplet.empty() and c_count[triplet.top().second] >= 2) triplet.pop();

        if (flg)
        {
            if (!triplet.empty()) Max(res[i + 3], tmpsum + triplet.top().first);
            if (!doublet.empty()) Max(res[i + 2], tmpsum + doublet.top().first);
            if (!pq2.empty()) Max(res[i + 1], tmpsum + pq2.top().first);
        }
        else
        {
            Max(res[i + 1], tmpsum + pq1.top().first);
            if (!se.empty()) Max(res[i], tmpsum + pq1.top().first - *se.begin());
        }

        plint pa;
        if (pq2.empty() or (!pq1.empty() and pq1.top().first >= pq2.top().first))
        {
            pa = pq1.top();
            pq1.pop();
        }
        else
        {
            pa = pq2.top();
            pq2.pop();
            se.insert(pa.first);
        }

        tmpsum += pa.first;
        c_count[pa.second]++;
        flg = (c_count[pa.second] >= 2);

        if (c_count[pa.second] == 2)
        {
            for (auto v : ma2[pa.second]) pq2.push(plint(v, pa.second));
        }
    }
    FOR(i, 1, N + 1) printf("%lld\n", res[i] / 2);
}

```