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

#define rep(i, N) for (int i = 0; i < (N); i++)
#define all(a) (a).begin(), (a).end()
#define pb push_back

using ll = long long;
using i_i = tuple<int, int>;
using ll_i = tuple<ll, int>;

int main() {
    int N, K;
    cin >> N >> K;
    vector<vector<int>> a(K);
    rep(i, N) {
        int k, x;
        scanf("%d%d", &k, &x);
        a[k - 1].pb(x);
    }
    vector<int> f(K);
    rep(k, K) {
        sort(all(a[k]));
        reverse(all(a[k]));
        f[k] = a[k][0] + a[k][1];
        a[k].erase(a[k].begin());
        a[k].erase(a[k].begin());
    }
    priority_queue<i_i> pq1;
    priority_queue<i_i> pq2;
    priority_queue<ll_i> pq3;
    priority_queue<i_i> pos;
    map<i_i, int> mp;
    rep(k, K) {
        pq2.push(i_i(f[k], k));
        for (int x: a[k]) {
            pq1.push(i_i(x, k));
            mp[i_i(x, k)]++;
        }
        if (!a[k].empty()) pq3.push(ll_i((ll)f[k] + a[k][0], k));
    }
    vector<ll> ans(N + 1, -1);
    int cnt = 0;
    ll sum = 0;
    int mi1 = INT_MAX;
    ll mi3 = LLONG_MAX;
    vector<bool> used2(K);
    vector<bool> used3(K);
    for (int i = 1; i <= N; i++) {
        int x1 = (pq1.empty() ? -1 : get<0>(pq1.top()) * 2);
        int x2 = (pq2.empty() ? -1 : get<0>(pq2.top()));
        if (x1 >= x2) {
            cnt++;
            sum += get<0>(pq1.top());
            mi1 = min(mi1, get<0>(pq1.top()));
            int k = get<1>(pq1.top());
            int x = get<0>(pq1.top());
            mp[i_i(x, k)]--;
            pq1.pop();
        }
        else if (cnt + 2 <= i) {
            cnt += 2;
            sum += get<0>(pq2.top());
            int k = get<1>(pq2.top());
            for (int x: a[k]) pos.push(i_i(x, k));
            if (!used2[k]) {
                used2[k] = true;
                mi3 = min(mi3, (ll)f[k]);
            }
            pq2.pop();
        }
        if (cnt == i) ans[i] = sum;
        else {
            while (!pos.empty()) {
                int x, k;
                tie(x, k) = pos.top();
                if (mp[i_i(x, k)] == 0) {
                    pos.pop();
                    continue;
                }
                ans[i] = max(ans[i], sum + x);
                break;
            }
            if (mi1 < INT_MAX && !pq2.empty()) ans[i] = max(ans[i], sum - mi1 + get<0>(pq2.top()));
            if (mi3 < LLONG_MAX) {
                while (!pq3.empty()) {
                    int k = get<1>(pq3.top());
                    if (used2[k]) {
                        pq3.pop();
                        continue;
                    }
                    ans[i] = max(ans[i], sum - mi3 + get<0>(pq3.top()));
                    break;
                }
            }
        }
    }
    for (int i = 1; i <= N; i++) printf("%lld\n", ans[i]);
}

### Prompt

Develop a solution in cpp to the problem described below:

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 all(a) (a).begin(), (a).end()
#define pb push_back

using ll = long long;
using i_i = tuple<int, int>;
using ll_i = tuple<ll, int>;

int main() {
    int N, K;
    cin >> N >> K;
    vector<vector<int>> a(K);
    rep(i, N) {
        int k, x;
        scanf("%d%d", &k, &x);
        a[k - 1].pb(x);
    }
    vector<int> f(K);
    rep(k, K) {
        sort(all(a[k]));
        reverse(all(a[k]));
        f[k] = a[k][0] + a[k][1];
        a[k].erase(a[k].begin());
        a[k].erase(a[k].begin());
    }
    priority_queue<i_i> pq1;
    priority_queue<i_i> pq2;
    priority_queue<ll_i> pq3;
    priority_queue<i_i> pos;
    map<i_i, int> mp;
    rep(k, K) {
        pq2.push(i_i(f[k], k));
        for (int x: a[k]) {
            pq1.push(i_i(x, k));
            mp[i_i(x, k)]++;
        }
        if (!a[k].empty()) pq3.push(ll_i((ll)f[k] + a[k][0], k));
    }
    vector<ll> ans(N + 1, -1);
    int cnt = 0;
    ll sum = 0;
    int mi1 = INT_MAX;
    ll mi3 = LLONG_MAX;
    vector<bool> used2(K);
    vector<bool> used3(K);
    for (int i = 1; i <= N; i++) {
        int x1 = (pq1.empty() ? -1 : get<0>(pq1.top()) * 2);
        int x2 = (pq2.empty() ? -1 : get<0>(pq2.top()));
        if (x1 >= x2) {
            cnt++;
            sum += get<0>(pq1.top());
            mi1 = min(mi1, get<0>(pq1.top()));
            int k = get<1>(pq1.top());
            int x = get<0>(pq1.top());
            mp[i_i(x, k)]--;
            pq1.pop();
        }
        else if (cnt + 2 <= i) {
            cnt += 2;
            sum += get<0>(pq2.top());
            int k = get<1>(pq2.top());
            for (int x: a[k]) pos.push(i_i(x, k));
            if (!used2[k]) {
                used2[k] = true;
                mi3 = min(mi3, (ll)f[k]);
            }
            pq2.pop();
        }
        if (cnt == i) ans[i] = sum;
        else {
            while (!pos.empty()) {
                int x, k;
                tie(x, k) = pos.top();
                if (mp[i_i(x, k)] == 0) {
                    pos.pop();
                    continue;
                }
                ans[i] = max(ans[i], sum + x);
                break;
            }
            if (mi1 < INT_MAX && !pq2.empty()) ans[i] = max(ans[i], sum - mi1 + get<0>(pq2.top()));
            if (mi3 < LLONG_MAX) {
                while (!pq3.empty()) {
                    int k = get<1>(pq3.top());
                    if (used2[k]) {
                        pq3.pop();
                        continue;
                    }
                    ans[i] = max(ans[i], sum - mi3 + get<0>(pq3.top()));
                    break;
                }
            }
        }
    }
    for (int i = 1; i <= N; i++) printf("%lld\n", ans[i]);
}

```