#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    vector<pair<ll,int>>jewels(n);
    vector<ll>colors[k+1];
    for (int i = 0; i < n; i++) {
        long long c, v;
        cin >> c >> v;
        colors[c].push_back(v * 2);
    }

    multiset<long long> top_one;
    multiset<long long> top_two;
    multiset<long long> top_three;
    for (int color = 1; color <= k; color++) {
        sort(colors[color].begin(), colors[color].end(), greater<long long>());
        int nc = colors[color].size();
        top_two.emplace(colors[color][0] + colors[color][1]);
        if (nc > 2) {
            top_three.emplace(colors[color][0] + colors[color][1] + colors[color][2]);
        }
        long long avg = (colors[color][0] + colors[color][1]) / 2;
        jewels.emplace_back(avg, color);
        jewels.emplace_back(avg, color);
        for (int i = 2; i < nc; i++) {
            jewels.emplace_back(colors[color][i], color);
        }
    }

    sort(jewels.begin(), jewels.end(), greater<pair<long long, int>>());


    long long sum = 0;
    vector<int> cnts(k + 1, 0);
    long long min_pair = 1LL<<60;
    long long min_one = 1LL<<60;
    for (int i = 0; i < n; i++) {
        long long color = jewels[i].second;
        long long value = jewels[i].first;
        long long ans = -2;
        if (cnts[color] == 0) {
            if (!top_one.empty()) {
                ans = max(ans, sum + *top_one.rbegin());
            }

            if (!top_two.empty()) {
                ans = max(ans, sum - min_one + *top_two.rbegin());
            }

            if (!top_three.empty()) {
                ans = max(ans, sum - min_pair + *top_three.rbegin());
            }


            // erase from top two and three
            top_two.erase(top_two.find(colors[color][0] + colors[color][1]));
            if (colors[color].size() > 2) {
                top_three.erase(top_three.find(colors[color][0] + colors[color][1] + colors[color][2]));
            }
        }
        else {
            ans = sum + value;
            if (cnts[color] == 1) {

                for (int i = 2; i < colors[color].size(); i++) {
                    top_one.emplace(colors[color][i]);
                }
                min_pair = min(min_pair, colors[color][0] + colors[color][1]);
            }
            else {
                top_one.erase(top_one.find(value));
                min_one = min(min_one, value);
            }
        }
        cnts[color]++;
        sum += value;
        cout << ans / 2 << endl;
    }
}

### Prompt

Please provide a cpp coded solution 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;
typedef long long ll;
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    vector<pair<ll,int>>jewels(n);
    vector<ll>colors[k+1];
    for (int i = 0; i < n; i++) {
        long long c, v;
        cin >> c >> v;
        colors[c].push_back(v * 2);
    }

    multiset<long long> top_one;
    multiset<long long> top_two;
    multiset<long long> top_three;
    for (int color = 1; color <= k; color++) {
        sort(colors[color].begin(), colors[color].end(), greater<long long>());
        int nc = colors[color].size();
        top_two.emplace(colors[color][0] + colors[color][1]);
        if (nc > 2) {
            top_three.emplace(colors[color][0] + colors[color][1] + colors[color][2]);
        }
        long long avg = (colors[color][0] + colors[color][1]) / 2;
        jewels.emplace_back(avg, color);
        jewels.emplace_back(avg, color);
        for (int i = 2; i < nc; i++) {
            jewels.emplace_back(colors[color][i], color);
        }
    }

    sort(jewels.begin(), jewels.end(), greater<pair<long long, int>>());


    long long sum = 0;
    vector<int> cnts(k + 1, 0);
    long long min_pair = 1LL<<60;
    long long min_one = 1LL<<60;
    for (int i = 0; i < n; i++) {
        long long color = jewels[i].second;
        long long value = jewels[i].first;
        long long ans = -2;
        if (cnts[color] == 0) {
            if (!top_one.empty()) {
                ans = max(ans, sum + *top_one.rbegin());
            }

            if (!top_two.empty()) {
                ans = max(ans, sum - min_one + *top_two.rbegin());
            }

            if (!top_three.empty()) {
                ans = max(ans, sum - min_pair + *top_three.rbegin());
            }


            // erase from top two and three
            top_two.erase(top_two.find(colors[color][0] + colors[color][1]));
            if (colors[color].size() > 2) {
                top_three.erase(top_three.find(colors[color][0] + colors[color][1] + colors[color][2]));
            }
        }
        else {
            ans = sum + value;
            if (cnts[color] == 1) {

                for (int i = 2; i < colors[color].size(); i++) {
                    top_one.emplace(colors[color][i]);
                }
                min_pair = min(min_pair, colors[color][0] + colors[color][1]);
            }
            else {
                top_one.erase(top_one.find(value));
                min_one = min(min_one, value);
            }
        }
        cnts[color]++;
        sum += value;
        cout << ans / 2 << endl;
    }
}

```