#include <cstdio>
#include <vector>
#include <set>
#include <algorithm>

using namespace std;

int f[100000];
long long ans[200001];
vector <long long> v[100000];

int main() {
    int n, k, c = 0, i, j;
    long long p = 0, q = 0;
    vector <pair<double, int>> w;
    multiset <long long> m1, m2;
    
    scanf("%d %d", &n, &k);
    
    for (i = 0; i < n; i++) {
        int x, y;
        
        scanf("%d %d", &x, &y);
        
        v[x - 1].push_back(y);
    }
    
    for (i = 0; i < k; i++) {
        sort(v[i].begin(), v[i].end());
        reverse(v[i].begin(), v[i].end());
        
        if (v[i].size() >= 3) m2.insert(v[i][0] + v[i][1] + v[i][2]);
        
        w.push_back(make_pair((v[i][0] + v[i][1]) / 2.0, i));
        w.push_back(make_pair((v[i][0] + v[i][1]) / 2.0, i));
        
        for (j = 2; j < v[i].size(); j++) w.push_back(make_pair(v[i][j], i));
    }
    
    sort(w.begin(), w.end());
    reverse(w.begin(), w.end());
    
    for (i = 1; i <= n; i++) ans[i] = -1;
    
    for (i = 0; i < w.size(); i++) {
        int x = w[i].second;
        
        f[x]++;
        
        if (f[x] == 2) {
            c += 2;
            ans[c] = ans[c - 2] + v[x][0] + v[x][1];
            
            if (p > 0) ans[c - 1] = max(ans[c - 1], ans[c] - p);
            if (m1.size() > 0) ans[c - 1] = max(ans[c - 1], ans[c - 2] + *m1.rbegin());
            if (q > 0 && m2.size() > 0) ans[c - 1] = max(ans[c - 1], ans[c - 2] - q + *m2.rbegin());
            
            q = v[x][0] + v[x][1];
            for (j = 2; j < v[x].size(); j++) m1.insert(v[x][j]);
            if (v[x].size() >= 3) m2.erase(m2.find(v[x][0] + v[x][1] + v[x][2]));
        } else if (f[x] > 2) {
            long long y = v[x][f[x] - 1];
            
            c++;
            ans[c] = ans[c - 1] + y;
            
            p = y;
            m1.erase(m1.find(y));
        }
    }
    
    for (i = 1; i <= n; i++) printf("%lld\n", ans[i]);
    
    return 0;
}

### 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 <cstdio>
#include <vector>
#include <set>
#include <algorithm>

using namespace std;

int f[100000];
long long ans[200001];
vector <long long> v[100000];

int main() {
    int n, k, c = 0, i, j;
    long long p = 0, q = 0;
    vector <pair<double, int>> w;
    multiset <long long> m1, m2;
    
    scanf("%d %d", &n, &k);
    
    for (i = 0; i < n; i++) {
        int x, y;
        
        scanf("%d %d", &x, &y);
        
        v[x - 1].push_back(y);
    }
    
    for (i = 0; i < k; i++) {
        sort(v[i].begin(), v[i].end());
        reverse(v[i].begin(), v[i].end());
        
        if (v[i].size() >= 3) m2.insert(v[i][0] + v[i][1] + v[i][2]);
        
        w.push_back(make_pair((v[i][0] + v[i][1]) / 2.0, i));
        w.push_back(make_pair((v[i][0] + v[i][1]) / 2.0, i));
        
        for (j = 2; j < v[i].size(); j++) w.push_back(make_pair(v[i][j], i));
    }
    
    sort(w.begin(), w.end());
    reverse(w.begin(), w.end());
    
    for (i = 1; i <= n; i++) ans[i] = -1;
    
    for (i = 0; i < w.size(); i++) {
        int x = w[i].second;
        
        f[x]++;
        
        if (f[x] == 2) {
            c += 2;
            ans[c] = ans[c - 2] + v[x][0] + v[x][1];
            
            if (p > 0) ans[c - 1] = max(ans[c - 1], ans[c] - p);
            if (m1.size() > 0) ans[c - 1] = max(ans[c - 1], ans[c - 2] + *m1.rbegin());
            if (q > 0 && m2.size() > 0) ans[c - 1] = max(ans[c - 1], ans[c - 2] - q + *m2.rbegin());
            
            q = v[x][0] + v[x][1];
            for (j = 2; j < v[x].size(); j++) m1.insert(v[x][j]);
            if (v[x].size() >= 3) m2.erase(m2.find(v[x][0] + v[x][1] + v[x][2]));
        } else if (f[x] > 2) {
            long long y = v[x][f[x] - 1];
            
            c++;
            ans[c] = ans[c - 1] + y;
            
            p = y;
            m1.erase(m1.find(y));
        }
    }
    
    for (i = 1; i <= n; i++) printf("%lld\n", ans[i]);
    
    return 0;
}

```