#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);
    }
    bool operator > (ii o) const {
        return tie(a, b) > tie(o.a, o.b);
    }
};

ll N, K, cnt[MAXN];
vector<ll> col[MAXN];

ll sumTwo(const vector<ll> &v) {
    return v[0]+v[1];
}

ll sumThree(const vector<ll> &v) {
    return v[0]+v[1]+v[2];
}

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<ii> arr;
    arr.push_back({INF, INF});
    set<ii> in[5], out[5];
    for (ll i = 1; i <= K; i++) {
        sort(col[i].begin(), col[i].end(), greater<ll>());
        col[i][0] = col[i][1] = sumTwo(col[i])/2;
        in[2].insert({sumTwo(col[i]), i});
        if (col[i].size() >= 3) in[3].insert({sumThree(col[i]), i});
        
        for (int j = 0; j < (int)col[i].size(); j++) {
            arr.push_back({col[i][j], i});
        }
    }
    sort(arr.begin(), arr.end(), greater<ii>());
    //~ for (auto i: arr) cerr << i.a << " " << i.b << "\n";
    
    ll sum = 0;
    for (ll i = 1; i <= N; i++) {
        if (in[2].find({sumTwo(col[arr[i].b]), arr[i].b}) == in[2].end()) {
            cout << (sum+arr[i].a)/2 << "\n";
        } else {
            ll ans = -2;
            if (!in[1].empty()) {
                ans = max(ans, sum+in[1].rbegin()->a);
            }
            if (!in[2].empty() && !out[1].empty()) {
                ans = max(ans, sum+out[1].rbegin()->a+in[2].rbegin()->a);
            }
            if (!in[3].empty() && !out[2].empty()) {
                ans = max(ans, sum+out[2].rbegin()->a+in[3].rbegin()->a);
            }
            cout << ans/2 << "\n";
        }
        int c = arr[i].b;
        cnt[c]++;
        sum += arr[i].a;
        if (cnt[c] == 1) {
            out[1].insert({-arr[i].a, c});
            //~ cerr << -arr[i].a << " " << c << " masuk\n";
            in[2].erase({sumTwo(col[c]), c});
            if (col[c].size() >= 3) in[3].erase({sumThree(col[c]), c});
        }
        if (cnt[c] == 2) {
            // valuenya pasti sama kayak yang in
            out[1].erase({-arr[i].a, c});
            out[2].insert({-sumTwo(col[c]), c});
        }
        if (cnt[c] >= 2) {
            if (cnt[c] >= 3) in[1].erase({col[c][cnt[c]-1], c});
            if ((int)col[c].size() > cnt[c]) in[1].insert({col[c][cnt[c]], c});
        }
        if (cnt[c] == 3) {
            out[2].erase({-sumTwo(col[c]), c});
        }
        if (cnt[c] >= 3) {
            if (cnt[c] >= 4) out[1].erase({-col[c][cnt[c]-2], c});
            out[1].insert({-col[c][cnt[c]-1], c});
        }
        if (cnt[c] >= 4) {
            if (cnt[c] >= 5) out[2].erase({-col[c][cnt[c]-2]-col[c][cnt[c]-3], c});
            out[2].insert({-col[c][cnt[c]-1]-col[c][cnt[c]-2], c});
        }
    }
}

### 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);
    }
    bool operator > (ii o) const {
        return tie(a, b) > tie(o.a, o.b);
    }
};

ll N, K, cnt[MAXN];
vector<ll> col[MAXN];

ll sumTwo(const vector<ll> &v) {
    return v[0]+v[1];
}

ll sumThree(const vector<ll> &v) {
    return v[0]+v[1]+v[2];
}

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<ii> arr;
    arr.push_back({INF, INF});
    set<ii> in[5], out[5];
    for (ll i = 1; i <= K; i++) {
        sort(col[i].begin(), col[i].end(), greater<ll>());
        col[i][0] = col[i][1] = sumTwo(col[i])/2;
        in[2].insert({sumTwo(col[i]), i});
        if (col[i].size() >= 3) in[3].insert({sumThree(col[i]), i});
        
        for (int j = 0; j < (int)col[i].size(); j++) {
            arr.push_back({col[i][j], i});
        }
    }
    sort(arr.begin(), arr.end(), greater<ii>());
    //~ for (auto i: arr) cerr << i.a << " " << i.b << "\n";
    
    ll sum = 0;
    for (ll i = 1; i <= N; i++) {
        if (in[2].find({sumTwo(col[arr[i].b]), arr[i].b}) == in[2].end()) {
            cout << (sum+arr[i].a)/2 << "\n";
        } else {
            ll ans = -2;
            if (!in[1].empty()) {
                ans = max(ans, sum+in[1].rbegin()->a);
            }
            if (!in[2].empty() && !out[1].empty()) {
                ans = max(ans, sum+out[1].rbegin()->a+in[2].rbegin()->a);
            }
            if (!in[3].empty() && !out[2].empty()) {
                ans = max(ans, sum+out[2].rbegin()->a+in[3].rbegin()->a);
            }
            cout << ans/2 << "\n";
        }
        int c = arr[i].b;
        cnt[c]++;
        sum += arr[i].a;
        if (cnt[c] == 1) {
            out[1].insert({-arr[i].a, c});
            //~ cerr << -arr[i].a << " " << c << " masuk\n";
            in[2].erase({sumTwo(col[c]), c});
            if (col[c].size() >= 3) in[3].erase({sumThree(col[c]), c});
        }
        if (cnt[c] == 2) {
            // valuenya pasti sama kayak yang in
            out[1].erase({-arr[i].a, c});
            out[2].insert({-sumTwo(col[c]), c});
        }
        if (cnt[c] >= 2) {
            if (cnt[c] >= 3) in[1].erase({col[c][cnt[c]-1], c});
            if ((int)col[c].size() > cnt[c]) in[1].insert({col[c][cnt[c]], c});
        }
        if (cnt[c] == 3) {
            out[2].erase({-sumTwo(col[c]), c});
        }
        if (cnt[c] >= 3) {
            if (cnt[c] >= 4) out[1].erase({-col[c][cnt[c]-2], c});
            out[1].insert({-col[c][cnt[c]-1], c});
        }
        if (cnt[c] >= 4) {
            if (cnt[c] >= 5) out[2].erase({-col[c][cnt[c]-2]-col[c][cnt[c]-3], c});
            out[2].insert({-col[c][cnt[c]-1]-col[c][cnt[c]-2], c});
        }
    }
}

```