#include <bits/stdc++.h>

#define F first
#define S second
#define prev azaza
#define MP make_pair
#define PB push_back

using namespace std;
typedef long long ll;
typedef long double ld;

const int max_n = 200111, inf = 1000111222;

int n, k;
ll ans[max_n];

vector<ll> col[max_n];
vector<pair<ll, int>> v;
bool has_ch_left[max_n];
bool np[max_n];
bool has[max_n];

multiset<ll> r1; // that has pair in the left
multiset<ll> r2; // over all
multiset<ll> r3; // over all
multiset<ll> l1; // over all
multiset<ll> l2; // that has no child in the left

ll set_mx(const multiset<ll>& ms) {
    if (ms.empty()) {
        exit(228);
    }
    auto it = ms.end();
    --it;
    return *it;
}

void add2(int k) {
    vector<ll>& c = col[k];
    r2.erase(r2.find(c[0] + c[1]));
    if (c.size() > 2) {
        r3.erase(r3.find(c[0] + c[1] + c[2]));
    }
    for (int i = 2; i < c.size(); ++i) {
        r1.insert(c[i]);
    }
    l2.insert(c[0] + c[1]);
}

void add1(int k, ll val) {
    vector<ll>& c = col[k];
    r1.erase(r1.find(val));
    l1.insert(val);
    /*if (has_ch_left[k] == 0) {
        has_ch_left[k] = 1;
        l2.erase(l2.find(col[k][0] + col[k][1]));
    }*/
}

int main()
{
    //freopen("input.txt", "r", stdin);
    cin >> n >> k;
    for (int i = 0; i < n; ++i) {
        int val, c;
        scanf("%d %d", &c, &val);
        val *= 2;
        --c;
        col[c].PB(val);
    }
    for (int i = 0; i < k; ++i) {
        sort(col[i].begin(), col[i].end(), greater<ll>());
        col[i][0] = col[i][1] = (col[i][0] + col[i][1]) / 2;
        r2.insert(2 * col[i][0]);
        if (col[i].size() > 2) {
            r3.insert(col[i][0] + col[i][1] + col[i][2]);
        }
        for (int val : col[i]) {
            v.emplace_back(val, i);
        }
    }
    sort(v.begin(), v.end(), greater<pair<ll, int>>());
    for (int i = 0; i < v.size(); ++i) {
        if (has[v[i].S] == 0) {
            has[v[i].S] = 1;
            np[i] = true;
        }
    }
    ll sum = 0;
    for (int i = 0; i < n; ++i) {
        ans[i] = -inf;
        if (np[i]) {
            if (!r1.empty()) {
                ans[i] = max(ans[i], sum + set_mx(r1));
            }
            if (!l1.empty() && !r2.empty()) {
                ans[i] = max(ans[i], sum - *l1.begin() + set_mx(r2));
            }
            if (!l2.empty() && !r3.empty()) {
                ans[i] = max(ans[i], sum - *l2.begin() + set_mx(r3));
            }
            sum += v[i].first;
            continue;
        }
        sum += v[i].first;
        ans[i] = sum;
        if (i && np[i - 1]) {
            add2(v[i].S);
        } else {
            add1(v[i].S, v[i].F);
        }
    }
    for (int i = 0; i < n; ++i) {
        if (ans[i] == -inf) {
            ans[i] = -2;
        }
        printf("%lld\n", ans[i] / 2);
    }

    return 0;
}



### Prompt

In cpp, your task is to solve 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>

#define F first
#define S second
#define prev azaza
#define MP make_pair
#define PB push_back

using namespace std;
typedef long long ll;
typedef long double ld;

const int max_n = 200111, inf = 1000111222;

int n, k;
ll ans[max_n];

vector<ll> col[max_n];
vector<pair<ll, int>> v;
bool has_ch_left[max_n];
bool np[max_n];
bool has[max_n];

multiset<ll> r1; // that has pair in the left
multiset<ll> r2; // over all
multiset<ll> r3; // over all
multiset<ll> l1; // over all
multiset<ll> l2; // that has no child in the left

ll set_mx(const multiset<ll>& ms) {
    if (ms.empty()) {
        exit(228);
    }
    auto it = ms.end();
    --it;
    return *it;
}

void add2(int k) {
    vector<ll>& c = col[k];
    r2.erase(r2.find(c[0] + c[1]));
    if (c.size() > 2) {
        r3.erase(r3.find(c[0] + c[1] + c[2]));
    }
    for (int i = 2; i < c.size(); ++i) {
        r1.insert(c[i]);
    }
    l2.insert(c[0] + c[1]);
}

void add1(int k, ll val) {
    vector<ll>& c = col[k];
    r1.erase(r1.find(val));
    l1.insert(val);
    /*if (has_ch_left[k] == 0) {
        has_ch_left[k] = 1;
        l2.erase(l2.find(col[k][0] + col[k][1]));
    }*/
}

int main()
{
    //freopen("input.txt", "r", stdin);
    cin >> n >> k;
    for (int i = 0; i < n; ++i) {
        int val, c;
        scanf("%d %d", &c, &val);
        val *= 2;
        --c;
        col[c].PB(val);
    }
    for (int i = 0; i < k; ++i) {
        sort(col[i].begin(), col[i].end(), greater<ll>());
        col[i][0] = col[i][1] = (col[i][0] + col[i][1]) / 2;
        r2.insert(2 * col[i][0]);
        if (col[i].size() > 2) {
            r3.insert(col[i][0] + col[i][1] + col[i][2]);
        }
        for (int val : col[i]) {
            v.emplace_back(val, i);
        }
    }
    sort(v.begin(), v.end(), greater<pair<ll, int>>());
    for (int i = 0; i < v.size(); ++i) {
        if (has[v[i].S] == 0) {
            has[v[i].S] = 1;
            np[i] = true;
        }
    }
    ll sum = 0;
    for (int i = 0; i < n; ++i) {
        ans[i] = -inf;
        if (np[i]) {
            if (!r1.empty()) {
                ans[i] = max(ans[i], sum + set_mx(r1));
            }
            if (!l1.empty() && !r2.empty()) {
                ans[i] = max(ans[i], sum - *l1.begin() + set_mx(r2));
            }
            if (!l2.empty() && !r3.empty()) {
                ans[i] = max(ans[i], sum - *l2.begin() + set_mx(r3));
            }
            sum += v[i].first;
            continue;
        }
        sum += v[i].first;
        ans[i] = sum;
        if (i && np[i - 1]) {
            add2(v[i].S);
        } else {
            add1(v[i].S, v[i].F);
        }
    }
    for (int i = 0; i < n; ++i) {
        if (ans[i] == -inf) {
            ans[i] = -2;
        }
        printf("%lld\n", ans[i] / 2);
    }

    return 0;
}



```