#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define FASTIO
#define ALL(v) (v).begin(), (v).end()
#define rep(i, l, r) for (int i = (l); i < (r); ++i)

#ifdef FASTIO
#define scanf abacaba
#define printf abacaba
#endif

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;


/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/

const int MAXN = 220000;

int n, k;

vector<int> go[MAXN];

vector<pair<ll, int>> vv;
ll ans[MAXN];
ll dp[MAXN];

int main() {
#ifdef FASTIO
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
	cin >> n >> k;
	for (int i = 0; i < n; ++i) {
		int c;
		ll v;
		cin >> c >> v;
		v *= 2;
		go[c].push_back(v);
	}
	for (int i = 0; i < MAXN; ++i) {
		if (go[i].empty())
			continue;
		sort(ALL(go[i]));
		ll sum = go[i].back();
		go[i].pop_back();
		sum += go[i].back();
		go[i].pop_back();
		sum /= 2;
		vv.emplace_back(sum, i);
	}
	sort(ALL(vv));
	reverse(ALL(vv));
	multiset<int> ss;
	for (int i = 0; i <= n; ++i)
		ans[i] = -2;
	int cur = 0;
	ll now = 0;
	dp[vv.size() - 1] = -1;
	for (int i = (int)vv.size() - 2; i >= 0; --i) {
		int k = vv[i + 1].second;
		dp[i] = dp[i + 1];
		if (!go[k].empty())
			dp[i] = max(dp[i], go[k].back() + vv[i + 1].first * 2);
	}
	multiset<ll> tak;
	if (dp[0] != -1)
		ans[3] = max(ans[cur], dp[0]);
	for (int i = 0; i < vv.size(); ++i) {
		cur += 2;
		now += 2 * vv[i].first;
		for (ll j: go[vv[i].second])
			ss.insert(j);
		if (!tak.empty()) {
			ll x = *tak.begin();
			ans[cur - 1] = max(ans[cur - 1], now - x);
			if (dp[i] != -1)
				ans[cur - 1 + 3] = max(ans[cur - 1 + 3], now - x + dp[i]);
		}
		while (true) {
			ans[cur] = max(ans[cur], now);
			if (dp[i] != -1)
				ans[cur + 3] = max(ans[cur + 3], now + dp[i]);
			if (!ss.empty()) {
				ll x = *ss.rbegin();
				ans[cur + 1] = max(ans[cur + 1], now + x);
				if (dp[i] != -1)
					ans[cur + 4] = max(ans[cur + 4], now + x + dp[i]);
			}
			if (!ss.empty() && (i + 1 == vv.size() || *ss.rbegin() > vv[i + 1].first)) {
				ll x = *ss.rbegin();
				ss.erase(ss.find(x));
				tak.insert(x);
				now += x;
				cur += 1;
			}
			else {
				break;
			}
		}
	}
	for (int i = 1; i <= n; ++i)
		cout << ans[i] / 2 << "\n";
	return 0;
}



### 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
#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define FASTIO
#define ALL(v) (v).begin(), (v).end()
#define rep(i, l, r) for (int i = (l); i < (r); ++i)

#ifdef FASTIO
#define scanf abacaba
#define printf abacaba
#endif

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;


/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/

const int MAXN = 220000;

int n, k;

vector<int> go[MAXN];

vector<pair<ll, int>> vv;
ll ans[MAXN];
ll dp[MAXN];

int main() {
#ifdef FASTIO
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
	cin >> n >> k;
	for (int i = 0; i < n; ++i) {
		int c;
		ll v;
		cin >> c >> v;
		v *= 2;
		go[c].push_back(v);
	}
	for (int i = 0; i < MAXN; ++i) {
		if (go[i].empty())
			continue;
		sort(ALL(go[i]));
		ll sum = go[i].back();
		go[i].pop_back();
		sum += go[i].back();
		go[i].pop_back();
		sum /= 2;
		vv.emplace_back(sum, i);
	}
	sort(ALL(vv));
	reverse(ALL(vv));
	multiset<int> ss;
	for (int i = 0; i <= n; ++i)
		ans[i] = -2;
	int cur = 0;
	ll now = 0;
	dp[vv.size() - 1] = -1;
	for (int i = (int)vv.size() - 2; i >= 0; --i) {
		int k = vv[i + 1].second;
		dp[i] = dp[i + 1];
		if (!go[k].empty())
			dp[i] = max(dp[i], go[k].back() + vv[i + 1].first * 2);
	}
	multiset<ll> tak;
	if (dp[0] != -1)
		ans[3] = max(ans[cur], dp[0]);
	for (int i = 0; i < vv.size(); ++i) {
		cur += 2;
		now += 2 * vv[i].first;
		for (ll j: go[vv[i].second])
			ss.insert(j);
		if (!tak.empty()) {
			ll x = *tak.begin();
			ans[cur - 1] = max(ans[cur - 1], now - x);
			if (dp[i] != -1)
				ans[cur - 1 + 3] = max(ans[cur - 1 + 3], now - x + dp[i]);
		}
		while (true) {
			ans[cur] = max(ans[cur], now);
			if (dp[i] != -1)
				ans[cur + 3] = max(ans[cur + 3], now + dp[i]);
			if (!ss.empty()) {
				ll x = *ss.rbegin();
				ans[cur + 1] = max(ans[cur + 1], now + x);
				if (dp[i] != -1)
					ans[cur + 4] = max(ans[cur + 4], now + x + dp[i]);
			}
			if (!ss.empty() && (i + 1 == vv.size() || *ss.rbegin() > vv[i + 1].first)) {
				ll x = *ss.rbegin();
				ss.erase(ss.find(x));
				tak.insert(x);
				now += x;
				cur += 1;
			}
			else {
				break;
			}
		}
	}
	for (int i = 1; i <= n; ++i)
		cout << ans[i] / 2 << "\n";
	return 0;
}



```