#include<cassert>
#include<set>
#include<stdio.h>
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<string.h>
using namespace std;

typedef long long LL;
typedef vector<int> VI;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define EACH(i,c) for(__typeof((c).begin()) i=(c).begin(),i##_end=(c).end();i!=i##_end;++i)
#define eprintf(...) fprintf(stderr, __VA_ARGS__)

template<class T> inline void amin(T &x, const T &y) { if (y<x) x=y; }
template<class T> inline void amax(T &x, const T &y) { if (x<y) x=y; }
template<class Iter> void rprintf(const char *fmt, Iter begin, Iter end) {
    for (bool sp=0; begin!=end; ++begin) { if (sp) putchar(' '); else sp = true; printf(fmt, *begin); }
    putchar('\n');
}

int N, K;

struct Jewel {
    LL value;
    int color;
    Jewel(LL value_, int color_) {
	value = value_;
	color=  color_;
    }
    bool operator<(const Jewel &y) const {
	return value != y.value ? value > y.value : color < y.color;
    }
};

VI G[100011];
const LL INF = 1LL<<60;
int C[100011];

void MAIN() {
    scanf("%d%d", &N, &K);
    REP (i, N) {
	int c, v;
	scanf("%d%d", &c, &v);
	c--;
	G[c].push_back(v);
    }

    vector<Jewel> J;
    multiset<Jewel> X, Y, Z;
    REP (k, K) if (!G[k].empty()) {
	VI &v = G[k];
	sort(v.begin(), v.end());
	reverse(v.begin(), v.end());
	REP (t, 2) J.emplace_back(v[0] + v[1], k);
	for (int a=2; a<(int)v.size(); a++) J.emplace_back(v[a]*2, k);

	Y.emplace(v[0]+v[1], k);
	if (v.size() >= 3u) {
	    Z.emplace(((LL)v[0]+v[1]+v[2])*2, k);
	}
    }
    sort(J.begin(), J.end());

    LL min_value = INF;
    LL min_pair = INF;
    LL sum = 0;
    for (int i=0; i<N; i++) {
	LL value = J[i].value;
	int color = J[i].color;
	LL ans = -2;

	if (C[color] == 0) {
	    if (!X.empty()) {
		amax(ans, sum + X.begin()->value);
	    }
	    if (!Y.empty()) {
		amax(ans, sum - min_value + Y.begin()->value * 2);
	    }
	    if (!Z.empty()) {
		amax(ans, sum - min_pair + Z.begin()->value);
	    }

	    Y.erase(Y.find(J[i]));
	    if (G[color].size() >= 3u) {
		Jewel key(((LL)G[color][0] + G[color][1] + G[color][2])*2, color);
		Z.erase(Z.find(key));
	    }
	} else {
	    ans = sum + value;;
	    if (C[color] == 1) {
		for (int p=2; p<(int)G[color].size(); p++) {
		    X.insert(Jewel(G[color][p]*2, color));
		}
		amin(min_pair, value * 2);
	    }
	    if (C[color] >= 2) {
		X.erase(X.find(J[i]));
		amin(min_value, value);
	    }
	}
	C[color]++;
	sum += value;

	printf("%lld\n", ans/2);
    }
}

int main() {
    int TC = 1;
//    scanf("%d", &TC);
    REP (tc, TC) MAIN();
    return 0;
}


### Prompt

Please formulate 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<cassert>
#include<set>
#include<stdio.h>
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<string.h>
using namespace std;

typedef long long LL;
typedef vector<int> VI;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define EACH(i,c) for(__typeof((c).begin()) i=(c).begin(),i##_end=(c).end();i!=i##_end;++i)
#define eprintf(...) fprintf(stderr, __VA_ARGS__)

template<class T> inline void amin(T &x, const T &y) { if (y<x) x=y; }
template<class T> inline void amax(T &x, const T &y) { if (x<y) x=y; }
template<class Iter> void rprintf(const char *fmt, Iter begin, Iter end) {
    for (bool sp=0; begin!=end; ++begin) { if (sp) putchar(' '); else sp = true; printf(fmt, *begin); }
    putchar('\n');
}

int N, K;

struct Jewel {
    LL value;
    int color;
    Jewel(LL value_, int color_) {
	value = value_;
	color=  color_;
    }
    bool operator<(const Jewel &y) const {
	return value != y.value ? value > y.value : color < y.color;
    }
};

VI G[100011];
const LL INF = 1LL<<60;
int C[100011];

void MAIN() {
    scanf("%d%d", &N, &K);
    REP (i, N) {
	int c, v;
	scanf("%d%d", &c, &v);
	c--;
	G[c].push_back(v);
    }

    vector<Jewel> J;
    multiset<Jewel> X, Y, Z;
    REP (k, K) if (!G[k].empty()) {
	VI &v = G[k];
	sort(v.begin(), v.end());
	reverse(v.begin(), v.end());
	REP (t, 2) J.emplace_back(v[0] + v[1], k);
	for (int a=2; a<(int)v.size(); a++) J.emplace_back(v[a]*2, k);

	Y.emplace(v[0]+v[1], k);
	if (v.size() >= 3u) {
	    Z.emplace(((LL)v[0]+v[1]+v[2])*2, k);
	}
    }
    sort(J.begin(), J.end());

    LL min_value = INF;
    LL min_pair = INF;
    LL sum = 0;
    for (int i=0; i<N; i++) {
	LL value = J[i].value;
	int color = J[i].color;
	LL ans = -2;

	if (C[color] == 0) {
	    if (!X.empty()) {
		amax(ans, sum + X.begin()->value);
	    }
	    if (!Y.empty()) {
		amax(ans, sum - min_value + Y.begin()->value * 2);
	    }
	    if (!Z.empty()) {
		amax(ans, sum - min_pair + Z.begin()->value);
	    }

	    Y.erase(Y.find(J[i]));
	    if (G[color].size() >= 3u) {
		Jewel key(((LL)G[color][0] + G[color][1] + G[color][2])*2, color);
		Z.erase(Z.find(key));
	    }
	} else {
	    ans = sum + value;;
	    if (C[color] == 1) {
		for (int p=2; p<(int)G[color].size(); p++) {
		    X.insert(Jewel(G[color][p]*2, color));
		}
		amin(min_pair, value * 2);
	    }
	    if (C[color] >= 2) {
		X.erase(X.find(J[i]));
		amin(min_value, value);
	    }
	}
	C[color]++;
	sum += value;

	printf("%lld\n", ans/2);
    }
}

int main() {
    int TC = 1;
//    scanf("%d", &TC);
    REP (tc, TC) MAIN();
    return 0;
}


```