Problem:
Score: 500 points

Problem Statement

New AtCoder City has an infinite grid of streets, as follows:

At the center of the city stands a clock tower. Let (0, 0) be the coordinates of this point.

A straight street, which we will call East-West Main Street, runs east-west and passes the clock tower. It corresponds to the x-axis in the two-dimensional coordinate plane.

There are also other infinitely many streets parallel to East-West Main Street, with a distance of 1 between them. They correspond to the lines \ldots, y = -2, y = -1, y = 1, y = 2, \ldots in the two-dimensional coordinate plane.

A straight street, which we will call North-South Main Street, runs north-south and passes the clock tower. It corresponds to the y-axis in the two-dimensional coordinate plane.

There are also other infinitely many streets parallel to North-South Main Street, with a distance of 1 between them. They correspond to the lines \ldots, x = -2, x = -1, x = 1, x = 2, \ldots in the two-dimensional coordinate plane.

There are N residential areas in New AtCoder City. The i-th area is located at the intersection with the coordinates (X_i, Y_i) and has a population of P_i. Each citizen in the city lives in one of these areas.

The city currently has only two railroads, stretching infinitely, one along East-West Main Street and the other along North-South Main Street.

M-kun, the mayor, thinks that they are not enough for the commuters, so he decides to choose K streets and build a railroad stretching infinitely along each of those streets.

Let the walking distance of each citizen be the distance from his/her residential area to the nearest railroad.

M-kun wants to build railroads so that the sum of the walking distances of all citizens, S, is minimized.

For each K = 0, 1, 2, \dots, N, what is the minimum possible value of S after building railroads?

Constraints

1 \leq N \leq 15

-10 \ 000 \leq X_i \leq 10 \ 000

-10 \ 000 \leq Y_i \leq 10 \ 000

1 \leq P_i \leq 1 \ 000 \ 000

The locations of the N residential areas, (X_i, Y_i), are all distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X_1 Y_1 P_1
X_2 Y_2 P_2
: : :
X_N Y_N P_N

Output

Print the answer in N+1 lines.

The i-th line (i = 1, \ldots, N+1) should contain the minimum possible value of S after building railroads for the case K = i-1.

Sample Input 1

3
1 2 300
3 3 600
1 4 800

Sample Output 1

2900
900
0
0

When K = 0, the residents of Area 1, 2, and 3 have to walk the distances of 1, 3, and 1, respectively, to reach a railroad.

Thus, the sum of the walking distances of all citizens, S, is 1 \times 300 + 3 \times 600 + 1 \times 800 = 2900.

When K = 1, if we build a railroad along the street corresponding to the line y = 4 in the coordinate plane, the walking distances of the citizens of Area 1, 2, and 3 become 1, 1, and 0, respectively.

Then, S = 1 \times 300 + 1 \times 600 + 0 \times 800 = 900.

We have many other options for where we build the railroad, but none of them makes S less than 900.

When K = 2, if we build a railroad along the street corresponding to the lines x = 1 and x = 3 in the coordinate plane, all citizens can reach a railroad with the walking distance of 0, so S = 0. We can also have S = 0 when K = 3.

The figure below shows the optimal way to build railroads for the cases K = 0, 1, 2.

The street painted blue represents the roads along which we build railroads.

Sample Input 2

5
3 5 400
5 3 700
5 5 1000
5 7 700
7 5 400

Sample Output 2

13800
1600
0
0
0
0

The figure below shows the optimal way to build railroads for the cases K = 1, 2.

Sample Input 3

6
2 5 1000
5 2 1100
5 5 1700
-2 -5 900
-5 -2 600
-5 -5 2200

Sample Output 3

26700
13900
3200
1200
0
0
0

The figure below shows the optimal way to build railroads for the case K = 3.

Sample Input 4

8
2 2 286017
3 1 262355
2 -2 213815
1 -3 224435
-2 -2 136860
-3 -1 239338
-2 2 217647
-1 3 141903

Sample Output 4

2576709
1569381
868031
605676
366338
141903
0
0
0

The figure below shows the optimal way to build railroads for the case K = 4.

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>

int main()
{
	int i, N, X[16], Y[16], P[16];
	scanf("%d", &N);
	for (i = 0; i < N; i++) scanf("%d %d %d", &(X[i]), &(Y[i]), &(P[i]));
	
	int j, k, dist[2][32768][16], num[32768] = {};
	const int bit[16] = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768};
	for (i = 0; i < bit[N]; i++) {
		for (k = 0; k < N; k++) if ((i & bit[k]) != 0) num[i]++;
		for (j = 0; j < N; j++) {
			dist[0][i][j] = abs(X[j]);
			dist[1][i][j] = abs(Y[j]);
			for (k = 0; k < N; k++) {
				if ((i & bit[k]) == 0) continue;
				if (abs(X[j] - X[k]) < dist[0][i][j]) dist[0][i][j] = abs(X[j] - X[k]);
				if (abs(Y[j] - Y[k]) < dist[1][i][j]) dist[1][i][j] = abs(Y[j] - Y[k]);
			}
		}
	}
	
	const long long sup = (long long)1 << 60;
	long long ans[16], tmp;
	for (i = 0, ans[N] = 0; i < N; i++) ans[i] = sup;
	for (i = 0; i < bit[N]; i++) {
		for (j = 0; j < bit[N]; j++) {
			if (num[i] + num[j] >= N - 4) continue;
			for (k = 0, tmp = 0; k < N; k++) tmp += (long long)P[k] * ((dist[0][i][k] < dist[1][j][k])? dist[0][i][k]: dist[1][j][k]);
			if (tmp < ans[num[i] + num[j]]) ans[num[i] + num[j]] = tmp;
		}
	}
	
	int l;
	long long min, max[16];
	for (k = 0; k < N; k++) max[k] = (long long)P[k] * ((abs(X[k]) < abs(Y[k]))? abs(X[k]): abs(Y[k]));
	for (i = 0; i < bit[N]; i++) {
		if (num[i] < N - 4) continue;
		for (j = 0; j < bit[N]; j++) {
			if ((i & j) != j) continue;
			for (k = 0, tmp = 0; k < N; k++) {
				if ((i & bit[k]) != 0) continue;
				for (l = 0, min = max[k]; l < N; l++) {
					if ((i & bit[l]) == 0) continue;
					if ((j & bit[l]) == 0 && (long long)P[k] * abs(X[k] - X[l]) < min) min = (long long)P[k] * abs(X[k] - X[l]);
					if ((j & bit[l]) != 0 && (long long)P[k] * abs(Y[k] - Y[l]) < min) min = (long long)P[k] * abs(Y[k] - Y[l]);
				}
				tmp += min;
			}
			if (tmp < ans[num[i]]) ans[num[i]] = tmp;
		}
	}
	
	for (i = 0; i <= N; i++) printf("%lld\n", ans[i]);
	fflush(stdout);
	return 0;
}