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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		long result[] = new long[N];
		City[] c = new City[N];
		Main main = new Main();
		for (int i = 0; i < N; i++) {
			result[i] = Long.MAX_VALUE;
			City temp = main.new City();
			temp.x = sc.nextInt();
			temp.y = sc.nextInt();
			temp.p = sc.nextLong();
			c[i] = temp;
		}
		int pattern = 1;
		for (int i = 0; i < N; i++) {
			pattern *= 3;
		}
		for (int i = 0; i < pattern - 1; i++) {
			int road = Main.roadCount(i);
			if (road >= N) {
				continue;
			}
			int p[] = Main.changePattern(i, N);
			long cost = Main.cost(c, p);
			if (result[road] > cost) {
				result[road] = cost;
			}
		}
		for (int i = 0; i < N; i++) {
			System.out.println(result[i]);
		}
		System.out.println(0);
	}

	public static int roadCount(int pattern) {
		int i = 0;
		while (pattern > 0) {
			if (pattern % 3 != 0) {
				i++;
			}
			pattern /= 3;
		}
		return i;
	}

	public static int[] changePattern(int pattern, int N) {
		int p[] = new int[N];
		int i = 0;
		while (pattern > 0) {
			p[i] = pattern % 3;
			i++;
			pattern /= 3;
		}
		return p;
	}

	public static long cost(City[] c, int[] pattern) {
		long result = 0;
		List<Integer> xRoad = new ArrayList<Integer>();
		xRoad.add(0);
		List<Integer> yRoad = new ArrayList<Integer>();
		yRoad.add(0);
		Collections.sort(xRoad);
		Collections.sort(yRoad);
		List<City> city = new ArrayList<City>();
		for (int i = 0; i < c.length; i++) {
			if (pattern[i] == 1) {
				// x road
				xRoad.add(c[i].x);
			} else if (pattern[i] == 2) {
				// y road
				yRoad.add(c[i].y);
			} else {
				city.add(c[i]);
			}
		}

		for (int i = 0; i < city.size(); i++) {
				long min = Long.MAX_VALUE;
				int m = 0;
				int a = xRoad.size()-1;
				int index =0;
				int re =a;
				while (m<=a) {
					index = (m+a)/2;
					if (xRoad.get(index)<city.get(i).x) {
						re = index;
						m = re+1;
					} else {
						a = index-1;
					}
				}
				if (re==xRoad.size()-1) {
					min = Math.abs(city.get(i).x - xRoad.get(re));
				} else {
					min = Math.min(Math.abs(city.get(i).x - xRoad.get(re)), Math.abs(city.get(i).x - xRoad.get(re+1)));
				}

				m = 0;
				a = yRoad.size()-1;
				index =0;
				re =a;
				while (m<=a) {
					index = (m+a)/2;
					if (yRoad.get(index)<city.get(i).y) {
						re = index;
						m = re+1;
					} else {
						a = index-1;
					}
				}
				if (re==yRoad.size()-1) {
					min = Math.abs(city.get(i).y - yRoad.get(re));
				} else {
					min = Math.min(Math.abs(city.get(i).y - yRoad.get(re)), Math.abs(city.get(i).y - yRoad.get(re+1)));
				}

				result += min * city.get(i).p;
		}
		return result;
	}

	public class City   {
		int x;
		int y;
		long p;
	}
}