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: Time Limit Exceeded)

import java.util.*;
import java.io.*;

public class Main {
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int n = sc.nextInt();
        int [] x = new int[n];
        int [] y = new int[n];
        long [] c = new long[n];
        for (int i = 0; i < n; i++) {
            x[i] = sc.nextInt(); y[i] = sc.nextInt(); c[i] = sc.nextInt();
        }
        long [] ans = new long[n + 1];
        Arrays.fill(ans, Long.MAX_VALUE / 2);
        for (int hor = 0; hor < (1 << n); hor++) {
            int bit = Integer.bitCount(hor);
            long [] sh = new long[bit];
            long [] cost = new long[bit];
            int cur = 0;
            for (int i = 0; i < n; i++) {
                if (((hor >> i) & 1) == 1) {
                    sh[cur] = x[i];
                    cost[cur] = c[i];
                    cur++;
                }
            }
            long [] h = new long[n + 1];
            Arrays.fill(h, Long.MAX_VALUE / 2);
            for (int mask = 0; mask < (1 << bit); mask++) {
                int put = Integer.bitCount(mask);
                TreeSet<Long> set = new TreeSet<>();
                set.add(0L);
                for (int j = 0; j < bit; j++) {
                    if (((mask >> j) & 1) == 1) {
                        set.add(sh[j]);
                    }
                }
                long ret = 0;
                for (int j = 0; j < bit; j++) {
                    if (set.contains(sh[j])) continue;
                    Long lower = set.lower(sh[j]);
                    Long upper = set.higher(sh[j]);
                    long min = Long.MAX_VALUE / 2;
                    if (lower != null) min = Math.min(cost[j] * (sh[j] - lower), min);
                    if (upper != null) min = Math.min(min, (upper - sh[j]) * cost[j]);
                    ret += min;
                }
                h[put] = Math.min(h[put], ret);
            }
            int vert = ((1 << n) - 1) ^ hor;
            bit = Integer.bitCount(vert);
            sh = new long[bit]; cost = new long[bit];
            cur = 0;
            for (int i = 0; i < n; i++) {
                if (((vert >> i) & 1) == 1) {
                    sh[cur] = y[i];
                    cost[cur] = c[i];
                    cur++;
                }
            }
            long [] v = new long[n + 1];
            Arrays.fill(v, Long.MAX_VALUE / 2);
            for (int mask = 0; mask < (1 <<bit); mask++) {
                int put = Integer.bitCount(mask);
                TreeSet<Long> set = new TreeSet<>();
                set.add(0L);
                for (int j = 0; j < bit; j++) {
                    if (((mask >> j) & 1) == 1) {
                        set.add(sh[j]);
                    }
                }
                long temp = 0;
                for (int j = 0; j < bit; j++) {
                    if (set.contains(sh[j])) continue;
                    Long lower = set.lower(sh[j]);
                    Long upper = set.higher(sh[j]);
                    long min = Long.MAX_VALUE / 2;
                    if (lower != null) min = Math.min(cost[j] * (sh[j] - lower), min);
                    if (upper != null) min = Math.min(min, (upper - sh[j]) * cost[j]);
                    temp += min;
                }
                v[put] = Math.min(v[put], temp);
            }
            for (int i = 0; i <= n; i++) {
                for (int j = 0; j <= n; j++) {
                    if (i + j > n) continue;
                    ans[i + j] = Math.min(ans[i + j], h[i] + v[j]);
                }
            }
        }
        for (int i = 0; i <= n; i++) out.println(ans[i]);
        out.close();
    }



    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }


    }

}