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: Accepted)

import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.InputStream;
import java.io.IOException;
import java.util.InputMismatchException;
import java.io.PrintWriter;
import java.util.Arrays;

public class Main {
    InputStream is;

    int __t__ = 1;
    int __f__ = 0;
    int __FILE_DEBUG_FLAG__ = __f__;
    String __DEBUG_FILE_NAME__ = "src/main/java/D2";

    FastScanner in;
    PrintWriter out;

    public void solve() {
        int n = in.nextInt();
        int[] x = new int[n];
        int[] y = new int[n];
        int[] p = new int[n];
        for (int i = 0; i < n; i++) {
            x[i] = in.nextInt();
            y[i] = in.nextInt();
            p[i] = in.nextInt();
        }

        long[][][] cal = new long[2][1<<n][n];
        for (int i = 0; i < 1 << n; i++) {
            for (int j = 0; j < n; j++) {
                cal[0][i][j] = Math.abs(x[j]);
                cal[1][i][j] = Math.abs(y[j]);
                for (int k = 0; k < n; k++) {
                    if ((i & (1 << k)) != 0) {
                        cal[0][i][j] = Math.min(cal[0][i][j], Math.abs(x[j] - x[k]));
                        cal[1][i][j] = Math.min(cal[1][i][j], Math.abs(y[j] - y[k]));
                    }
                }
            }
        }

        /*
        long[] res = new long[1<<n];
        Arrays.fill(res, Long.MAX_VALUE);
        int[] three = new int[n+1];
        three[0] = 1;
        for (int i = 0; i < n; i++) {
            three[i+1] = three[i] * 3;
        }
        int all = three[n];
        for (int bits = 0; bits < all; bits++) {
            int zero = 0;
            int one = 0;
            int bi = 0;
            for (int i = 0; i < n; i++) {
                int b = (bits / three[i]) % 3;
                if (b == 0) {
                    zero |= (1 << i);
                } else if (b == 1) {
                    one |= (1 << i);
                }
                if (b != 2) {
                    bi |= (1 << i);
                }
            }
            long sum = 0;
            for (int i = 0; i < n; i++) {
                sum += Math.min(cal[0][zero][i], cal[1][one][i]) * p[i];
            }
            res[bi] = Math.min(res[bi], sum);
        }

        long[] res2 = new long[n+1];
        Arrays.fill(res2, Long.MAX_VALUE);
        for (int i = 0; i < 1 << n; i++) {
            int b = Integer.bitCount(i);
            res2[b] = Math.min(res2[b], res[i]);
        }
        for (int i = 0; i <= n; i++) {
            System.out.println(res2[i]);
        }

         */

        long[] res2 = new long[n+1];
        Arrays.fill(res2, Long.MAX_VALUE);
        for (int bits = 0; bits < 1 << n; bits++) {
            int cnt = Integer.bitCount(bits);
            for (int one = bits; one >= 0; one--) {
                one &= bits;
                int zero = bits - one;
                long sum = 0;
                for (int i = 0; i < n; i++) {
                    sum += Math.min(cal[0][zero][i], cal[1][one][i]) * p[i];
                }
                res2[cnt] = Math.min(res2[cnt], sum);
            }
        }

        for (int i = 0; i <= n; i++) {
            System.out.println(res2[i]);
        }
    }

    public void run() {
        if (__FILE_DEBUG_FLAG__ == __t__) {
            try {
                is = new FileInputStream(__DEBUG_FILE_NAME__);
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
            System.out.println("FILE_INPUT!");
        } else {
            is = System.in;
        }
        in = new FastScanner(is);
        out = new PrintWriter(System.out);

        solve();
    }

    public static void main(final String[] args) {
        new Main().run();
    }
}


class FastScanner {
    private InputStream stream;
    private byte[] buf = new byte[1024];
    private int curChar;
    private int numChars;

    public FastScanner(InputStream stream) {
        this.stream = stream;
        // stream = new FileInputStream(new File("dec.in"));
    }

    int read() {
        if (numChars == -1)
            throw new InputMismatchException();
        if (curChar >= numChars) {
            curChar = 0;
            try {
                numChars = stream.read(buf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (numChars <= 0)
                return -1;
        }
        return buf[curChar++];
    }

    public boolean isSpaceChar(int c) {
        return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
    }

    public boolean isEndline(int c) {
        return c == '\n' || c == '\r' || c == -1;
    }

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

    public int[] nextIntArray(int n) {
        int[] array = new int[n];
        for (int i = 0; i < n; i++)
            array[i] = nextInt();

        return array;
    }

    public int[][] nextIntMap(int n, int m) {
        int[][] map = new int[n][m];
        for (int i = 0; i < n; i++) {
            map[i] = nextIntArray(m);
        }
        return map;
    }

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

    public long[] nextLongArray(int n) {
        long[] array = new long[n];
        for (int i = 0; i < n; i++)
            array[i] = nextLong();

        return array;
    }

    public long[][] nextLongMap(int n, int m) {
        long[][] map = new long[n][m];
        for (int i = 0; i < n; i++) {
            map[i] = nextLongArray(m);
        }
        return map;
    }

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

    public double[] nextDoubleArray(int n) {
        double[] array = new double[n];
        for (int i = 0; i < n; i++)
            array[i] = nextDouble();

        return array;
    }

    public double[][] nextDoubleMap(int n, int m) {
        double[][] map = new double[n][m];
        for (int i = 0; i < n; i++) {
            map[i] = nextDoubleArray(m);
        }
        return map;
    }

    public String next() {
        int c = read();
        while (isSpaceChar(c))
            c = read();
        StringBuilder res = new StringBuilder();
        do {
            res.appendCodePoint(c);
            c = read();
        } while (!isSpaceChar(c));
        return res.toString();
    }

    public String[] nextStringArray(int n) {
        String[] array = new String[n];
        for (int i = 0; i < n; i++)
            array[i] = next();

        return array;
    }

    public String nextLine() {
        int c = read();
        while (isEndline(c))
            c = read();
        StringBuilder res = new StringBuilder();
        do {
            res.appendCodePoint(c);
            c = read();
        } while (!isEndline(c));
        return res.toString();
    }

    public int[][] nextPackedIntArrays(int packN, int size) {
        int[][] res = new int[packN][size];
        for (int i = 0; i < size; i++) {
            for (int j = 0; j < packN; j++) {
                res[j][i] = nextInt();
            }
        }
        return res;
    }
}