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

enum XY {
  X, Y;
}

class Solver {
  final int n;
  final int[][] xyps;

  Solver(int n, int[][] xyps) {
    this.n = n;
    this.xyps = xyps;
  }

  /*
  private long solve(int k) {
    for (int bitset = 0; bitset < (1 << k); bitset++) {
      for (int flagset = 0; flagset < )
    }
    return 0;
  }
  */

  public List<Long> solve() {
    long[] answers = new long[n + 1];
    int[] indexes = new int[n];
    XY[] xys = new XY[n];
    int[] vs = new int[n];
    Arrays.fill(answers, Long.MAX_VALUE);
    for (int bitset = 0; bitset < (1 << n); bitset++) {
      int bitcount = Integer.bitCount(bitset);
      {
        int indexesIndex = 0;
        for (int i = 0; i < n; i++) {
          if (((bitset >> i) & 1) == 1) {
            indexes[indexesIndex] = i;
            indexesIndex++;
          }
        }          
      }


      for (int flagset = 0; flagset < (1 << bitcount); flagset++) {
        {
          for (int i = 0; i < bitcount; i++) {
            if (((flagset >> i) & 1) == 1) {
              xys[i] = XY.X;
              vs[i] = xyps[indexes[i]][0];
            } else {
              xys[i] = XY.Y;
              vs[i] = xyps[indexes[i]][1];
            }
          }  
        }

        long steps = 0;
        for (int i = 0; i < n; i++) {
          long minStep = Math.min(Math.abs(xyps[i][0]), Math.abs(xyps[i][1]));
          for (int j = 0; j < bitcount; j++) {
            if (xys[j] == XY.X) {
              minStep = Math.min(minStep, Math.abs(xyps[i][0] - vs[j]));
            } else {
              minStep = Math.min(minStep, Math.abs(xyps[i][1] - vs[j]));              
            }
          }
          steps += minStep * (long) xyps[i][2];
        }

        answers[bitcount] = Math.min(answers[bitcount], steps);
      }
    }

    List<Long> answerList = new ArrayList<>();
    for (int i = 0; i <= n; i++) {
      answerList.add(answers[i]);
    }
    return answerList;
  }
}

public class Main {
  private static void execute(ContestReader reader, ContestWriter out) {
    int n = reader.nextInt();
    int[][] xyps = reader.nextInt(n, 3);
    out.printList(new Solver(n, xyps).solve());
  }
  
  public static void main(String[] args) {
    ContestReader reader = new ContestReader(System.in);
    ContestWriter out = new ContestWriter(System.out);
    execute(reader, out);
    out.flush();
  }
}

class ContestWriter extends PrintWriter {
  ContestWriter(PrintStream printStream) {
    super(printStream);
  }

  public void printList(List<? extends Object> list) {
    for (Object object : list) {
      println(object);
    }
  }

  public void printListOneLine(List<? extends Object> list) {
    List<String> stringList = new ArrayList<>();
    for (Object object : list) {
      stringList.add(object.toString());
    }
    println(String.join(" ", stringList));
  }
}

class ContestReader {
  private static final int BUFFER_SIZE = 1024;
  
  private final InputStream stream;
  private final byte[] buffer;
  private int pointer;
  private int bufferLength;
  
  ContestReader(InputStream stream) {
    this.stream = stream;
    this.buffer = new byte[BUFFER_SIZE];
    this.pointer = 0;
    this.bufferLength = 0;
  }
  
  private boolean hasNextByte() {
    if (pointer < bufferLength) {
      return true;
    }
    
    pointer = 0;
    try {
      bufferLength = stream.read(buffer);
    } catch (IOException e) {
      throw new RuntimeException(e);
    }
    return bufferLength > 0;
  }
  
  private int readByte() {
    if (hasNextByte()) {
      return buffer[pointer++];
    } else {
      return -1;
    }
  }
  
  private static boolean isPrintableChar(int c) {
    return 33 <= c && c <= 126;
  }
  
  public boolean hasNext() {
    while (hasNextByte() && !isPrintableChar(buffer[pointer])) {
      pointer++;
    }
    return hasNextByte();
  }
  
  public String next() {
    if (!hasNext()) {
      throw new NoSuchElementException();
    }
    StringBuilder sb = new StringBuilder();
    while(true) {
      int b = readByte();
      if (!isPrintableChar(b)) {
        break;
      }
      sb.appendCodePoint(b);
    }
    return sb.toString();
  }
  
  public String nextLine() {
    if (!hasNext()) {
      throw new NoSuchElementException();
    }
    StringBuilder sb = new StringBuilder();
    while(true) {
      int b = readByte();
      if (!isPrintableChar(b) && b != 0x20) {
        break;
      }
      sb.appendCodePoint(b);
    }
    return sb.toString();
  }
  
  public char nextChar() {
    return next().charAt(0);
  }
  
  public int nextInt() {
    if (!hasNext()) {
      throw new NoSuchElementException();
    }
    
    int n = 0;
    boolean minus = false;
    
    {
      int b = readByte();
      if (b == '-') {
        minus = true;
      } else if ('0' <= b && b <= '9') {
        n = b - '0';
      } else {
        throw new NumberFormatException();
      }
    }
    
    while(true){
      int b = readByte();
      if ('0' <= b && b <= '9') {
        n *= 10;
        n += b - '0';
      } else if (b == -1 || !isPrintableChar(b)) {
        return minus ? -n : n;
      } else {
        throw new NumberFormatException();
      }
    }
  }
  
  public long nextLong() {
    if (!hasNext()) {
      throw new NoSuchElementException();
    }
    
    long n = 0;
    boolean minus = false;
    
    {
      int b = readByte();
      if (b == '-') {
        minus = true;
      } else if ('0' <= b && b <= '9') {
        n = b - '0';
      } else {
        throw new NumberFormatException();
      }
    }
    
    while(true){
      int b = readByte();
      if ('0' <= b && b <= '9') {
        n *= 10;
        n += b - '0';
      } else if (b == -1 || !isPrintableChar(b)) {
        return minus ? -n : n;
      } else {
        throw new NumberFormatException();
      }
    }
  }
  
  public double nextDouble() {
    return Double.parseDouble(next());
  }
  
  public String[] next(int n) {
    String[] array = new String[n];
    for (int i = 0; i < n; i++) {
      array[i] = next();
    }
    return array;
  }
  
  public String[] nextLine(int n) {
    String[] array = new String[n];
    for (int i = 0; i < n; i++) {
      array[i] = nextLine();
    }
    return array;
  }
  
  public char[] nextChar(int n) {
    char[] array = new char[n];
    for (int i = 0; i < n; i++) {
      array[i] = nextChar();
    }
    return array;
  }
  
  public int[] nextInt(int n) {
    int[] array = new int[n];
    for (int i = 0; i < n; i++) {
      array[i] = nextInt();
    }
    return array;
  }
  
  public long[] nextLong(int n) {
    long[] array = new long[n];
    for (int i = 0; i < n; i++) {
      array[i] = nextLong();
    }
    return array;
  }
  
  public double[] nextDouble(int n) {
    double[] array = new double[n];
    for (int i = 0; i < n; i++) {
      array[i] = nextDouble();
    }
    return array;
  }
  
  public char[] nextCharArray() {
    return next().toCharArray();
  }
  
  public String[][] next(int n, int m) {
    String[][] matrix = new String[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        matrix[i][j] = next();
      }
    }
    return matrix;
  }
  
  public int[][] nextInt(int n, int m) {
    int[][] matrix = new int[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        matrix[i][j] = nextInt();
      }
    }
    return matrix;
  }
  
  public char[][] nextChar(int n, int m) {
    char[][] matrix = new char[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        matrix[i][j] = nextChar();
      }
    }
    return matrix;
  }
  
  public long[][] nextLong(int n, int m) {
    long[][] matrix = new long[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        matrix[i][j] = nextLong();
      }
    }
    return matrix;
  }
  
  public double[][] nextDouble(int n, int m) {
    double[][] matrix = new double[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        matrix[i][j] = nextDouble();
      }
    }
    return matrix;
  }
  
  public char[][] nextCharArray(int n) {
    char[][] matrix = new char[n][];
    for (int i = 0; i < n; i++) {
      matrix[i] = next().toCharArray();
    }
    return matrix;
  }
}

class MyAssert {
  public static void myAssert(boolean flag, String message) {
    if (!flag) {
      throw new RuntimeException(message);
    }
  }
 
  public static void myAssert(boolean flag) {
    myAssert(flag, "");
  }
}