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# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using CompLib.Util;

public class Program
{
    private int N;
    private long[] X, Y, P;

    private long[][] Cost;

    private long[] Ans;

    private long Sum;

    public void Solve()
    {
        var sc = new Scanner();
        N = sc.NextInt();
        X = new long[N];
        Y = new long[N];
        P = new long[N];
        for (int i = 0; i < N; i++)
        {
            X[i] = sc.NextInt();
            Y[i] = sc.NextInt();
            P[i] = sc.NextInt();
        }

        Sum = 0;
        Cost = new long[N + 1][];
        for (int i = 0; i <= N; i++)
        {
            Cost[i] = new long[N];
        }

        for (int i = 0; i < N; i++)
        {
            Cost[0][i] = Math.Min(Math.Abs(X[i]), Math.Abs(Y[i])) * P[i];
            Sum += Cost[0][i];
        }

        Ans = new long[N + 1];
        for (int i = 0; i <= N; i++)
        {
            Ans[i] = long.MaxValue;
        }

        Go(0, 0);

        Console.WriteLine(string.Join("\n", Ans));
    }

    void Go(int n, int k)
    {
        Ans[k] = Math.Min(Ans[k], Sum);
        if (k < N && n < N)
        {
            // 何もしない
            Array.Copy(Cost[n], Cost[n + 1], N);
            Go(n + 1, k);
            long diff = 0;
            // 縦
            for (int i = 0; i < N; i++)
            {
                Cost[n + 1][i] = Math.Min(Cost[n][i], Math.Abs(X[n] - X[i]) * P[i]);
                diff += Cost[n + 1][i] - Cost[n][i];
            }

            Sum += diff;
            Go(n + 1, k + 1);
            Sum -= diff;
            diff = 0;
            // 横
            for (int i = 0; i < N; i++)
            {
                Cost[n + 1][i] = Math.Min(Cost[n][i], Math.Abs(Y[n] - Y[i]) * P[i]);
                diff += Cost[n + 1][i] - Cost[n][i];
            }

            Sum += diff;
            Go(n + 1, k + 1);
            Sum -= diff;
        }
    }

    public static void Main(string[] args) => new Program().Solve();
}

namespace CompLib.Util
{
    using System;
    using System.Linq;

    class Scanner
    {
        private string[] _line;
        private int _index;
        private const char Separator = ' ';

        public Scanner()
        {
            _line = new string[0];
            _index = 0;
        }

        public string Next()
        {
            while (_index >= _line.Length)
            {
                _line = Console.ReadLine().Split(Separator);
                _index = 0;
            }

            return _line[_index++];
        }

        public int NextInt() => int.Parse(Next());
        public long NextLong() => long.Parse(Next());
        public double NextDouble() => double.Parse(Next());
        public decimal NextDecimal() => decimal.Parse(Next());
        public char NextChar() => Next()[0];
        public char[] NextCharArray() => Next().ToCharArray();

        public string[] Array()
        {
            _line = Console.ReadLine().Split(Separator);
            _index = _line.Length;
            return _line;
        }

        public int[] IntArray() => Array().Select(int.Parse).ToArray();
        public long[] LongArray() => Array().Select(long.Parse).ToArray();
        public double[] DoubleArray() => Array().Select(double.Parse).ToArray();
        public decimal[] DecimalArray() => Array().Select(decimal.Parse).ToArray();
    }
}