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 System.Collections;
using System.Collections.Generic;
using System.Diagnostics;
using System.Globalization;
using System.IO;
using System.Text;
using System.Linq;

using E = System.Linq.Enumerable;
using System.Runtime.InteropServices.ComTypes;
using System.Data.SqlTypes;
using System.Numerics;

internal partial class Solver {
    public void Run() {
        var n = ni();
        var x = new int[n];
        var y = new int[n];
        var p = new int[n];

        for (int i = 0; i < n; i++) {
            x[i] = ni();
            y[i] = ni();
            p[i] = ni();
        }

        var ans = E.Repeat(long.MaxValue, n + 1).ToArray();

        long[] cost = new long[n];
        for (int i = 0; i < n; i++) {
            cost[i] = Math.Min(Math.Abs(x[i]), Math.Abs(y[i]));
        }
        void dfs(int pos, int k, long[] currentCost) {
            if (pos == n) {
                //check
                long sub = 0;
                for (int i = 0; i < n; i++) {
                    sub += currentCost[i] * p[i];
                }
                ans[k] = Math.Min(ans[k], sub);
                return;
            }
            dfs(pos + 1, k, currentCost); // 0
            var next = currentCost.ToArray();
            for (int i = 0; i < n; i++) {
                next[i] = Math.Min(next[i], Math.Abs(x[pos] - x[i]));
            }
            dfs(pos + 1, k + 1, next);
            next = currentCost.ToArray();
            for (int i = 0; i < n; i++) {
                next[i] = Math.Min(next[i], Math.Abs(y[pos] - y[i]));
            }
            dfs(pos + 1, k + 1, next);
        }

        dfs(0, 0, cost);

        cout.WriteLine(ans.JoinToString("\n"));
    }

    public static ulong NextPermutationBit(ulong x) {
        if (x == 0) {
            return 0;
        }

        ulong rightOne = (ulong)((long)x & -(long)x);
        ulong nextHigherOneBit = x + rightOne;
        return nextHigherOneBit | ((x ^ nextHigherOneBit) / rightOne) >> 2;
    }

    private static IEnumerable<ulong> EnumerateFlags(int n, int k) {
        ulong x = (1UL << k) - 1;
        ulong max = 1UL << n;
        while (x < max) {
            yield return x;
            x = NextPermutationBit(x);
            if (x == 0) {
                yield break;
            }
        }
    }
}

// PREWRITEN CODE BEGINS FROM HERE

static public class StringExtensions {
    static public string JoinToString<T>(this IEnumerable<T> source, string separator = " ") {
        return string.Join(separator, source);
    }
}

internal partial class Solver : Scanner {
    public static void Main() {
#if LOCAL
        byte[] inputBuffer = new byte[1000000];
        var inputStream = Console.OpenStandardInput(inputBuffer.Length);
        using (var reader = new StreamReader(inputStream, Console.InputEncoding, false, inputBuffer.Length)) {
            Console.SetIn(reader);
            new Solver(Console.In, Console.Out).Run();
        }
#else
        Console.SetOut(new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false });
        new Solver(Console.In, Console.Out).Run();
        Console.Out.Flush();
#endif
    }

#pragma warning disable IDE0052
    private readonly TextReader cin;
    private readonly TextWriter cout;
    private readonly TextWriter cerr;
#pragma warning restore IDE0052

    public Solver(TextReader reader, TextWriter writer)
        : base(reader) {
        cin = reader;
        cout = writer;
        cerr = Console.Error;
    }

    public Solver(string input, TextWriter writer)
        : this(new StringReader(input), writer) {
    }

#pragma warning disable IDE1006
#pragma warning disable IDE0051
    private int ni() { return NextInt(); }
    private int[] ni(int n) { return NextIntArray(n); }
    private long nl() { return NextLong(); }
    private long[] nl(int n) { return NextLongArray(n); }
    private double nd() { return NextDouble(); }
    private double[] nd(int n) { return NextDoubleArray(n); }
    private string ns() { return Next(); }
    private string[] ns(int n) { return NextArray(n); }
#pragma warning restore IDE1006
#pragma warning restore IDE0051
}

#if DEBUG
internal static class LinqPadExtension {
    public static string TextDump<T>(this T obj) {
        if (obj is IEnumerable) return (obj as IEnumerable).Cast<object>().JoinToString().Dump();
        else return obj.ToString().Dump();
    }
    public static T Dump<T>(this T obj) {
        return LINQPad.Extensions.Dump(obj);
    }
}
#endif

public class Scanner {
    private readonly TextReader Reader;
    private readonly CultureInfo ci = CultureInfo.InvariantCulture;

    private readonly char[] buffer = new char[2 * 1024];
    private int cursor = 0, length = 0;
    private string Token;
    private readonly StringBuilder sb = new StringBuilder(1024);

    public Scanner()
        : this(Console.In) {
    }

    public Scanner(TextReader reader) {
        Reader = reader;
    }

    public int NextInt() { return checked((int)NextLong()); }
    public long NextLong() {
        var s = Next();
        long r = 0;
        int i = 0;
        bool negative = false;
        if (s[i] == '-') {
            negative = true;
            i++;
        }
        for (; i < s.Length; i++) {
            r = r * 10 + (s[i] - '0');
#if DEBUG
            if (!char.IsDigit(s[i])) throw new FormatException();
#endif
        }
        return negative ? -r : r;
    }
    public double NextDouble() { return double.Parse(Next(), ci); }
    public string[] NextArray(int size) {
        string[] array = new string[size];
        for (int i = 0; i < size; i++) {
            array[i] = Next();
        }

        return array;
    }
    public int[] NextIntArray(int size) {
        int[] array = new int[size];
        for (int i = 0; i < size; i++) {
            array[i] = NextInt();
        }

        return array;
    }

    public long[] NextLongArray(int size) {
        long[] array = new long[size];
        for (int i = 0; i < size; i++) {
            array[i] = NextLong();
        }

        return array;
    }

    public double[] NextDoubleArray(int size) {
        double[] array = new double[size];
        for (int i = 0; i < size; i++) {
            array[i] = NextDouble();
        }

        return array;
    }

    public string Next() {
        if (Token == null) {
            if (!StockToken()) {
                throw new Exception();
            }
        }
        var token = Token;
        Token = null;
        return token;
    }

    public bool HasNext() {
        if (Token != null) {
            return true;
        }

        return StockToken();
    }

    private bool StockToken() {
        while (true) {
            sb.Clear();
            while (true) {
                if (cursor >= length) {
                    cursor = 0;
                    if ((length = Reader.Read(buffer, 0, buffer.Length)) <= 0) {
                        break;
                    }
                }
                var c = buffer[cursor++];
                if (33 <= c && c <= 126) {
                    sb.Append(c);
                } else {
                    if (sb.Length > 0) break;
                }
            }

            if (sb.Length > 0) {
                Token = sb.ToString();
                return true;
            }

            return false;
        }
    }
}