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: Time Limit Exceeded)

using System;
using System.Collections.Generic;

namespace msol
{
    class Program
    {
        static int[] ReadNumList()
        {
            var line = Console.ReadLine().Split(' ');
            if (line.Length == 1 && string.IsNullOrEmpty(line[0]))
            {
                return new int[0];
            }
            var numList = new int[line.Length];
            for (var i = 0; i < line.Length; ++i)
            {
                numList[i] = int.Parse(line[i]);
            }
            return numList;
        }
        static void Main()
        {
            var n = int.Parse(Console.ReadLine());
            var vils = new int[n][];
            for (var i = 0; i < n; ++i)
            {
                vils[i] = ReadNumList();
            }
            // 候補座標リスト
            var xSet = new HashSet<int>();
            var ySet = new HashSet<int>();
            for (var i = 0; i < vils.Length; ++i)
            {
                if (vils[i][0] != 0)
                {
                    xSet.Add(vils[i][0]);
                }
                if (vils[i][1] != 0)
                {
                    ySet.Add(vils[i][1]);
                }
            }
            var lineDefList = new List<int>();
            lineDefList.AddRange(xSet);
            lineDefList.AddRange(ySet);
            var lineDefs = lineDefList.ToArray();
            var lineDefXs = new List<int>(xSet).ToArray();
            var lineDefYs = new List<int>(ySet).ToArray();
            // 前処理
            var sForAllX = new long[(int)Math.Pow(2, xSet.Count)][];
            for (var i = 0; i < sForAllX.Length; ++i)
            {
                sForAllX[i] = CalcSsForX(i, lineDefXs, vils);
            }
            var sForAllY = new long[(int)Math.Pow(2, ySet.Count)][];
            for (var i = 0; i < sForAllY.Length; ++i)
            {
                sForAllY[i] = CalcSsForY(i, lineDefYs, vils);
            }
            // 計算
            for (var k = 0; k <= n; ++k)
            {
                Console.WriteLine(MinS(k, sForAllX, sForAllY, lineDefXs.Length, lineDefYs.Length));
            }
        }
        static long[] CalcSsForX(int xBit, int[] lineDefXs, int[][] vils)
        {
            var ss = new long[vils.Length];
            for (var v = 0; v < vils.Length; ++v)
            {
                var dist = Math.Min(Math.Abs(vils[v][0]), Math.Abs(vils[v][1]));
                for (var l = 0; l < lineDefXs.Length; ++l)
                {
                    if (((xBit >> l) & 1) == 1)
                    {
                        dist = Math.Min(dist, Math.Abs(vils[v][0] - lineDefXs[l]));
                    }
                }
                ss[v] = (long)dist * vils[v][2];
            }
            return ss;
        }
        static long[] CalcSsForY(int yBit, int[] lineDefYs, int[][] vils)
        {
            var ss = new long[vils.Length];
            for (var v = 0; v < vils.Length; ++v)
            {
                var dist = Math.Min(Math.Abs(vils[v][0]), Math.Abs(vils[v][1]));
                for (var l = 0; l < lineDefYs.Length; ++l)
                {
                    if (((yBit >> l) & 1) == 1)
                    {
                        dist = Math.Min(dist, Math.Abs(vils[v][1] - lineDefYs[l]));
                    }
                }
                ss[v] = (long)dist * vils[v][2];
            }
            return ss;
        }
        static long MinS(int k, long[][] sForAllX, long[][] sForAllY, int xLineCount, int yLineCount)
        {
            if (xLineCount <= k || yLineCount <= k)
            {
                return 0;
            }
            var min = long.MaxValue;
            foreach (var lines in Subset(xLineCount + yLineCount, k))
            {
                var xBit = 0;
                var yBit = 0;
                for (var l = 0; l < lines.Length; ++l)
                {
                    if (lines[l] < xLineCount)
                    {
                        xBit += (int)Math.Pow(2, lines[l]);
                    }
                    else
                    {
                        yBit += (int)Math.Pow(2, lines[l] - xLineCount);
                    }
                }
                min = Math.Min(min, CalcS(xBit, yBit, sForAllX, sForAllY));
            }
            return min;
        }
        static long CalcS(int xBit, int yBit, long[][] sForAllX, long[][] sForAllY)
        {
            var s = 0L;
            for (var v = 0; v < sForAllX[0].Length; ++v)
            {
                s += Math.Min(sForAllX[xBit][v], sForAllY[yBit][v]);
            }
            return s;
        }
        // nCrの組み合わせリストを出力
        static int[][] Subset(int n, int k)
        {
            var result = new List<int[]>();
            Subset_(result, n, k, new int[k], 0);
            return result.ToArray();
        }
        static void Subset_(List<int[]> result, int n, int k, int[] current, int d)
        {
            if (d == k)
            {
                result.Add((int[])current.Clone());
                return;
            }
            var start = d == 0 ? 0 : current[d - 1] + 1;
            for (var i = start; i <= n - k + d; ++i)
            {
                current[d] = i;
                Subset_(result, n, k, current, d + 1);
            }
        }
    }
}