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.Text;
using System.Diagnostics;
using System.Numerics;
using System.Collections.Generic;

class Myon

{
    public Myon() { }
    public static int Main()
    {
        cin = new Scanner();
        new Myon().calc();
        return 0;
    }
    static Scanner cin;


    void calc()
    {
        int N = cin.nextInt();
        int[] X = new int[N];
        int[] Y = new int[N];
        long[] P = new long[N];

        for (int i = 0; i < N; i++)
        {
            X[i] = cin.nextInt();
            Y[i] = cin.nextInt();
            P[i] = cin.nextInt();
        }

        long[,] XDist = new long[1 << N, N];
        long[,] YDist = new long[1 << N, N];

        long MAX = long.MaxValue / 2;

        for (int i = 0; i < (1 << N); i++)
        {
            for (int j = 0; j < N; j++)
            {
                XDist[i, j] = Math.Abs(X[j]) * P[j];
                YDist[i, j] = Math.Abs(Y[j]) * P[j];
                for (int k = 0; k < N; k++)
                {
                    if(((i >> k) & 1) == 1)
                    {
                        XDist[i, j] = Math.Min(XDist[i, j], Math.Abs(X[j] - X[k]) * P[j]) ;
                        YDist[i, j] = Math.Min(YDist[i, j], Math.Abs(Y[j] - Y[k]) * P[j]) ;
                    }
                }
            }
        }

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

        for (int i = 0; i < (1 << N); i++)
        {
            int bi = bitCount(i);
            for (int j = i; j >= 0; j--)
            {
                j &= i;
                int k = i - j;
                long sum = 0;
                for (int p = 0; p < N; p++)
                {
                    sum += Math.Min(XDist[k, p], YDist[j, p]);
                }
                ans[bi] = Math.Min(ans[bi], sum);
            }
        }

        for (int i = 0; i <= N; i++)
        {
            Console.WriteLine(ans[i]);
        }



    }

    int bitCount(long x)
    {
        x = (x & 0x5555555555555555) + (x >> 1 & 0x5555555555555555);
        x = (x & 0x3333333333333333) + (x >> 2 & 0x3333333333333333);
        x = (x & 0x0f0f0f0f0f0f0f0f) + (x >> 4 & 0x0f0f0f0f0f0f0f0f);
        x = (x & 0x00ff00ff00ff00ff) + (x >> 8 & 0x00ff00ff00ff00ff);
        x = (x & 0x0000ffff0000ffff) + (x >> 16 & 0x0000ffff0000ffff);
        return (int)((x & 0x00000000ffffffff) + (x >> 32 & 0x00000000ffffffff));
    }





}


class Scanner
{
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner()
    {
        s = new string[0];
        i = 0;
    }

    public string next()
    {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        if (s.Length == 0) return next();
        i = 0;
        return s[i++];
    }

    public int nextInt()
    {
        return int.Parse(next());
    }
    public int[] ArrayInt(int N, int add = 0)
    {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextInt() + add;
        }
        return Array;
    }

    public long nextLong()
    {
        return long.Parse(next());
    }

    public long[] ArrayLong(int N, long add = 0)
    {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextLong() + add;
        }
        return Array;
    }

    public double nextDouble()
    {
        return double.Parse(next());
    }


    public double[] ArrayDouble(int N, double add = 0)
    {
        double[] Array = new double[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextDouble() + add;
        }
        return Array;
    }
}