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;
using System.Linq;
using static Scanner;

class Program
{
    static void Main() => new Program().Solve();

    public void Solve()
    {
        int n = ReadInt();
        long[] x = new long[n];
        long[] y = new long[n];
        long[] d = new long[n];
        long[] p = new long[n];
        for (int i = 0; i < n; i++)
        {
            x[i] = ReadInt();
            y[i] = ReadInt();
            d[i] = Math.Min(Math.Abs(x[i]), Math.Abs(y[i]));
            p[i] = ReadInt();
        }
        Console.WriteLine(GetSum(n, p, d));
        bool check = false;
        for (int k = 1; k <= n; k++)
        {
            if (check)
            {
                Console.WriteLine(0);
                continue;
            }
            long min = long.MaxValue;
            for (long i = (1 << k) - 1; i <= (1 << (n * 2)); i = MyMath.NextCombination(i))
            {
                long[] dd = new long[n];
                Array.Copy(d, dd, n);
                for (int j = 0; j < n; j++)
                {
                    if ((i & (1 << j)) != 0)
                    {
                        for (int l = 0; l < n; l++)
                        {
                            dd[l] = Math.Min(dd[l], Math.Abs(x[l] - x[j]));
                        }
                    }
                }
                for (int j = n; j < n * 2; j++)
                {
                    if ((i & (1 << j)) != 0)
                    {
                        for (int l = 0; l < n; l++)
                        {
                            dd[l] = Math.Min(dd[l], Math.Abs(y[l] - y[j - n]));
                        }
                    }
                }
                min = Math.Min(min, GetSum(n, p, dd));
            }
            Console.WriteLine(min);
            if (min == 0) check = true;
        }
    }

    long GetSum(int n, long[] p, long[] d)
    {
        long sum = 0;
        for (int i = 0; i < n; i++)
        {
            sum += p[i] * d[i];
        }
        return sum;
    }
}

static class Scanner
{
    static Queue<string> que = new Queue<string>();
    public static string Read() { if (!que.Any()) foreach (var s in Console.ReadLine().Split()) que.Enqueue(s); return que.Dequeue(); }
    public static int ReadInt() => int.Parse(Read());
    public static IEnumerable<string> Read(long n) { for (long i = 0; i < n; i++) yield return Read(); }
    public static IEnumerable<int> ReadInt(long n) => Read(n).Select(int.Parse);
}

static class MyMath
{
    public static long NextCombination(long bit)
    {
        long x = bit & -bit;
        long y = x + bit;
        return y | (((bit & ~y) / x) >> 1);
    }
}