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# (Mono-mcs 6.8.0.105) (judge verdict: Wrong Answer)

using System;
using System.Linq;
using System.Collections.Generic;
using System.IO;
using static System.Console;

namespace atcorder2
{
    class Program
    {
        public static long Read()
        {
            return long.Parse(Console.ReadLine());
        }

        public static long[] Reads()
        {
            return Console.ReadLine().Split().Select(long.Parse).ToArray();
        }


        public static long gcd(long x, long y)
        {
            if (y < x) gcd(y, x);
            while (y != 0)
            {
                var z = x % y;
                x = y;
                y = z;
            }
            return x;
        }
        public static long keta(long x)
        {
            var z = 1;
            while (x > 0)
            {
                x /= 10;
                z++;
            }
            return z;
        }
        static void Main(string[] args)
        {
            var sw = new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
            Console.SetOut(sw);

            var n = Read();
            long[] x = new long[n], y = new long[n], p = new long[n];
            long ans = 0;
            var d = new Dictionary<long, long>();
            for (int i = 0; i < n; i++)
            {
                var xyp = Reads();
                x[i] = xyp[0];
                y[i] = xyp[1];
                p[i] = xyp[2];
                d.Add(i, Math.Min(Math.Abs(x[i]), Math.Abs(y[i])) * p[i]);
                ans += d[i];
            }

            WriteLine(ans);


            for (int k = 0; k < n; k++)
            {

                ans = 0;
                long ma = 0;
                for (int i = 0; i < n; i++)
                {
                    for (int j = 0; j < n; j++)
                    {
                        var a = Math.Min(Math.Abs(x[j] - x[i]), Math.Abs(y[j])) * p[j];
                        var b = Math.Min(a, d[i]);
                        ma += b;
                    }
                    if (ma < ans)
                    {
                        for (int j = 0; j < n; j++)
                        {
                            var a = Math.Min(Math.Abs(x[j] - x[i]), Math.Abs(y[j])) * p[j];
                            d[i] = Math.Min(a, d[i]);
                            
                        }
                        ans = ma;
                    }
                    
                    
                }
                for (int i = 0; i < n; i++)
                {
                    for (int j = 0; j < n; j++)
                    {
                        var a = Math.Min(Math.Abs(x[j]), Math.Abs(y[j] - y[i])) * p[j];
                        var b = Math.Min(a, d[i]);
                        ma += b;
                    }
                    if (ma < ans)
                    {
                        for (int j = 0; j < n; j++)
                        {
                            var a = Math.Min(Math.Abs(x[j] - x[i]), Math.Abs(y[j])) * p[j];
                            d[i] = Math.Min(a, d[i]);

                        }
                        ans = ma;
                    }
                 
                }
                WriteLine(ans);


            }
















            /*


                for (int i = 0; i < n; i++)
            {
                lmx[i] = 1000000;
            }
            for (int i = 0; i < n; i++)
            {
                lmy[i] = 1000000;
            }


           
            for (int k = 0; k < n; k++)
            {
                long memo = 0;
              
                for (int i = -10000; !lmx.Contains(i) && i <= 10000; i++)
                {
                    for (int j = 0; j < n; j++)
                    {
                        memo += Math.Min(Math.Abs(x[j] - i),Math.Abs(y[j]))* p[j];
                    }
                    if (memo < ans)
                    {
                        ans = memo;
                        lmx[k] = i;
                    }
                }
                for (int i = -10000; !lmy.Contains(i) && i <= 10000; i++)
                {
                    for (int j = 0;j < n; j++)
                    {
                        memo += Math.Min(Math.Abs(y[j] - i), Math.Abs(x[j])) * p[j];
                    }
                    if (memo < ans)
                    {
                        ans = memo;
                        lmy[k] = i;
                    }
                }
                WriteLine(ans);

            }
            */
            Console.Out.Flush();
        }
    }
}