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.Linq;
using System.Collections.Generic;
using System.Numerics;
using static System.Math;
using static System.Console;
using static System.Linq.Enumerable;
using static System.Numerics.BitOperations;


namespace msolutions2020
{
    class SolverE
    {
        Dictionary<ulong,long>[] costX;
        Dictionary<ulong,long>[] costY;
        static int N;
        static Village[] V;
        static void Main()
        {
            new SolverE().Solve();
        }

        public void Solve()
        {
            N = Get.Int();
            V = new Village[N];
            
            foreach(int i in Range(0,N))
            {
                (int x,int y,int p)vil = Get.Tuple<int,int,int>();
                V[i].x = vil.x;
                V[i].y = vil.y;
                V[i].p = vil.p;
            }


            // 事前辞書の構築 X

            long pow2_N = IntPow(2,(uint)N);
            costX = new Dictionary<ulong,long>[V.Length];
            for(int i=0; i<V.Length; i++)
            {
                costX[i] = new Dictionary<ulong,long>(60000);
                
            }
            for(int vil=0 ; vil < V.Length ; vil++)
            {
                for(int ptnX = 0; ptnX < pow2_N; ptnX++)
                {
                    ulong ptnKey = 0;
                    for(int i=0; i<N; i++)
                    {
                        ptnKey = (ptnX >> i & 1)==1 ? (ptnKey | ((ulong)1 << 2*i)) : ptnKey;
                    }
                    costX[vil].Add(ptnKey, Dim1Pattern.DistX(ptnX, V[vil], V) * V[vil].p);
                }
            }

            // 事前辞書の構築 Y

            costY = new Dictionary<ulong,long>[V.Length];
            for(int i=0; i<V.Length; i++)
            {
                costY[i] = new Dictionary<ulong,long>(60000);
            }
            for(int vil=0 ; vil < V.Length ; vil++)
            {
                for(int ptnY = 0; ptnY < pow2_N; ptnY++)
                {
                    ulong ptnKey = 0;
                    for(int i=0; i<N; i++)
                    {
                        ptnKey = (ptnY >> i & 1)==1 ? (ptnKey | ((ulong)1 << 2*i+1)) : ptnKey;
                    }
                    costY[vil].Add(ptnKey, Dim1Pattern.DistY(ptnY, V[vil], V) * V[vil].p);
                }
            }

            // 事前辞書の構築 ここまで
            // メインループ開始

            Pattern pattern = new Pattern(N);
            long[] ans = Repeat(long.MaxValue, N+1).ToArray();
            do
            {
                int t = BitOperations.PopCount(pattern.pattern);
                ans[t] = Min(ans[t], cost(pattern.pattern));

            }while(pattern.Next() != null);

            foreach(var a in ans)
            {
                WriteLine(a);
            }
                      
        }

        const long xmask = 0b_01010101010101010101010101010101;
        const long ymask = 0b_10101010101010101010101010101010;
        long cost(ulong ptn)
        {

            int vLength = V.Length;
            long cost = 0;
            
            for (int v = 0; v < vLength; v++)
            {
                if( ((ptn >> (v *2)) & 0b11) > 0 ) continue; // 集落上に道があるならゼロコスト
                cost += Min(costX[v][ptn & xmask], costY[v][ptn & ymask]);
            }
            return cost;
        }

        struct Village
        {
            public int x;
            public int y;
            public int p;
        }

        class Pattern
        {
            public ulong pattern;
            int N;

            public Pattern(int _n)
            {
                pattern = 0;
                N = _n;
            }

            public Pattern Next()
            {
                pattern++;
                
                for (int i = 0; i < N ; i++)
                {

                    if(this[i] > 2) // 繰り上がる
                    {
                        this[i] = 0;
                        this[i+1]++;
                        
                    }else break; // 繰り上がらない

                    if(this[N] == 1) {return null;} // 終了
                }
                return this;
            }
         

            public ulong this[int i]
            {
                get { return (pattern >> (i*2)) & 0b11;}
                set { pattern = (pattern & (ulong)~(0b11 << (i*2))) | value << (i*2);}
            }
        }
  

        class Dim1Pattern
        {

            public static long DistX(int pattern, Village v, Village [] Vils)
            {
                int dist = Abs(v.x);
                for(int i=0; i<Vils.Length; i++)
                {
                    if(((pattern >> i) & 1) == 1)
                    {
                        dist = Min(dist, Abs(Vils[i].x - v.x));
                    }
                }
                return dist;
            }

            public static long DistY(int pattern, Village v, Village [] Vils)
            {
                int dist = Abs(v.y);
                for(int i=0; i<Vils.Length; i++)
                {
                    if(((pattern >> i) & 1) == 1)
                    {
                        dist = Min(dist, Abs(Vils[i].y - v.y));
                    }
                }
                return dist;
            }

        }

        static long IntPow(int x, uint pow)
        {
            long ret = 1;
            while ( pow != 0 )
            {
                if ( (pow & 1) == 1 )
                    ret *= x;
                x *= x;
                pow >>= 1;
            }
            return ret;
        }

        static class Get
        {
            public static string Str() => ReadLine().Trim();
            public static int Int() => int.Parse(Str());
            public static long Long() => long.Parse(Str());
            public static double Double() => double.Parse(Str());
            public static string[] Strs() => Str().Split(' ', StringSplitOptions.RemoveEmptyEntries);
            public static int[] Ints() => Strs().Select(int.Parse).ToArray();
            public static long[] Longs() => Strs().Select(long.Parse).ToArray();
            public static double[] Doubles() => Strs().Select(double.Parse).ToArray();
            static bool TypeEq<T,U>() => typeof(T).Equals(typeof(U));
            static T TypeConv<T,U>(U u) => (T)Convert.ChangeType(u, typeof(T));
            static T TypeConv<T>(string s) => TypeEq<T, int>() ?   TypeConv<T, int>(int.Parse(s))
                                        : TypeEq<T, long>() ?       TypeConv<T, long>(long.Parse(s))
                                        : TypeEq<T, double>() ?     TypeConv<T, double>(long.Parse(s))
                                        : TypeConv<T, string>(s);
            public static (T,U) Tuple<T,U>() {string[] strs = Strs(); T t = TypeConv<T>(strs[0]); U u = TypeConv<U>(strs[1]); return(t,u);}
            public static (T,U,V) Tuple<T,U,V>() {string[] strs = Strs(); T t = TypeConv<T>(strs[0]); U u = TypeConv<U>(strs[1]); V v = TypeConv<V>(strs[2]); return(t,u,v);}
        }
    }
}