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: Wrong Answer)

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


namespace msolutions2020
{
    class SolverE
    {
        static Dictionary<int,int>[] distDictX;
        static Dictionary<int,int>[] distDictY;
        static void Main()
        {
            Solve();
        }

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

            distDictX = new Dictionary<int, int>[V.Length];
            for(int i=0; i<V.Length; i++){ distDictX[i] = new Dictionary<int,int>();}
            for(int vil=0 ; vil < V.Length ; vil++)
            {
                for(int ptnX = 0; ptnX < IntPow(2,(uint)N); ptnX++)
                {
                    Dim1Pattern ptn = new Dim1Pattern(ptnX, N);
                    distDictX[vil].Add(ptnX, ptn.DistX(V[vil], V));
                }
            }

            distDictY = new Dictionary<int, int>[V.Length];
            for(int i=0; i<V.Length; i++){ distDictY[i] = new Dictionary<int,int>();}
            for(int vil=0 ; vil < V.Length ; vil++)
            {
                for(int ptnY = 0; ptnY < IntPow(2,(uint)N); ptnY++)
                {
                    Dim1Pattern ptn = new Dim1Pattern(ptnY, N);
                    distDictY[vil].Add(ptnY, ptn.DistY(V[vil], V));
                }
            }

            

            // var distDictX = new Dictionary<Pattern, int>[V.Length];
            // for (int vil=0; vil < V.Length; vil++)
            // {
            //     for (int ptnX = 0; ptnX < IntPow(2,(uint)N); ptnX++)
            //     {
            //         Pattern ptn = new Pattern(N);
            //         for (int i = 0; i < N; i++)
            //         {
            //             ptn.pattern[i] = (ptnX >> i) & 1;
            //         }
                    
            //         distDictX[vil].Add(ptn, cost(ptn, V, N));
            //     }
            // }
            // for (int ptnX = 0; ptnX < IntPow(2,(uint)N); ptnX++)
            // {
            //     Pattern ptn = new Pattern(N);
            //     for (int i = 0; i < N; i++)
            //     {
            //         ptn.pattern[i] = (ptnX >> i) & 1;
            //     }
            //     int minD
            //     distDictX.Add(ptn, cost(ptn, V, N));
            // }


            
            for (int k = 0; k <= N; k++)
            {
                long minCost = long.MaxValue;
                var ptn = new Pattern(N);
                while(ptn.next() != null)
                {
                    minCost = Min(minCost, cost(ptn, V, k));
                }
                WriteLine(minCost);
            }          
        }

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

        static long cost(Pattern ptn, (int x, int y, int p)[] V, int k)
        {
            if(ptn.trainNum > k ) return long.MaxValue;
            int pLength = ptn.pattern.Length -1;
            int vLength = V.Length;
            long cost = 0;
            for (int v = 0; v < vLength; v++)
            {
                // long minDist = Min(Abs(V[v].x), Abs(V[v].y));
                // for (int p = 0; p < pLength; p++)
                // {
                //     var dist = int.MaxValue;
                //     if(ptn[p] == 1) dist = Abs(V[v].x - V[p].x);
                //     else if(ptn[p] == 2) dist = Abs(V[v].y - V[p].y);
                //     minDist = Min(minDist, dist);
                // }
                int ptx = 0;
                for(int i=0 ; i<ptn.pattern.Length-1; i++)
                {
                    ptx += ptn[i] == 1 ?(1<<i) : 0;
                }
                Dim1Pattern ptnx = new Dim1Pattern(ptx, V.Length);
                
                int pty = 0;
                for(int i=0 ; i<ptn.pattern.Length-1; i++)
                {
                    pty += ptn[i] == 2 ? (1<<i) : 0;
                }
                Dim1Pattern ptny = new Dim1Pattern(pty, V.Length);

                // WriteLine(ptnx.pattern);
                // foreach(var d in distDictX[0])
                // {
                //     WriteLine($"{d.Key} -> {d.Value}");
                // }
                cost +=Min(distDictX[v][ptnx.pattern], distDictY[v][ptny.pattern]) * V[v].p;
            }
            // WriteLine(string.Join(", ", ptn.pattern) + $"  train:{ptn.trainNum}   cost:{cost}");
            return cost;
        }


        class Pattern : IEqualityComparer<Pattern>
        {
            
            public int[] pattern;
            public int trainNum = 0;
            public Pattern(int N)
            {
                pattern = Repeat(0,N+1).ToArray();
                pattern[0] = -1;
            }

            public int[] next()
            {
                pattern[0]++;
                if(pattern[0] == 1) trainNum++;
                for (int i = 0; i < pattern.Length ; i++)
                {
                    if(pattern[i] <= 2) // 繰り上がらない
                    {
                        break;
                    }
                    if(pattern[i] > 2) // 繰り上がる
                    {
                        pattern[i] = 0; trainNum--;
                        pattern[i+1]++;
                        if(pattern[i+1]==1) trainNum++;
                        if(pattern.Last() == 1) return null;
                    }
                }
                return pattern;
            }

            public int this[int i]
            {
                get{ return pattern[i]; }
            }

            public bool Equals(Pattern ptn1 , Pattern ptn2){
                return ptn1.pattern.SequenceEqual(ptn2.pattern); 
            }

            public int GetHashCode(Pattern ptn)
            {
                return (ptn[0].GetHashCode() + ptn[ptn.pattern.Length].GetHashCode() + ptn[ptn.pattern.Length/2].GetHashCode()).GetHashCode();
            }
        }

        class Dim1Pattern : IEqualityComparer<Dim1Pattern>
        {
            public int pattern;
            int N;

            public Dim1Pattern(int _pattern, int n)
            {
                pattern = _pattern;
                N = n;
            }

            

            public int DistX((int x, int y, int p)v, (int x, int y, int p)[] 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 int DistY((int x, int y, int p)v, (int x, int y, int p)[] 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;
            }

            public bool Equals(Dim1Pattern p1, Dim1Pattern p2)
            {
                return p1.pattern == p2.pattern;
            }
            
            public int GetHashCode(Dim1Pattern p)
            {
                return p.pattern.GetHashCode();
            }

        }


        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);}
        }
    }
}