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;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using static System.Console;
using static System.Math;
 
public class Solve{
    static public int mod = 1000000007;
    static public string al = "abcdefghijklmnopqrstuvwxyz";
    public static void Main(){
        // •ûj
        //
        var n = rint();
        var x = new long[n];
        var y = new long[n];
        var p = new long[n];
        for(int i=0;i<n;i++){
            var a = inta();
            x[i] = a[0];
            y[i] = a[1];
            p[i] = a[2];
        }
        for(int i=0;i<=n;i++){
            var bit = new int[n];
            long min = 1000000000000000000;
            while(bit[n-1] < 3){
                var count = 0;
                var xs = new List<long>();
                var ys = new List<long>();
                for(int j=0;j<n;j++){
                    if(bit[j] == 1){
                        xs.Add(x[j]);
                    }else if(bit[j] == 2){
                        ys.Add(y[j]);
                    }else{
                        continue;
                    }
                    count++;
                }
                if(count == i){
                    xs.Add(0);
                    ys.Add(0);
                    long dis = 0;
                    for(int j=0;j<n;j++){
                        long mind = 1000000000000000000;
                        for(int k=0;k<xs.Count;k++){
                            mind = Min(mind,Abs(x[j]-xs[k]));
                        }
                        for(int k=0;k<ys.Count;k++){
                            mind = Min(mind,Abs(y[j]-ys[k]));
                        }
                        dis += mind * p[j];
                    }
                    min = Min(min,dis);
                }
                
                bit[0]++;
                for(int j=0;j<n-1;j++){
                    if(bit[j] > 2){
                        bit[j] = 0;
                        bit[j+1]++;
                    }
                }
            }
            WriteLine(min);
        }
        
	
    }
    public static void swap(ref int a,ref int b){int temp = a;a= b;b = temp;}
    static void charswap(ref char a,ref char b){char temp = a;a= b;b = temp;}
    static int ncr(int n,int r){if(n<r)return 0;r = Min(r,n-r);long nn = 1;for(int i=n-r+1;i<=n;i++){nn = nn*i%mod;}long rr = 1;for(int i=1;i<=r;i++){rr = rr*i%mod;}rr = square((int)rr,mod-2);nn = nn * rr %mod;return (int)nn;}
    // a^b mod
    static int square(int a,int b){string binary = Convert.ToString(b,2);int bileng = binary.Length;long a_power = a;long value = 1;for(int i=bileng-1;i>=0;i--){if(binary[i] == '1'){value = value*a_power%mod;}a_power = a_power*a_power%mod;}return (int)value;}
    static int square2(int a,int b){long output = 1;var list = new List<long>();int sh = 1;long n = a;list.Add(a);while(sh < b){sh *= 2;n = n*n%mod;list.Add(n);}for(int i=list.Count-1;i>=0;i--){if(b > sh){b -= sh;sh /= 2;output = output*list[i]%mod;}}return (int)output;}
    //Šeí“Çæ
    static string rstr(){ return ReadLine(); }
    static int rint(){ return int.Parse(ReadLine()); }
    static long rlong(){ return long.Parse(ReadLine()); }
    static string[] stra(){ return ReadLine().Split(' '); }
    static char[] chara(){ string[] a=stra();string b="";for(int i=0;i<a.Length;i++){b+=a[i];}return b.ToCharArray();}
    static int[] inta(){ string[] read_str_array = ReadLine().Split(' '); int[] return_int_array = new int[read_str_array.Length]; for(int countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_int_array[countup_i] = int.Parse(read_str_array[countup_i]); } return return_int_array; }
    static int[,] inta2(int num_array,int in_array){ int[,] int_array2 = new int[num_array,in_array]; for(int i=0;i<num_array;i++){ int[] temp_array = inta(); for(int j=0;j<in_array;j++){ int_array2[i,j] = temp_array[j]; } } return int_array2; }
    static long[] longa(){ string[] read_str_array = ReadLine().Split(' '); long[] return_long_array = new long[read_str_array.Length]; for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_long_array[countup_i] = long.Parse(read_str_array[countup_i]); } return return_long_array; }
    static double[] doublea(){ string[] read_str_array = ReadLine().Split(' '); double[] return_double_array = new double[read_str_array.Length]; for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_double_array[countup_i] = double.Parse(read_str_array[countup_i]); } return return_double_array; }
    // -----------------------------
    static long GCD(long a,long b){ if(a < b){ long temp = a; a = b; b = temp; } if(a % b == 0){ return b; } else{ long temp = b; b = a%b; a = temp; return GCD(a,b); } }
    static long LCM(long a,long b){ return a * b / GCD(a,b); }
    static void WriteArray(int[,] a,int b,int c){for(int i=0;i<b;i++){for(int j=0;j<c;j++){if(j!=0) Write(" ");Write(a[i,j]);}WriteLine();}}
}