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.Collections.Generic;
using System.Collections;
using System.Collections.Specialized;
using System.Linq;
using System.Text;
using System.IO;
using System.Reflection;
using static System.Math;
using System.Numerics;
static class Program{
	const int mod=(int)1e9+7;
	const double eps=1e-11;
	static void Main(){
		Sc sc=new Sc();
		var n=sc.I;
		int m=(int)Fp(3,n);
		var a=new long[n][];
		var ans=new long[n+1];
		for(int i = 0;i<n;i++) {
			a[i]=sc.La;
			ans[i+1]=long.MaxValue;
		}
		var dp=new long[m][];
		var g=new int[m];
		var pa=new long[m];
		dp[0]=new long[n];
		for(int i = 0;i<n;i++) {ans[0]+=dp[0][i]=Min(Abs(a[i][0]),Abs(a[i][1]))*a[i][2];}
		int bb=1;
		var h=new long[]{0,a[0][1]};
		for(int i = 1,j=0;i<m;i++) {
			if(i==bb<<1){
				bb<<=1;
				h[0]=a[j][0];
				h[1]=0;
			}
			else if((bb&1)==0&&i==(bb>>1)*3){
				j++;
				bb=(bb>>1)*3;
				h[0]=0;
				h[1]=a[j][1];
			}
			if(g[i-bb]==-1){
				g[i]=-1;
				continue;
			}
			dp[i]=new long[n];
			g[i]=g[i-bb]+1;
			for(int k = 0;k<n;k++) {pa[i]+=dp[i][k]=Min(dp[i-bb][k],Min(Abs(a[k][0]-h[0]),Abs(a[k][1]-h[1]))*a[k][2]);}
			if(pa[i]==pa[i-bb]){
				g[i]=-1;
				continue;
			}
			ans[g[i]]=Min(ans[g[i]],pa[i]);
		}
		for(int i = 1;i<=n;i++) {ans[i]=Min(ans[i],ans[i-1]);}
		Console.WriteLine("{0}",String.Join("\n",ans));
	}
	static long Fp(long x,long e){
		long r=1;
		while(e>0){
			if((e&1)>0){r*=x;r%=mod;}
			x=(x*x)%mod;
			e>>=1;
		}
		return r;
	}
}
public class Sc{
	public int I{get{return int.Parse(Console.ReadLine());}}
	public long L{get{return long.Parse(Console.ReadLine());}}
	public double D{get{return double.Parse(Console.ReadLine());}}
	public string S{get{return Console.ReadLine();}}
	public int[] Ia{get{return Array.ConvertAll(Console.ReadLine().Split(),int.Parse);}}
	public long[] La{get{return Array.ConvertAll(Console.ReadLine().Split(),long.Parse);}}
	public double[] Da{get{return Array.ConvertAll(Console.ReadLine().Split(),double.Parse);}}
	public string[] Sa{get{return Console.ReadLine().Split();}}
	public object[] Oa{get{return Console.ReadLine().Split();}}
	public int[] Ia2{get{return Array.ConvertAll(("0 "+Console.ReadLine()+" 0").Split(),int.Parse);}}
	public int[] Ia3(string a,string b){return Array.ConvertAll((a+Console.ReadLine()+b).Split(),int.Parse);}
	public int[] Ia3(int a){return Array.ConvertAll((Console.ReadLine()+" "+a.ToString()).Split(),int.Parse);}
	public long[] La2{get{return Array.ConvertAll(("0 "+Console.ReadLine()+" 0").Split(),long.Parse);}}
	public long[] La3(string a,string b){return Array.ConvertAll((a+Console.ReadLine()+b).Split(),long.Parse);}
	public long[] La3(int a){return Array.ConvertAll((Console.ReadLine()+" "+a.ToString()).Split(),long.Parse);}
	public double[] Da2{get{return Array.ConvertAll(("0 "+Console.ReadLine()+" 0").Split(),double.Parse);}}
	public double[] Da3(string a,string b){return Array.ConvertAll((a+Console.ReadLine()+b).Split(),double.Parse);}
	public T[] Arr<T>(int n,Func<T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f();}return a;}
	public T[] Arr<T>(int n,Func<int,T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f(i);}return a;}
	public T[] Arr<T>(int n,Func<string[],T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f(Console.ReadLine().Split());}return a;}
	public T[] Arr<T>(int n,Func<int,string[],T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f(i,Console.ReadLine().Split());}return a;}
}