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 Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)

import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Deque;
import java.util.NoSuchElementException;


public class Main {
	public static void main(String[] args) {
		for(int i=0;i<1;i++) {
			slover();
			out.flush();
		}
	}
	static FastScanner sc = new FastScanner();
	static PrintWriter out = new PrintWriter(System.out);
	//StringBuffer S = new StringBuffer(sc.next());
		//String hoge2 = str.reverse().toString();
		//map.containsKey(A)
		//Map<Integer, Integer> map = new HashMap<Integer, Integer>(N);
		/*for ( キーのデータ型 key : マップの名前.keySet() ) {
			データのデータ型 data = マップの名前.get( key );

			// keyやdataを使った処理;
		}*/
		//int i = Integer.parseInt(s);
		//Queue<String> qq=new ArrayDeque<>(); //add,poll,peek BFSは前から実行される
		//Deque<String> dd=new ArrayDeque<>();//push後ろに入れる,poll(pop)後ろからとる,peek addは先頭に入るからバグ注意
		//stackdfsは後ろから実行される
		//ArrayList<Integer> cc = new ArrayList<>(N);
		//cc.contains(tmp)
		//Arrays.asList(c).contains("a")
		//list.set(1,"walk");//1 1 2 3 5
		//PriorityQueue<Integer> r=new PriorityQueue<Integer>();//add poll

		private static int INF =1000000007;



		private static class D{
		
			ArrayList<Integer> X=new ArrayList<>();
			ArrayList<Integer> Y=new ArrayList<>();
			int t=0;
			int index=0;
			D(ArrayList<Integer> X,ArrayList<Integer> Y,int t,int index){
				this.X=X;;
				this.Y=Y;
				this.t=t;
				this.index=index;
			}

		}



	//Set<Long> a=new HashSet<Long>();
	
	
	private static ArrayList<Long> divisors2(long t) {
		ArrayList<Long> c=new ArrayList<>();
		for(int i=1;i*i<=t;i++) {
			if(t%i==0) {
				if(i*i!=t) {
					c.add(t/i);
				}
				
				c.add((long)i);
			}
		}
		Collections.sort(c);
		Collections.reverse(c);
		return c;
	}
	
	//Integer.toBinaryString(i);
	//Integer.toString(i, 2);
//	//Integer.parseInt(bin, 2);
	//bitset
	private static int mod=1000000007;
	private static int slover()   {
		//int N=sc.nextInt();
		//String S=sc.next();
		//char a[]=sc.next().toCharArray();
		
		int N=sc.nextInt();
		int x[]=new int[N];
		int y[]=new int[N];
		int p[]=new int[N];
		
		for(int i=0;i<N;i++) {
			x[i]=sc.nextInt();
			y[i]=sc.nextInt();
			p[i]=sc.nextInt();
		}
		
		for(int i=0;i<N;i++) {
			//iはkの値
			long Fans=100000000000000000L;
			
			//X,Yにi個足す
			Deque<D> qq=new ArrayDeque<>(); //add,poll,peek BFSは前から実行される
			qq.add(new D(new ArrayList<Integer>(),new ArrayList<Integer>(),0,0));
			qq.peek().X.add(0);
			qq.peek().Y.add(0);
			
			while(!qq.isEmpty()) {
				D c=qq.poll();
			
				if(c.t==i) {
				long ans1=0;
				//System.out.println(i);
				//System.out.println(c.X);
				//System.out.println(c.Y);
				for(int t=0;t<N;t++) {
					int minxy=mod;
					for(int v=0;v<c.X.size();v++) {
						minxy=Math.min(Math.abs(x[t]-c.X.get(v)), minxy);
						//System.out.println(x[t]-c.X.get(v));
					}
					for(int v=0;v<c.Y.size();v++) {
						minxy=Math.min(Math.abs(y[t]-c.Y.get(v)), minxy);
					}
					//System.out.println(t+" "+minxy+" "+x[t]+" "+y[t]);
					ans1+=(long)minxy*p[t];
				}
				Fans=Math.min(Fans, ans1);
				if(Fans==0)break;
				continue;
				
				}
				if(!(c.index<N))continue;
				qq.add(new D(c.X,c.Y,c.t,c.index+1));
				
				ArrayList<Integer> f1=new ArrayList<>(c.X);
				ArrayList<Integer> f2=new ArrayList<>(c.Y);
				f1.add(x[c.index]);
				qq.add(new D(f1,c.Y,c.t+1,c.index+1));
				
				f2.add(y[c.index]);
				qq.add(new D(c.X,f2,c.t+1,c.index+1));
				
			}
			
		p(Fans);	
			
			
		}
		p(0);
		return 0;
	}
	//速度が足りないときは、前計算を1回だけにしたり、longをintに変えたりするといい
	private static long modNcR2(int n,int r) {
		long x_b=1;
		
		for(int i=0;i<r;i++) {
			x_b*=(n-i);
			x_b%=mod;
		}
		
		long y=1;
		for(int i=1;i<=r;i++) {
			y*=i;
			y%=mod;
		}
		return  (x_b*modPow((int) y,mod-2,mod)%mod);
		
	}
	private static int modPow2(String S,int mod) {
		//iのt乗をO(log t)で返す
		int a=2;
		int res=1;
		int N=S.length()-1;
		for(int i=0;i<S.length();i++) {
			if(S.charAt(N-i)=='1') {
				res=res*a%mod;
			}
			a=a*a%mod;
		}
		return res;
	}
	
	
	
	
	private static long modPow(int i,int t,long mod) {
		//iのt乗をO(log t)で返す
		long a=i;
		long res=1;
		while(t!=0) {
			//p(t);
			if((1&t)==1) {
				res=res*a%mod;
			}
			a=a*a%mod;
			t=t>>1;
		}
		return res;
	}

	public static long gcd(long num1,long num2) {
        if(num2==0) return num1;
        else return gcd(num2,num1%num2);
    }
	public static long lcm(long num1,long num2) {
		return num1*num2/gcd(num1,num2);
	}
	//O(N^0.5)
	private static boolean isPrime(long t) {
		if(t<2)return false;
		for(int i=2;i*i<=t;i++) {
			if(t%i==0)return false;
		}
		return true;
	}
	private static ArrayList<Long> divisors(long t) {
		ArrayList<Long> c=new ArrayList<>();
		for(int i=1;i*i<=t;i++) {
			if(t%i==0) {
				if(i*i!=t) {
					c.add(t/i);
				}
				c.add((long)i);
			}
		}
		return c;
	}
	private static void bubunwa() {
		int N=sc.nextInt();
		int K=sc.nextInt();
		int a[]=sc.nextIntArray(N, false);

		boolean dp[] =new boolean[K+1];

		Arrays.fill(dp, false);
		dp[0]=true;
		for(int i=0;i<N;i++) {
			for(int x=K-a[i];x>=0;x--) {
				if(dp[x])dp[x+a[i]]=true;
			}
		}

		p(dp[K] ? "Yes":"No");
	}
	
	private static String bitToString(int i) {
		
			String T="";
			for(int j=0;j<2;j++) {
				//if(i>>j==0)break;
				if((1&i>>j)==1) {
					T+="1";
				}else T+="0";
			}
			return (i+" "+T);

		

	}
/*********************************************************************/
	//target以下までの値のindexを返す
	//target以上が欲しいときは返り値+1すればいい
	//0-indexで個数を求めるときはさらにindex+1する必要あり
	private static int lower_bound(long a[],long target) {
	//p(target+ " "+Arrays.toString(a));
	if(a[0]>target)return -1;
	//最後の値がtarget以下であるときは最後のindexを返す。
	//target以上が欲しいときは注意する必要あり
	if(a[a.length-1]<=target) return a.length-1;

	int S=-1;
	int E=a.length;

	while(S<E) {

		int G=(S+E)/2;
		if(G+1==a.length) {
			G--;
		}
		//未満にしたいときは=の場所を変える
		if(a[G]<=target&&a[G+1]>target) {
			return G;

		}else if(a[G]<=target){
			S=G;
		}else if(a[G]>target) {
			E=G;
		}

	}


	return -1;
	}
	static String nextPermutation(String s) {
		ArrayList<Character> list=new ArrayList<>();
		for(int i=0;i<s.length();i++) {
			list.add(s.charAt(i));
		}
		int pivotPos=-1;
		char pivot=0;
		for(int i=list.size()-2;i>=0;i--) {
			if(list.get(i)<list.get(i+1)) {
				pivotPos=i;
				pivot=list.get(i);
				break;
			}
		}
		if(pivotPos==-1&&pivot==0) {
			return "Final";
		}
		int L=pivotPos+1,R=list.size()-1;



		int minPos=-1;
		char min =Character.MAX_VALUE;

		for(int i=R;i>=L;i--) {
			if(pivot<list.get(i)) {
				if(list.get(i)<min) {
					min=list.get(i);
					minPos=i;
				}
			}
		}

		Collections.swap(list, pivotPos, minPos);
		Collections.sort(list.subList(L, R+1));


		StringBuilder sb=new StringBuilder();
		for(int i=0;i<list.size();i++) {
			sb.append(list.get(i));
		}
		return sb.toString();
	}
private static long[][] com;
private static void nCr() {
	int MAX = 3001;
	com= new long[MAX][MAX];
	for(int i = 0; i < MAX; i++)
	    com[i][0] = 1;
	for(int i = 1; i < MAX; i++) {
	    for(int j = 1; j <= i; j++) {
	        com[i][j] = com[i-1][j-1] + com[i-1][j];
	        com[i][j] %= mod;
	    }
	}
}


	static void p(String ans) {out.println(ans);};
	static void p(int ans) {out.println(ans);};
	static void p() {out.println();};
	static void p(long ans) {out.println(ans);};
	static void p(double ans) {out.println(ans);};

	static void pn(String ans) {out.print(ans);};
	static void pn(int ans) {out.print(ans);};
	static void pn() {out.print("");};
	static void pn(long ans) {out.print(ans);};
	static void pn(double ans) {out.print(ans);};

	static class FastScanner {
	    private final InputStream in = System.in;
	    private final byte[] buffer = new byte[1024];
	    private int ptr = 0;
	    private int buflen = 0;

	    private boolean hasNextByte() {
	      if (ptr < buflen) {
	        return true;
	      } else {
	        ptr = 0;
	        try {
	          buflen = in.read(buffer);
	        } catch (IOException e) {
	          e.printStackTrace();
	        }
	        if (buflen <= 0) {
	          return false;
	        }
	      }
	      return true;
	    }

	    private int readByte() {
	      if (hasNextByte()) return buffer[ptr++];
	      else return -1;
	    }

	    private static boolean isPrintableChar(int c) {
	      return 33 <= c && c <= 126;
	    }

	    private void skipUnprintable() {
	      while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
	    }

	    public boolean hasNext() {
	      skipUnprintable();
	      return hasNextByte();
	    }

	    public String next() {
	      if (!hasNext()) throw new NoSuchElementException();
	      StringBuilder sb = new StringBuilder();
	      int b = readByte();
	      while (isPrintableChar(b)) {
	        sb.appendCodePoint(b);
	        b = readByte();
	      }
	      return sb.toString();
	    }

	    public long nextLong() {
	      if (!hasNext()) throw new NoSuchElementException();
	      long n = 0;
	      boolean minus = false;
	      int b = readByte();
	      if (b == '-') {
	        minus = true;
	        b = readByte();
	      }
	      if (b < '0' || '9' < b) {
	        throw new NumberFormatException();
	      }
	      while (true) {
	        if ('0' <= b && b <= '9') {
	          n *= 10;
	          n += b - '0';
	        } else if (b == -1 || !isPrintableChar(b)) {
	          return minus ? -n : n;
	        } else {
	          throw new NumberFormatException();
	        }
	        b = readByte();
	      }
	    }

	    public int nextInt() {
	      return (int) nextLong();
	    }

	    public double nextDouble(){
	    	return Double.parseDouble(next());
	    }

	    public int[] nextIntArray(int N, boolean oneBased) {
	      if (oneBased) {
	        int[] array = new int[N + 1];
	        for (int i = 1; i <= N; i++) {
	          array[i] = sc.nextInt();
	        }
	        return array;
	      } else {
	        int[] array = new int[N];
	        for (int i = 0; i < N; i++) {
	          array[i] = sc.nextInt();
	        }
	        return array;
	      }
	    }

	    public long[] nextLongArray(int N, boolean oneBased) {
	      if (oneBased) {
	        long[] array = new long[N + 1];
	        for (int i = 1; i <= N; i++) {
	          array[i] = sc.nextLong();
	        }
	        return array;
	      } else {
	        long[] array = new long[N];
	        for (int i = 0; i < N; i++) {
	          array[i] = sc.nextLong();
	        }
	        return array;
	      }
	    }
	  }
}