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.IO;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCoder2 {
	class Program {
		static void Main(string[] args) {

			//出力の都度更新をやめる
			var sw = new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
			Console.SetOut(sw);


			Ans();

			//更新する
			Console.Out.Flush();
			
		}

 public static void Ans() {

			var inputArray = retreiveLongArray();
			var inputCount = 1;

			// 不正な入力は処理しない
			if (inputArray.Length != inputCount) {
				Console.WriteLine("invalid input");
				return;
			}
			var N = (int)inputArray[0];

			var info = new long[N,3];

			for (int i = 0; i < N; i++) {
				var inp = retreiveLongArray();
				info[i, 0] = inp[0];
				info[i, 1] = inp[1];
				info[i, 2] = inp[2];
			}

			//i番目の町のY方向に道を通した場合、各町がX方向に移動してその町にたどり着く距離を求める

			var xRoot = new Dictionary<int, List<long>>();

			for (int i = -1; i < N; i++) {
				//-1番目の町が、原点にあるとする
				var x = i == -1 ? 0: info[i, 0];
				var ixRoot = new List<long>();
				for (int j = 0; j < N; j++) {
					ixRoot.Add(Math.Abs(info[j, 0] - x) * info[j, 2]);
				}
				xRoot.Add(i, ixRoot);
			}

			//i番目の町のX方向に道を通した場合、各町がY方向に移動してその町にたどり着く距離を求める
			var yRoot = new Dictionary<int, List<long>>();

			for (int i = -1; i < N; i++) {
				//-1番目の町が、原点にあるとする
				var y = i == -1 ? 0 : info[i, 1];
				var iyRoot = new List<long>();
				for (int j = 0; j < N; j++) {
					iyRoot.Add(Math.Abs(info[j, 1] - y)* info[j, 2]);
				}
				yRoot.Add(i, iyRoot);
			}

			var minRoot = Enumerable.Repeat(long.MaxValue,N+1).ToArray();//0本からN本、それぞれ鉄道を設置した場合の最小コストを格納

			//すべての鉄道の設置方法を試す
			var patterns = Math.Pow(3, N);

			for (long i = 0; i < patterns; i++) {
				var ableRoot = new List<List<long>>();
				ableRoot.Add(xRoot[-1]);
				ableRoot.Add(yRoot[-1]);

				var tetsudosu = 0;
				var cases = i;
				for (var j = 0; j < N; j++) {//各集落ごとに鉄道を引かないかx方向か、y方向か
					var tesutdoPattern = cases % 3; ;

					if (tesutdoPattern == 1) {//x方向に路線を引く
						ableRoot.Add(xRoot[j]);
						tetsudosu++;
					}
					if (tesutdoPattern == 2) {//y方向に路線を引く
						ableRoot.Add(yRoot[j]);
						tetsudosu++;
					}
					cases /= 3;
				}
				

				//各町の最小値を求める
				var eachMin = Enumerable.Repeat(long.MaxValue, N).ToArray();

				foreach (var root in ableRoot) {
					for (int j = 0; j < N; j++) {
						eachMin[j] = Math.Min(eachMin[j], root[j]);
					}
				}
				var cost = eachMin.Sum();
				//最小値がこの鉄道数に対して、最小か調べる
				minRoot[tetsudosu] = Math.Min(minRoot[tetsudosu], cost);
			}

			for (int i = 0; i < N+1; i++) {
				Console.WriteLine(minRoot[i]);
			}

	
		}

		/// <summary>
		/// スペースで区切られた文字列配列の入力を受け取る
		/// </summary>
		/// <returns></returns>
		private static string[] retreiveStringArray() {
			var inputArray = Console.ReadLine().Split(' ');
			return inputArray;
		}

		/// <summary>
		/// スペースで区切られた整数配列の入力を受け取る
		/// </summary>
		/// <returns></returns>
		private static long[] retreiveLongArray() {
			var inputArray = Console.ReadLine().Split(' ');
			var longArray = inputArray.Select(s => long.Parse(s));
			return longArray.ToArray();
		}


		/// <summary>
		/// 2組みずつの入力データ(数値)からTuppleListを作成する
		/// </summary>
		/// <param name="N"></param>
		/// <returns></returns>
		private static List<Tuple<long, long>> createLongTuppleList(long N) {
			var inpList = new List<Tuple<long, long>>();

			for (int i = 0; i < N; i++) {
				var inp = retreiveLongArray();
				var item1 = inp[0];
				var item2 = inp[1];
				inpList.Add(new Tuple<long, long>(item1, item2));
			}
			return inpList;
		}
	}
}