Problem: 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.

Examples

Input

3
1 2 300
3 3 600
1 4 800


Output

2900
900
0
0


Input

5
3 5 400
5 3 700
5 5 1000
5 7 700
7 5 400


Output

13800
1600
0
0
0
0


Input

6
2 5 1000
5 2 1100
5 5 1700
-2 -5 900
-5 -2 600
-5 -5 2200


Output

26700
13900
3200
1200
0
0
0


Input

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


Output

2576709
1569381
868031
605676
366338
141903
0
0
0

Solution (in JAVA):

Solution:
import java.io.*;
import java.math.*;
import java.util.*;
 
public class Main {
	static final long MOD = 1000000007L;
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter pw = new PrintWriter(System.out);
		
		int N = sc.ni();
		long[][] nums = new long[N][3];
		for (int i = 0; i < N; i++) {
			nums[i] = new long[] {sc.nl(),sc.nl(),sc.nl()};
		}
		long[] ans = new long[N+1];
		Arrays.fill(ans, Long.MAX_VALUE);
		
		long[][] Xcost = new long[1<<N][N];
		long[][] Ycost = new long[1<<N][N];
		int[] Xroads = new int[1<<N];
		int[] Yroads = new int[1<<N];
		for (int b = 0; b < (1<<N); b++) {
			HashSet<Long> Xcovered = new HashSet<Long>();
			HashSet<Long> Ycovered = new HashSet<Long>();
			for (int i = 0; i < N; i++) {
				if ((b&(1<<i))==0) continue;
				Xcovered.add(nums[i][0]);
				Ycovered.add(nums[i][1]);
			}
			Xroads[b] = Xcovered.size();
			Yroads[b] = Ycovered.size();
			
			for (int j = 0; j < N; j++) {
				Xcost[b][j] = nums[j][2]*Math.abs(nums[j][0]);
				Ycost[b][j] = nums[j][2]*Math.abs(nums[j][1]);
				for (int i = 0; i < N; i++) {
					if ((b&(1<<i))==0) continue;
					Xcost[b][j] = Math.min(Xcost[b][j],nums[j][2]*Math.abs(nums[j][0]-nums[i][0]));
					Ycost[b][j] = Math.min(Ycost[b][j],nums[j][2]*Math.abs(nums[j][1]-nums[i][1]));
				}
			}
		}
		
		//build vertical roads through bit b
		for (int b = 0; b < (1<<N); b++) {
			int size = N-Integer.bitCount(b);
			int[] val = new int[size];
			int ind = 0;
			for (int i = 0; i < N; i++) {
				if ((b&(1<<i))>0) continue;
				val[ind] = (1<<i);
				ind++;
			}
			for (int y = 0; y < (1<<size); y++) {
				int bit = 0;
				for (int i = 0; i < size; i++) {
					if ((y&(1<<i))==0) continue;
					bit += val[i];
				}
				int uncovered = (1<<N)-1-b-bit;
				int roads = Xroads[b]+Yroads[bit];
				long cost = 0;
				for (int i = 0; i < N; i++) {
					if ((uncovered&(1<<i))==0) continue;
					cost += Math.min(Xcost[b][i],Ycost[bit][i]);
				}
				ans[roads] = Math.min(ans[roads],cost);
			}
		}
		
		for (int i = 1; i <= N; i++) {
			ans[i] = Math.min(ans[i], ans[i-1]);
		}
		for (long a: ans)
			pw.println(a);
		pw.close();
	}
	
    static class FastScanner { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastScanner() { 
            br = new BufferedReader(new InputStreamReader(System.in)); 
        } 
  
        String next() { 
            while (st == null || !st.hasMoreElements()) { 
                try { 
                    st = new StringTokenizer(br.readLine());
                } catch (IOException  e) { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        }
        
        int ni() { 
            return Integer.parseInt(next()); 
        }
  
        long nl() { 
            return Long.parseLong(next()); 
        } 
  
        double nd() { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() {
            String str = ""; 
            try { 
                str = br.readLine(); 
            } catch (IOException e) {
                e.printStackTrace(); 
            } 
            return str;
        }
    }
}