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.util.*;

public class Main {


    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] x = new int[n];
        int[] y = new int[n];
        long[] p = new long[n];
        long[] res = new long[n+1];
        Arrays.fill(res, Long.MAX_VALUE);
        for(int i=0; i<n; i++) {
            x[i] = sc.nextInt();
            y[i] = sc.nextInt();
            p[i] = sc.nextInt();
        }
        int pow = 1;
        for(int i=0; i<n; i++) {
            pow *= 3;
        }

        long[][] xmin = new long[n][1<<n];
        long[][] ymin = new long[n][1<<n];

        for(int i=0; i<(1<<n); i++) {
            for(int j=0; j<n; j++) {
                xmin[j][i] = Math.abs(x[j]) * p[j];
                ymin[j][i] = Math.abs(y[j]) * p[j];
                for(int k=0; k<n; k++) {
                    if((i>>k)%2==1) {
                        xmin[j][i] = Math.min(xmin[j][i], Math.abs(x[j]-x[k])*p[j]);
                        ymin[j][i] = Math.min(ymin[j][i], Math.abs(y[j]-y[k])*p[j]);
                    }
                }
            }
        }

        for(int i=0; i<pow; i++) {
            long sum = 0;
            int[] nums = new int[n];
            int ii = i;
            int xi = 0;
            int yi = 0;
            int bits = 0;
            for(int j=0; j<n; j++) {
                int num = ii%3;
                if(num == 1) {
                    xi += (1<<j);
                    bits++;
                } else if (num == 2) {
                    yi += (1<<j);
                    bits++;
                }
                ii/=3;
            }
            for(int j=0; j<n; j++) {
                sum += Math.min(xmin[j][xi], ymin[j][yi]);
            }
            res[bits] = Math.min(res[bits], sum );
        }

        for(int i=0; i<=n; i++) {
            System.out.println(res[i]);
        }


    }
}