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.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Scanner;
import java.util.TreeSet;

public class Main {
    public static void main(String[] args) {
        Main main = new Main();
        main.solve();
    }
    public void solve() {
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int N = scan.nextInt();
        Group[] groups = new Group[N];
        for (int i = 0; i < N; i++) {
            int X = scan.nextInt();
            int Y = scan.nextInt();
            long P = scan.nextLong();
            groups[i] = new Group(X, Y, P);
        }
        int limit = pow(2, N);
        long[][] distX = new long[limit][N];
        long[][] distY = new long[limit][N];
        for (int i = 0; i < limit; i++) {
            for (int j = 0; j < N; j++) {
                distX[i][j] = Math.abs(groups[j].x);
                distY[i][j] = Math.abs(groups[j].y);
                for (int k = 0; k < N; k++) {
                    if (((i >> k) & 1) == 1) {
                        distX[i][j] = Math.min(distX[i][j], Math.abs(groups[j].x - groups[k].x));
                        distY[i][j] = Math.min(distY[i][j], Math.abs(groups[j].y - groups[k].y));
                    }
                }
            }
        }
        long[] ans = new long[N+1];
        Arrays.fill(ans, Long.MAX_VALUE);
        limit = pow(3, N);
        for (int i = 0; i < limit; i++) {
            int posX = 0;
            int posY = 0;
            int t = i;
            int c = 1;
            int count = 0;
            while (0 < t) {
                if (t % 3 == 1) {
                    posX += c;
                    count += 1;
                }
                if (t % 3 == 2) {
                    posY += c;
                    count += 1;
                }
                t /= 3;
                c *= 2;
            }
            long current = 0;
            for (int j = 0; j < N; j++) {
                current += Math.min(distX[posX][j], distY[posY][j])*groups[j].p;
            }
            ans[count] = Math.min(ans[count], current);
        }
        for (int i = 0; i <= N; i++) {
            System.out.println(ans[i]);
        }
    }
    private int pow(int base, int count) {
        if (count == 0) {
            return 1;
        }
        if (count % 2 == 1) {
            return base * pow(base, count - 1);
        }
        int t = pow(base, count / 2);
        return t * t;
    }
    class Group {
        int x;
        int y;
        long p;
        Group(int x, int y, long p) {
            this.x = x;
            this.y = y;
            this.p = p;
        }
    }
}