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.Scanner;
import java.util.stream.IntStream;

public class Main {
	private static int[] numbers2 = IntStream.range(0, 16).map(i -> (int) Math.pow(2, i)).toArray();

	public static void main(String[] args) {
		var sc = new Scanner(System.in);
		var n = sc.nextInt();
		var x = new int[n];
		var y = new int[n];
		var p = new int[n];
		for (var i = 0; i < n; i++) {
			x[i] = sc.nextInt();
			y[i] = sc.nextInt();
			p[i] = sc.nextInt();
		}
		sc.close();

		var mins = new long[n + 1];
		for (var i = 0; i <= n; i++) {
			mins[i] = Long.MAX_VALUE;
		}
		check(x, y, p, mins);
		for (var i = 0; i <= n; i++) {
			System.out.println(mins[i]);
		}
	}

	private static void check(int[] x, int[] y, int[] p, long[] mins) {
		var xdists = new int[x.length][x.length];
		var ydists = new int[x.length][x.length];
		for (var i = 0; i < x.length; i++) {
			for (var j = 0; j < x.length; j++) {
				xdists[i][j] = Math.abs(x[i] - x[j]);
				ydists[i][j] = Math.abs(y[i] - y[j]);
			}
		}

		var xmins = new int[x.length][(int) Math.pow(2, x.length)];
		var ymins = new int[x.length][(int) Math.pow(2, x.length)];
		for (var i = 0; i < x.length; i++) {
			var values2 = new int[x.length + 1];
			for (var j = 0; j < (int) Math.pow(2, x.length); j++) {
				xmins[i][j] = Math.abs(x[i]);
				ymins[i][j] = Math.abs(y[i]);
				for (var k = 0; k < x.length; k++) {
					if (values2[k] == 1) {
						xmins[i][j] = Math.min(xmins[i][j], xdists[i][k]);
						ymins[i][j] = Math.min(ymins[i][j], ydists[i][k]);
					}
				}
				incValues(values2, 2);
			}
		}

		var values3 = new int[x.length + 1];
		for (var i = 0; i < Math.pow(3, x.length); i++) {
			var count = 0;
			var xnumber = 0;
			var ynumber = 0;
			for (var j = 0; j < x.length; j++) {
				if (values3[j] == 1) {
					xnumber += numbers2[j];
					count++;
				} else if (values3[j] == 2) {
					ynumber += numbers2[j];
					count++;
				}
			}

			var sum = 0L;
			for (var j = 0; j < x.length; j++) {
				sum += (long) Math.min(xmins[j][xnumber], ymins[j][ynumber]) * p[j];
			}
			mins[count] = Math.min(mins[count], sum);
			incValues(values3, 3);
		}
	}

	public static void incValues(int[] values, int b) {
		values[0]++;
		for (var i = 0; i < values.length; i++) {
			if (values[i] < b) {
				break;
			}
			values[i] = 0;
			values[i + 1]++;
		}
	}
}