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 {
    static int[] xArr;
    static int[] yArr;
    static long[] cArr;
    static long[][] xValues;
    static long[][] yValues;
    static int n;
	public static void main (String[] args) {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		xArr = new int[n];
		yArr = new int[n];
		cArr = new long[n];
		long[] base = new long[n];
		for (int i = 0; i < n; i++) {
		    xArr[i] = sc.nextInt();
		    yArr[i] = sc.nextInt();
		    cArr[i] = sc.nextInt();
		    base[i] = Math.min(Math.abs(xArr[i]), Math.abs(yArr[i])) * cArr[i];
		}
		xValues = new long[1 << n][];
		yValues = new long[1 << n][];
		xValues[0] = base;
		yValues[0] = base;
		for (int i = 0; i < n; i++) {
		    setX(base, 0, i);
		    setY(base, 0, i);
		}
		long[] ans = new long[n + 1];
		Arrays.fill(ans, Long.MAX_VALUE);
		for (int i = 0; i < (1 << n); i++) {
		    for (int j = 0; j < (1 << n); j++) {
		        if ((i & j) != 0) {
		            continue;
		        }
		        int idx = getIdx(i | j);
		        long sum = 0;
		        for (int k = 0; k < n; k++) {
		            sum += Math.min(xValues[i][k], yValues[j][k]);
		        }
		        ans[idx] = Math.min(ans[idx], sum);
		    }
		}
		StringBuilder sb = new StringBuilder();
		for (int i = 0; i <= n; i++) {
		    sb.append(ans[i]).append("\n");
		}
		System.out.print(sb);
	}
	
	static int getIdx(int x) {
	    int count = 0;
	    while (x > 0) {
	        count += x % 2;
	        x /= 2;
	    }
	    return count;
	}
	
	static void setX(long[] base, int mask, int idx) {
	    mask = (mask | (1 << idx));
	    if (xValues[mask] != null) {
	        return;
	    }
	    xValues[mask] = (long[])base.clone();
	    for (int i = 0; i < n; i++) {
	        xValues[mask][i] = Math.min(xValues[mask][i], Math.abs(xArr[i] - xArr[idx]) * cArr[i]);
	    }
	    for (int i = 0; i < n; i++) {
	        if ((mask & (1 << i)) != 0) {
	            continue;
	        }
	        setX(xValues[mask], mask, i);
	    }
	}

	static void setY(long[] base, int mask, int idx) {
	    mask = (mask | (1 << idx));
	    if (yValues[mask] != null) {
	        return;
	    }
	    yValues[mask] = (long[])base.clone();
	    for (int i = 0; i < n; i++) {
	        yValues[mask][i] = Math.min(yValues[mask][i], Math.abs(yArr[i] - yArr[idx]) * cArr[i]);
	    }
	    for (int i = 0; i < n; i++) {
	        if ((mask & (1 << i)) != 0) {
	            continue;
	        }
	        setY(yValues[mask], mask, i);
	    }
	}
	
}