Problem:
Score: 500 points

Problem Statement

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.

Sample Input 1

3
1 2 300
3 3 600
1 4 800

Sample Output 1

2900
900
0
0

When K = 0, the residents of Area 1, 2, and 3 have to walk the distances of 1, 3, and 1, respectively, to reach a railroad.

Thus, the sum of the walking distances of all citizens, S, is 1 \times 300 + 3 \times 600 + 1 \times 800 = 2900.

When K = 1, if we build a railroad along the street corresponding to the line y = 4 in the coordinate plane, the walking distances of the citizens of Area 1, 2, and 3 become 1, 1, and 0, respectively.

Then, S = 1 \times 300 + 1 \times 600 + 0 \times 800 = 900.

We have many other options for where we build the railroad, but none of them makes S less than 900.

When K = 2, if we build a railroad along the street corresponding to the lines x = 1 and x = 3 in the coordinate plane, all citizens can reach a railroad with the walking distance of 0, so S = 0. We can also have S = 0 when K = 3.

The figure below shows the optimal way to build railroads for the cases K = 0, 1, 2.

The street painted blue represents the roads along which we build railroads.

Sample Input 2

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

Sample Output 2

13800
1600
0
0
0
0

The figure below shows the optimal way to build railroads for the cases K = 1, 2.

Sample Input 3

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

Sample Output 3

26700
13900
3200
1200
0
0
0

The figure below shows the optimal way to build railroads for the case K = 3.

Sample Input 4

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

Sample Output 4

2576709
1569381
868031
605676
366338
141903
0
0
0

The figure below shows the optimal way to build railroads for the case K = 4.

Solution in Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {

    static int gcd (int a, int b) {return b>0?gcd(b,a%b):a;}
    static long mod = 1000000007;

    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];
        int[] P = new int[N];
        int[] nowMin = new int[N];

        for(int i=0;i<N;i++){
            X[i] = sc.nextInt();
            Y[i] = sc.nextInt();
            P[i] = sc.nextInt();
            nowMin[i] = 10000001;
        }

        long ans=0;
        for(int i=0;i<N;i++){
            ans += Math.min(Math.abs(X[i]),Math.abs(Y[i])) * P[i];
            nowMin[i] =  Math.min(Math.abs(X[i]),Math.abs(Y[i]));
        }
        System.out.println(ans);

        for(int i=0;i<N;i++){
            long minAns = -1;
            long minNum = 100000000;
            for(int j = -10000;j<10001;j++){
                long tmpansNS =0;
                long tmpansWE =0;
                Map<Long,Integer> valMap = new HashMap();
                for(int k=0;k<N;k++){
                    int tmpMinNS = Math.min(nowMin[k],Math.abs(X[k]-j));
                    int tmpMinWE = Math.min(nowMin[k],Math.abs(Y[k]-j));
                    tmpansNS += tmpMinNS * P[k];
                    valMap.put(tmpansNS,j);
                    tmpansWE += tmpMinWE * P[k];
                    valMap.put(tmpansWE,j+100000);

                }
                if(minAns == -1){
                    minAns = Math.min(tmpansNS,tmpansWE);
                    minNum = valMap.get(minAns);
                } else {
                    if(minAns > Math.min(minAns,Math.min(tmpansNS,tmpansWE))){
                        minAns = Math.min(minAns,Math.min(tmpansNS,tmpansWE));
                        minNum = valMap.get(minAns);
                    }

                }

            }
            System.out.println(minAns);
            if(minNum > 50000){
                minNum -= 100000;
                for(int l=0;l<N;l++){
                    nowMin[l] = Math.min(Math.abs(Y[l]-(int)minNum),nowMin[l]);
                }
            } else {
                for(int l=0;l<N;l++){
                    nowMin[l] = Math.min(Math.abs(X[l]-(int)minNum),nowMin[l]);
                }
            }
        }


    }

}