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 Python (3.8.2) (judge verdict: Time Limit Exceeded)

# coding: utf-8
import sys
#from operator import itemgetter
sysread = sys.stdin.buffer.readline
read = sys.stdin.buffer.read
#from heapq import heappop, heappush
from collections import defaultdict
sys.setrecursionlimit(10**7)
import math
from itertools import product, accumulate, combinations, product
#import bisect# lower_bound etc
#import numpy as np
#from copy import deepcopy
#from collections import deque


INF = 1 << 50

def generate_dist_arr(X, Y, P):
    N = len(X)
    dist_arr_x = [[] for _ in range(1<<N)]
    dist_arr_y = [[] for _ in range(1<<N)]
    for j in range(1 << N):  # x-y-lines
        x_lines_list = [0]
        y_lines_list = [0]
        for k in range(N):
            if j >> k & 1:
                x_lines_list.append(X[k])
                y_lines_list.append(Y[k])

        dist_xx, dist_yy = [INF] * N, [INF] * N

        for line in x_lines_list:
            for idx, x in enumerate(X):
                dist_xx[idx] = min(abs(line - x) * P[idx], dist_xx[idx])
        dist_arr_x[j] = dist_xx

        for line in y_lines_list:
            for idx, y in enumerate(Y):
                dist_yy[idx] = min(abs(line - y) * P[idx], dist_yy[idx])
        dist_arr_y[j] = dist_yy
    return dist_arr_x, dist_arr_y


def run():
    N = int(sysread())
    X, Y, P = [], [], []
    ans = [INF] * N

    for i in range(1, N+1):
        x,y,p = map(int, sysread().split())
        X.append(x)
        Y.append(y)
        P.append(p)

    ans = [INF] * (N + 1)

    dist_arr_x , dist_arr_y = generate_dist_arr(X,Y,P)

    for i in range(1 << N):# x-y-directions
        dist_towns_x = [INF] * (N + 1)
        dist_towns_y = [INF] * (N + 1)
        x_town_list = []
        y_town_list = []
        for k in range(N):
            if i >> k & 1:
                x_town_list.append(k)
            else:
                y_town_list.append(k)


        for j in range(1 << N):  # x-lines
            x_lines_list = [0]
            for k in range(N):
                if j >> k & 1:
                    x_lines_list.append(X[k])

            dists_base = dist_arr_x[j]
            dist_ = []
            for k in x_town_list:
                dist_.append(dists_base[k])

            tmp = len(x_lines_list)-1
            dist_towns_x[tmp] = min(sum(dist_), dist_towns_x[tmp])

        for j in range(1 << N):  # y-lines
            y_lines_list = [0]
            for k in range(N):
                if j >> k & 1:
                    y_lines_list.append(Y[k])

            dists_base = dist_arr_y[j]
            dist_ = []
            for k in y_town_list:
                dist_.append(dists_base[k])

            tmp = len(y_lines_list) - 1
            dist_towns_y[tmp] = min(sum(dist_), dist_towns_y[tmp])

        #print(bin(i))
        #print(dist_towns_x)
        #print(dist_towns_y)

        for i in range(N + 1):
            for j in range(N + 1 - i):
                val = dist_towns_x[i] + dist_towns_y[j]
                ans[i + j] = min(ans[i + j], val)


    for a in ans:
        print(a)


if __name__ == "__main__":
    run()