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: Runtime Error)

import sys

sys.setrecursionlimit(1000000)

import itertools


def t_slowsolve(Xs, Ys, maxcost, XYPs, k):
    candidate = []
    for x in list(Xs):
        candidate.append(("X", x))
    for y in list(Ys):
        candidate.append(("Y", y))

    total_cost = maxcost
    for lines in itertools.combinations(candidate, k):
        cost = 0
        for x, y, p in XYPs:
            c = p * min(abs(x), abs(y))
            for l in lines:
                if l[0] == "X":
                    c = min(c, p * min(abs(x), abs(y), abs(l[1] - x)))
                elif l[0] == "Y":
                    c = min(c, p * min(abs(x), abs(y), abs(l[1] - y)))
            cost += c
        total_cost = min(total_cost, cost)
    return total_cost

def t_solve(maxcost, XYPs, candidate, current_lines):
    total_cost = maxcost
    selected_line = ""
    for lines in itertools.combinations(candidate, 1):
        cost = 0
        for x, y, p in XYPs:
            c = p * min(abs(x), abs(y))
            for l in current_lines + list(lines):
                if l[0] == "X":
                    c = min(c, p * min(abs(x), abs(y), abs(l[1] - x)))
                elif l[0] == "Y":
                    c = min(c, p * min(abs(x), abs(y), abs(l[1] - y)))
            cost += c
        if cost < total_cost:
            total_cost = min(total_cost, cost)
            selected_line = lines[0]
    return total_cost, selected_line

def solve(N, XYPs):
    # print(N, ",", XYPs)
    Xs = set(xyps[0] for xyps in XYPs)
    Ys = set(xyps[1] for xyps in XYPs)
    maxcost = sum(min(abs(xyps[0]), abs(xyps[1])) * xyps[2] for xyps in XYPs)
    ans = []
    ans.append(str(maxcost))
    current_lines = []
    candidate = []
    for x in list(Xs):
        candidate.append(("X", x))
    for y in list(Ys):
        candidate.append(("Y", y))

    for k in range(1, N + 1):
        if k >= len(Xs) or k >= len(Ys):
            ans.append("0")
        else:
            cost, selected = t_solve(maxcost, XYPs, candidate, current_lines)
            current_lines.append(selected)
            ans.append(str(cost))
        # print(ans)
    return "\n".join(ans)
#
# import random
#
# for i in range(10):
#     solve(15, [(random.randint(-10, 10),random.randint(-100, 10),random.randint(1, 100)) for i in range(15)])
#
# print("DONE")
#
# assert (solve(3, [(1, 2, 300), (3, 3, 600), (1, 4, 800)]) == "\n".join(
#     ["2900",
#      "900",
#      "0",
#      "0",
#      ]))
# assert (solve(5, [(3, 5, 400), (5, 3, 700), (5, 5, 1000), (5, 7, 700), (7, 5, 400)]) == "\n".join(
#     ["13800",
#      "1600",
#      "0",
#      "0",
#      "0",
#      "0"
#      ]
# ))
# assert (solve(6, [(2, 5, 1000), (5, 2, 1100), (5, 5, 1700), (-2, -5, 900), (-5, -2, 600), (-5, -5, 2200)]) == "\n".join(
#     [
#         "26700",
#         "13900",
#         "3200",
#         "1200",
#         "0",
#         "0",
#         "0"
#     ]))
# assert (solve(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)]) == "\n".join(["2576709",
#                                                                    "1569381",
#                                                                    "868031",
#                                                                    "605676",
#                                                                    "366338",
#                                                                    "141903",
#                                                                    "0",
#                                                                    "0",
#                                                                    "0"
#                                                                    ]))
# print("OK")

if __name__ == "__main__":
    N = int(input())
    XYPs = [tuple(map(int, input().split(" "))) for _ in range(N)]

    print(solve(N, XYPs))