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 PyPy3 (7.3.0) (judge verdict: Time Limit Exceeded)

def main():
    from sys import setrecursionlimit, stdin, stderr
    from os import environ
    from collections import defaultdict, deque, Counter
    from math import ceil, floor
    from itertools import accumulate, combinations, combinations_with_replacement, product
    setrecursionlimit(10**6)
    dbg = (lambda *something: stderr.write("\033[92m{}\033[0m".format(str(something)+'\n'))) if 'TERM_PROGRAM' in environ else lambda *x: 0
    input = lambda: stdin.readline().rstrip()
    LMIIS = lambda: list(map(int,input().split()))
    II = lambda: int(input())
    P = 10**9+7
    INF = 10**12

    N = II()
    positions = deque()
    for i in range(N):
        positions.append(LMIIS())
    
    #縦、横の最短距離
    v = []
    h = []
    for i in range(2**N):
        v.append([INF]*N)
        h.append([INF]*N)
    for i in range(2**N):
        v_lines = deque([0])
        h_lines = deque([0])
        for j in range(N):
            if 2 ** j & i:
                v_lines.append(positions[j][0])
                h_lines.append(positions[j][1])
        for j,(x,y,p) in enumerate(positions):
            min_x = INF
            min_y = INF
            for x_c in v_lines:
                min_x = min(min_x, abs(x_c - x)*p)
            for y_c in h_lines:
                min_y = min(min_y, abs(y_c - y)*p)
            v[i][j] = min_x
            h[i][j] = min_y

    ans = [INF]*(N+1)
    #for pattern in product(range(3),repeat=N):
    n_bit_mask = (1<<N)-1
    for pattern_v in range(2**N):
        pattern_h = h_mask = n_bit_mask^pattern_v
        pattern_h += 1
        while pattern_h > 0:
            pattern_h -= 1
            pattern_h &= h_mask
            pattern = pattern_h|pattern_v
            key = 0
            for i in range(N):
                if 2**i & pattern:
                    key += 1


        # for p in pattern:
        #     pattern_v *= 2
        #     pattern_h *= 2
        #     if p == 1:
        #         pattern_v += 1
        #     elif p == 2:
        #         pattern_h += 1
        #     else:a
        #         key -= 1
            sum_diff = 0
            for i in range(N):
                sum_diff += min(v[pattern_v][i],h[pattern_h][i])

            ans[key] = min(ans[key], sum_diff)
    print('\n'.join(map(str,ans)))





            



    

    



    
main()