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 PYTHON3):

Solution:
#import random


n = int(input())
info = [list(map(int, input().split())) for i in range(n)]
INF = 10 ** 13

#n = 15
#info = []
#for i in range(n):
#    info.append([random.randint(-10000, 10000), random.randint(-10000, 10000), random.randint(1, 1000000)])


costs = [INF] * n
for i, (x, y, c) in enumerate(info):
    costs[i] = c * min(abs(x), abs(y))

OFFSET = n
x_dir = [INF] * ((1 << n) * OFFSET)
y_dir = [INF] * ((1 << n) * OFFSET)


for bit_state in range(1 << n):
    for i, (xi, yi, c) in enumerate(info):
        x_dir[bit_state * OFFSET + i] = min(costs[i], x_dir[bit_state * OFFSET + i])
        y_dir[bit_state * OFFSET + i] = min(costs[i], y_dir[bit_state * OFFSET + i])
        if not((bit_state >> i) & 1):
            continue
        for j, (xj, yj, c) in enumerate(info):
            x_dir[bit_state * OFFSET + j] = min(x_dir[bit_state * OFFSET + j], c * abs(yi - yj))
            y_dir[bit_state * OFFSET + j] = min(y_dir[bit_state * OFFSET + j], c * abs(xi - xj))


ans = [INF] * (n + 1)
for bit_state in range((1 << n) - 1):
    x_bit = bit_state
    cnt = bin(bit_state).count("1")
    while x_bit >= 0:
        x_bit &= bit_state
        # bit_stateの下位集合sub_stateの列挙
        # 処理はじめ
        y_bit = bit_state - x_bit
        tmp = 0
        for i in range(n):
            tmp += min(x_dir[x_bit * OFFSET + i], y_dir[y_bit * OFFSET + i])
        ans[cnt] = min(ans[cnt], tmp)
        # 処理おわり
        x_bit -= 1

ans[-1] = 0
for res in ans:
    print(res)