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:
# 解説と 提出 #15122924 を参考に作成.
# 遅い. PyPyで出すと通る. はず.

# import sys
# sys.setrecursionlimit(10 ** 6)
# import bisect
# from collections import deque
# from decorator import stop_watch
#
#
# @stop_watch
def solve(N, X, Y, P):
    inf = 10 ** 18
    cnb_N = 2 ** N
    # 前計算
    # # 集落の部分集合毎の、Y軸X軸へのそれぞれの最短経路
    to_x = [[0] * N for _ in range(cnb_N)]
    to_y = [[0] * N for _ in range(cnb_N)]
    for cn in range(cnb_N):
        for n in range(N):
            to_x[cn][n] = abs(X[n]) * P[n]
            to_y[cn][n] = abs(Y[n]) * P[n]
            for n2 in range(N):
                if cn >> n2 & 1:
                    to_x[cn][n] = min(to_x[cn][n], abs(X[n] - X[n2]) * P[n])
                    to_y[cn][n] = min(to_y[cn][n], abs(Y[n] - Y[n2]) * P[n])

    # 実計算
    # # 各集落の集合に対して、Y軸を取った場合とX軸を取った場合で最短となる経路を算出
    ans = [inf] * (N + 1)
    for cn in range(cnb_N):
        count = 0
        for i in range(N):
            count += 1 if cn >> i & 1 else 0
        cn2 = cn
        while cn2 >= 0:
            cn2 = cn2 & cn
            subsum = 0
            for n in range(N):
                if not (cn >> n & 1):
                    subsum += min(to_x[cn - cn2][n], to_y[cn2][n])
            ans[count] = min(ans[count], subsum)
            cn2 -= 1

    for n in range(N + 1):
        print(ans[n])


if __name__ == '__main__':
    # S = input()
    N = int(input())
    # N, M = map(int, input().split())
    X, Y, P = [], [], []
    for _ in range(N):
        tmp = [int(i) for i in input().split()]
        X.append(tmp[0])
        Y.append(tmp[1])
        P.append(tmp[2])
    solve(N, X, Y, P)