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:
# writer解(c++)を写経した．answerのfor文のところ何やってるかさっぱりわかりませんが…
# https://atcoder.jp/contests/m-solutions2020/submissions/15122924
import sys
input = sys.stdin.readline
INF = 10**13


def main():
    N = int(input())
    X = []
    Y = []
    P = []
    for _ in range(N):
        a, b, c = (int(i) for i in input().split())
        X.append(a)
        Y.append(b)
        P.append(c)
    X_dist = [[0]*N for _ in range(1 << N)]
    Y_dist = [[0]*N for _ in range(1 << N)]

    # bitという道路群を建設したときの，各集落の評価距離
    # bit 001001 とかの1が立っている集落に道を作る
    for bit in range(1 << N):
        for i in range(N):
            X_dist[bit][i] = P[i]*abs(X[i])
            Y_dist[bit][i] = P[i]*abs(Y[i])
            for k in range(N):
                if bit & (1 << k):
                    X_dist[bit][i] = \
                        min(X_dist[bit][i], P[i]*abs(X[i] - X[k]))
                    Y_dist[bit][i] = \
                        min(Y_dist[bit][i], P[i]*abs(Y[i] - Y[k]))

    # これはなに？
    answer = [INF] * (N+1)
    for i in range(1 << N):
        cnt = 0
        for j in range(N):
            if (i >> j) & 1:
                cnt += 1
        j = i
        while j >= 0:
            j &= i
            cost = 0
            for k in range(N):
                if not((i >> k) & 1):
                    cost += min(X_dist[j][k], Y_dist[i - j][k])
            answer[cnt] = min(answer[cnt], cost)
            j -= 1

    for ans in answer:
        print(ans)


if __name__ == '__main__':
    main()