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:
inf = 10**15


def main():
    import sys
    input = sys.stdin.readline

    N = int(input())
    X = []
    Y = []
    P = []
    for _ in range(N):
        x, y, p = map(int, input().split())
        X.append(x)
        Y.append(y)
        P.append(p)

    dx = [[inf] * N for _ in range(1<<N)]
    dy = [[inf] * N for _ in range(1<<N)]
    for i in range(1 << N):
        Xline = [0]
        Yline = [0]
        for j in range(N):
            if i>>j&1:
                Xline.append(X[j])
                Yline.append(Y[j])
        for j in range(N):
            for x, y in zip(Xline, Yline):
                dx[i][j] = min(dx[i][j], abs(x - X[j]))
                dy[i][j] = min(dy[i][j], abs(y - Y[j]))

    ans = [inf] * (N+1)
    beki = [1<<j for j in range(N)]
    for i in range(3**(N-1)):
        M = i
        k = 0
        ix = iy = 0
        J = []
        for j in range(N-1):
            M, m = divmod(M, 3)
            if m == 0:
                J.append(j)
            elif m == 1:
                k += 1
                ix += beki[j]
            else:
                k += 1
                iy += beki[j]
        for x in range(3):
            kk = k
            ix_ = ix
            iy_ = iy
            if x == 0:
                J.append(N-1)
            elif x == 1:
                kk += 1
                ix_ += beki[N-1]
            else:
                kk += 1
                iy_ += beki[N-1]
            cost = 0
            for j in J:
                p = P[j]
                cost += min(dx[ix_][j], dy[iy_][j]) * p
            ans[kk] = min(ans[kk], cost)
    for c in ans:
        print(c)


if __name__ == '__main__':
    main()