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 sys
def input(): return sys.stdin.readline().rstrip()


def main():
    n = int(input())
    li = []
    for _ in range(n):
        xi, yi, pi = map(int, input().split())
        li.append([xi, yi, pi])
    li.sort(key=lambda x: x[2], reverse=True)
    xx = [[abs(li[j][0]) for j in range(n)] for i in range(2**n)]
    yy = [[abs(li[j][1]) for j in range(n)] for i in range(2**n)]

    for i in range(2**n):
        X = [0]
        rest = []
        for j in range(n):
            if i >> j & 1 == 1:
                X.append(li[j][0])
                xx[i][j] = 0
            else:
                rest.append(j)
        for j in rest:
            res = 10**15
            for k in X:
                res = min(res, abs(li[j][0]-k))
            xx[i][j] = res

    for i in range(2**n):
        Y = [0]
        rest = []
        for j in range(n):
            if i >> j & 1 == 1:
                Y.append(li[j][1])
                yy[i][j] = 0
            else:
                rest.append(j)
        for j in rest:
            res = 10**15
            for k in Y:
                res = min(res, abs(li[j][1]-k))
            yy[i][j] = res

    index_list = [2**q for q in range(n)]
    klist = [10**15]*(n+1)
    klist[n] = 0
    xx=tuple(xx)
    yy=tuple(yy)

    for xyr2 in range(3**(n-1)):
        xyr = xyr2
        k, xopen, yopen = 0, 0, 0
        for index in range(n-1):
            xyr, num = divmod(xyr, 3)
            k += (num+1)//2
            xopen += (num % 2)*index_list[index]
            yopen += (num//2)*index_list[index]
        for app in range(3):
            k2, xopen2, yopen2 = k, xopen, yopen
            k2 += (app+1)//2
            xopen2 += (app % 2)*index_list[n-1]
            yopen2 += (app//2)*index_list[n-1]

            cost = 0
            for w in range(n):
                cost += min(xx[xopen2][w], yy[yopen2][w])*li[w][2]

            klist[k2] = min(klist[k2], cost)

    for g in range(n+1):
        print(klist[g])


if __name__ == '__main__':
    main()