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
input = sys.stdin.readline


N = int(input())
XYP = [list(map(int, input().split())) for _ in range(N)]

INF = 10**18

def scores(Xs):
    L = len(Xs)
    dp = [[INF]*(L+1) for _ in range(L+1)]
    # k = 0
    init = 0
    for x, p in Xs:
        init += abs(x)*p
    if L == 0:
        return [0]
    # k = 1
    for ix, (x, p) in enumerate(Xs):
        dp[ix][0] = init
        score = 0
        for jx, (y, q) in enumerate(Xs):
            score += min(abs(y)*q, abs(x-y)*q)
        dp[ix][1] = score
    
    # k > 1
    for k in range(1, L):
        for ix, (x, p) in enumerate(Xs):
            for iy in range(ix+1, L):
                y, q = Xs[iy]
                tmp = dp[ix][k]
                for iz in range(ix+1, L):
                    z, r = Xs[iz]
                    new = abs(z-y)
                    pre = min(abs(z), abs(z-x))
                    if new < pre:
                        tmp += (new - pre)*r
                dp[iy][k+1] = min(dp[iy][k+1], tmp)
    
    ret = [INF]*(L+1)
    for k in range(L+1):
        for j in range(L+1):
            ret[k] = min(dp[j][k], ret[k])

    return ret

ans = [INF]*(N+1)
for bit in range(1<<N):
    Xs = []
    Ys = []
    for i, (x, y, p) in enumerate(XYP):
        if (1<<i)&bit:
            Xs.append((x, p))
        else:
            Ys.append((y, p))

    Xs.sort()
    Ys.sort()
    dpx = scores(Xs)
    dpy = scores(Ys)
    for i, x in enumerate(dpx):
        for j, y in enumerate(dpy):
            ans[i+j] = min(ans[i+j], x+y)

print(*ans, sep="\n")