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 = float("inf")

N, *XYP = map(int, open(0).read().split())
XYP = list(zip(*[iter(XYP)] * 3))

def solve(XP):
    M = len(XP)
    XP.sort()

    dp = [[INF] * (M + 1) for _ in range(M + 1)]
    dp[0][0] = 0

    for i, (x, p) in enumerate(XP):
        dp[i + 1][0] = dp[i][0] + abs(x) * p

    for l, dpl in enumerate(dp[:M]):
        for r in range(l + 1, M + 1):
            val = INF
            for lx, _ in XP[l:r]:
                cur = 0
                for x, p in XP[l:r]:
                    cur += p * min(abs(x - lx), abs(x))
                val = min(val, cur)

            for i in range(1, M + 1):
                dp[r][i] = min(
                    dp[r][i],
                    val + dpl[i - 1]
                )

    res = [INF] * (N + 1)
    res[:M + 1] = dp[M]
    return res


A = [INF] * (N + 1)
for state in range(1 << N):
    XP, YP = [], []
    for i, (x, y, p) in enumerate(XYP):
        if (state >> i) & 1:
            XP.append((x, p))
        else:
            YP.append((y, p))

    X, Y = solve(XP), solve(YP)
    for i in range(N + 1):
        for j in range(N + 1 - i):
            A[i + j] = min(A[i + j], X[i] + Y[j])

for a in A:
    print(a)