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

n, *towns = map(int, sys.stdin.buffer.read().split())
xxx = towns[0::3]
yyy = towns[1::3]
ppp = towns[2::3]

INF = 10 ** 18
ans = [INF] * (n + 1)

precalc_dist_x = []
precalc_dist_y = []
for bit in range(1 << n):
    dists_x = []
    dists_y = []
    for i in range(n):
        x = xxx[i]
        y = yyy[i]
        dist_x = abs(x)
        dist_y = abs(y)

        t = bit
        j = 0
        while t:
            if t & 1:
                dist_x = min(dist_x, abs(xxx[j] - x))
                dist_y = min(dist_y, abs(yyy[j] - y))
            t >>= 1
            j += 1
        dists_x.append(dist_x * ppp[i])
        dists_y.append(dist_y * ppp[i])
    precalc_dist_x.append(dists_x)
    precalc_dist_y.append(dists_y)

for bit in range(1 << n):
    k = (bit & 0x5555) + (bit >> 1 & 0x5555)
    k = (k & 0x3333) + (k >> 2 & 0x3333)
    k = (k & 0x0f0f) + (k >> 4 & 0x0f0f)
    k = (k & 0x00ff) + (k >> 8 & 0x00ff)
    vertical = bit
    while vertical:
        cost = 0
        pcx = precalc_dist_x[vertical]
        pcy = precalc_dist_y[vertical ^ bit]
        for i in range(n):
            cost += min(pcx[i], pcy[i])
        ans[k] = min(ans[k], cost)
        vertical = (vertical - 1) & bit
    cost = 0
    pcx = precalc_dist_x[0]
    pcy = precalc_dist_y[bit]
    for i in range(n):
        cost += min(pcx[i], pcy[i])
    ans[k] = min(ans[k], cost)

print('\n'.join(map(str, ans)))