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:
#!/usr/bin/env python3
import sys
rl = sys.stdin.buffer.readline
INF = sys.maxsize

N = int(rl())
PS = [tuple(map(int, rl().split())) for i in range(N)]

def bitcount(n):
  n = n - ((n >> 1) & 0x5555555555555555)
  n = (n & 0x3333333333333333) + ((n >> 2) & 0x3333333333333333)
  n = (n + (n >> 4)) & 0x0f0f0f0f0f0f0f0f
  n = n + (n >> 8)
  n = n + (n >> 16)
  n = n + (n >> 32)
  return n & 0x0000007f

def subsets(s):
  t = s
  while t >= 0:
    t &= s
    yield t
    t -= 1

# (x|y)cost[i][j] := 頂点の集合 i に路線が引かれているとき、j 番目の集落から路線までの最小距離
xcost = [[INF] * N for _ in range(1 << N)]
ycost = [[INF] * N for _ in range(1 << N)]
for i in range(1 << N):
  for j in range(N):
    xcost[i][j] = abs(PS[j][0])
    ycost[i][j] = abs(PS[j][1])
    for k in range(N):
      if ((i >> k) & 1) == 1:
        xcost[i][j] = min(xcost[i][j], abs(PS[j][0] - PS[k][0]))
        ycost[i][j] = min(ycost[i][j], abs(PS[j][1] - PS[k][1]))

# 計算結果の距離を歩行距離(= 距離 * 人口)にする
for i in range(1 << N):
  for j in range(N):
    xcost[i][j] = xcost[i][j] * PS[j][2]
    ycost[i][j] = ycost[i][j] * PS[j][2]

# x, y 座標に引かれた線路の引き方の組み合わせの中から最もコストが小さくなるものを求める
ans = [INF] * (N + 1)
for i in range(1 << N):
  size = bitcount(i)
  for j in subsets(i):
    cost = 0
    for k in range(N):
      # i に含まれていない頂点を合計する(i に含まれている頂点の歩行距離は常に０なので)
      if ((i >> k) & 1) == 0:
        cost += min(xcost[j][k], ycost[i-j][k])

    ans[size] = min(ans[size], cost)

# 結果出力
for a in ans:
  print(a)