Problem:
Score: 500 points

Problem Statement

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.

Sample Input 1

3
1 2 300
3 3 600
1 4 800

Sample Output 1

2900
900
0
0

When K = 0, the residents of Area 1, 2, and 3 have to walk the distances of 1, 3, and 1, respectively, to reach a railroad.

Thus, the sum of the walking distances of all citizens, S, is 1 \times 300 + 3 \times 600 + 1 \times 800 = 2900.

When K = 1, if we build a railroad along the street corresponding to the line y = 4 in the coordinate plane, the walking distances of the citizens of Area 1, 2, and 3 become 1, 1, and 0, respectively.

Then, S = 1 \times 300 + 1 \times 600 + 0 \times 800 = 900.

We have many other options for where we build the railroad, but none of them makes S less than 900.

When K = 2, if we build a railroad along the street corresponding to the lines x = 1 and x = 3 in the coordinate plane, all citizens can reach a railroad with the walking distance of 0, so S = 0. We can also have S = 0 when K = 3.

The figure below shows the optimal way to build railroads for the cases K = 0, 1, 2.

The street painted blue represents the roads along which we build railroads.

Sample Input 2

5
3 5 400
5 3 700
5 5 1000
5 7 700
7 5 400

Sample Output 2

13800
1600
0
0
0
0

The figure below shows the optimal way to build railroads for the cases K = 1, 2.

Sample Input 3

6
2 5 1000
5 2 1100
5 5 1700
-2 -5 900
-5 -2 600
-5 -5 2200

Sample Output 3

26700
13900
3200
1200
0
0
0

The figure below shows the optimal way to build railroads for the case K = 3.

Sample Input 4

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

Sample Output 4

2576709
1569381
868031
605676
366338
141903
0
0
0

The figure below shows the optimal way to build railroads for the case K = 4.

Solution in Python (3.8.2) (judge verdict: Wrong Answer)

import sys
readline=sys.stdin.readline

N=int(readline())
X = [None] * N
Y = [None] * N
P = [None] * N
for i in range(N):
  X[i],Y[i],P[i] = map(int,readline().split())

# 必ずいずれかの都市と重なるように作るのが合理的。
# 従い、縦15通り横15通りの最大30パターンを毎回試せばよい
# 縦、横ともに、どの筋に道路が通っているかのリスト（最大15）を持ち、
# 最も近い道路を検索できるようにする(小さいのでソートや二分探索不要)

xroad = [0]
yroad = [0]

def calc_S(xroad, yroad):
  S = 0
  for i in range(N):
    # 都市iに関して
    # X軸で最も近い道路を探す
    mindist = 10 ** 10
    for x in range(len(xroad)):
      if abs(X[i] - xroad[x]) < mindist:
        mindist = abs(X[i] - xroad[x])
    # Y軸で最も近い道路を探す
    for y in range(len(yroad)):
      if abs(Y[i] - yroad[y]) < mindist:
        mindist = abs(Y[i] - yroad[y])
    S += mindist * P[i]
  return S

# 道路を指定したときの各都市からの距離を返す
def calc_segment(road):
  R = [None] * N
  #print("road",road)
  for i in range(N):
    mindist = 10 ** 10
    for p in range(len(road)):
      if abs(X[i] - road[p]) < mindist:
        mindist = abs(X[i] - road[p])
    R[i] = mindist * P[i]
  #print("R",R)
  return R

# 初期状態
#print(calc_S(xroad, yroad))

# それぞれ算出しておく
# X軸方向に0～15本引いた場合の各i都市のX軸の値
# Y軸方向に0～15本引いた場合の各i都市のX軸の値

# ループ内で、本数の和がKになるように各組み合わせを試す

# X軸方向
x_pairs = [[] for i in range(N + 1)]
for i in range(2 ** N):
  xroad = [0]
  #print("bin(i)",bin(i))
  for j in range(N):
    if (i >> j) & 1:
      xroad.append(X[j])
    xr = calc_segment(xroad)
    num = bin(i).count("1")
    #print("xr",xr,"x_pairs",x_pairs)
    x_pairs[num] += [xr]

#print("x_pairs",x_pairs)

y_pairs = [[] * N for i in range(N + 1)]
for i in range(2 ** N):
  yroad = [0]
  for j in range(N):
    if (i >> j) & 1:
      yroad.append(X[j])
    yr = calc_segment(yroad)
    num = bin(i).count("1")
    y_pairs[num] += [yr]

for K in range(N + 1):
  # 和がKになるように、x_pairsとy_pairsから配列を取得する
  ans = 10 ** 10
  for x in range(K + 1):
    y = K - x
    x_list = x_pairs[x]
    y_list = y_pairs[y]
    dist = 0
    for xl in x_list:
      for yl in x_list:
        for i in range(N):
          dist += min(xl[i], yl[i])
    if ans > dist:
      ans = dist
  print(ans)