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 Ruby (2.7.1) (judge verdict: Time Limit Exceeded)

def gs() gets.chomp end
def gi() gets.to_i end
def gsmi() gets.chomp.split.map(&:to_i) end
def debug(k,v) puts "#{k}: #{v}" end
def desc(ar) ar.sort!{|x,y|y<=>x} end
def min(a,b) a==-1 ? b : (a<=b ? a : b) end
def max(a,b) a>=b ? a : b end
def sum(ar) ar.inject(:+) end

n = gi #15
xyps = Array.new(n){gsmi}
ans = Array.new(n+1){-1}

xss = Array.new(2**n) do |num|
  i = 0
  xls = [0]
  for j in (0...n)
    if (num>>j)&1==1
      i+=1
      xls << xyps[j][0]
    end
    break if num>>j==0
  end
  xyps.map do |x,y,p|
    min = 10000
    xls.each do |xl|
      min = min(min, (xl-x).abs)
    end
    min * p
  end
end

yss = Array.new(2**n) do |num|
  i = 0
  yls = [0]
  for j in (0...n)
    if (num>>j)&1==1
      i+=1
      yls << xyps[j][1]
    end
    break if num>>j==0
  end
  xyps.map do |x,y,p|
    min = 10000
    yls.each do |y1|
      min = min(min, (y1-y).abs)
    end
    min * p
  end
end

(3**n).times do |num|
  i = 0
  xls = [] # 2
  yls = []
  for j in (0...n)
    if (num/(3**j))%3==1
      xls << j
      i+=1
    elsif (num/(3**j))%3==2
      yls << j
      i+=1
    end
    break if (num/(3**j))==0
  end
  xi = xls.sum{|xl| 2**xl }
  yi = yls.sum{|yl| 2**yl }
  sum = n.times.to_a.sum do |j|
    min(xss[xi][j], yss[yi][j])
  end
  #p [num,i,xss[xi],yss[yi],sum]
  ans[i] = min(ans[i], sum)
end
puts ans