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)

require 'set'; require 'prime'; require 'pp'
INF=Float::INFINITY; MOD=10**9+7
n = gets.chomp.to_i
x = []
y = []
q = []
n.times do
  a,b,c = gets.chomp.split.map(&:to_i)
  x.push a
  y.push b
  q.push c
end

@tate = Array.new(n+1){Array.new}
@yoko = Array.new(n+1){Array.new}

2.times do |set_count|
  is_yoko = set_count == 0
  [false, true].repeated_permutation(n) do |bits|
    candidates = []
    n.times do |i|
      if bits[i]
        candidates.push is_yoko ? y[i] : x[i]
      end
    end

    ar = Array.new(n)
    n.times do |i|
      if bits[i]
        ar[i] = 0
      else
        if candidates.empty?
          min = INF
        else
          if is_yoko
            ate = y[i]
          else
            ate = x[i]
          end
          index = candidates.bsearch_index{|x| x >= ate}
          if index.nil?
            min = (candidates.last - ate).abs
          elsif index == 0
            min = (candidates.first - ate).abs
          else
            min = [(candidates[index] - ate).abs, (candidates[index-1] - ate).abs].min
          end
        end

        # min = INF
        # candidates.each do |can|
        #   if is_yoko
        #     min = [min, (can - y[i]).abs].min
        #   else
        #     min = [min, (can - x[i]).abs].min
        #   end
        # end

        if is_yoko
          ar[i] = [min, x[i].abs].min * q[i]
        else
          ar[i] = [min, y[i].abs].min * q[i]
        end
      end
    end
    bit_count = bits.count{|e| e}
    if is_yoko
      @yoko[bit_count].push ar
    else
      @tate[bit_count].push ar
    end
  end
end

@yoko.each_with_index do |yoko, i|
  # p ['yoko', i, yoko]
end
@tate.each_with_index do |yoko, i|
  # p ['tate', i, yoko]
end
@yoko.uniq!
@tate.uniq!

# メインループ
n.times do |k|
# [1].each do |k|
  ans = INF
  (k+1).times do |i|
    yoko_num = i
    tate_num = k - i

    @yoko[yoko_num].each do |yoko|
      @tate[tate_num].each do |tate|
        cost = 0
        n.times do |j|
          cost += [yoko[j], tate[j]].min
        end
        ans = [ans, cost].min
      end
    end
    # p [k, cost]
  end
  puts ans
end
puts 0