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 Lua (LuaJIT 2.1.0) (judge verdict: Wrong Answer)

local mmi, mma = math.min, math.max
local mab = math.abs
local mfl, mce = math.floor, math.ceil
local bls, brs = bit.lshift, bit.rshift
local bxor = bit.bxor

local n = io.read("*n")
local x, y, p = {}, {}, {}
local flag = {}
for i = 1, n do
  x[i], y[i], p[i] = io.read("*n", "*n", "*n")
  flag[i] = false
end

local function grayCode(x)
  return bxor(x, brs(x, 1))
end

local function findright(idxes, ofst)
  for i = ofst, #idxes do
    if flag[idxes[i]] then
      return i
    end
  end
  return false
end

local function getmincosts(ik, pos, val, idx_pos, idx_val)
  local retary = {}
  for i = 1, ik + 1 do retary[i] = 0 end
  for i = 1, #pos do flag[i] = false end
  for linecnt = 0, ik do
    if #pos <= linecnt then
      break
    end
    if 0 < linecnt then
      flag[idx_val[linecnt]] = true
    end
    local retval = 0
    local leftpos = false
    local rightpos = findright(idx_pos, 1)
    for i = 1, #idx_pos do
      local idx_i = idx_pos[i]
      if flag[idx_i] then
        leftpos = rightpos
        if rightpos then
          rightpos = findright(idx_pos, rightpos + 1)
        end
      else
        local cand = mab(pos[idx_i])
        if leftpos then
          cand = mmi(cand, mab(pos[idx_i] - pos[idx_pos[leftpos]]))
        end
        if rightpos then
          cand = mmi(cand, mab(pos[idx_i] - pos[idx_pos[rightpos]]))
        end
        retval = retval + cand * val[idx_i]
      end
    end
    retary[linecnt + 1] = retval
  end
  return retary
end

local retik = {}

local g_ik = 0
local xs, ys = {}, {}
local function add_func(idx)
  local xv = x[idx]
  local yv = y[idx]
  xs[xv] = xs[xv] - p[idx]
  if xs[xv] == 0 then xs[xv] = nil end
  if not ys[yv] then
    ys[yv] = p[idx]
  else
    ys[yv] = ys[yv] + p[idx]
  end
end
local function rm_func(idx)
  local xv = x[idx]
  local yv = y[idx]
  ys[yv] = ys[yv] - p[idx]
  if ys[yv] == 0 then ys[yv] = nil end
  if not xs[xv] then
    xs[xv] = p[idx]
  else
    xs[xv] = xs[xv] + p[idx]
  end
end
local function work_func()
  local cand = 0
  local xpos, xval, xidx_pos, xidx_val = {}, {}, {}, {}
  for k, v in pairs(xs) do
    table.insert(xpos, k)
    table.insert(xval, v)
  end
  for i = 1, #xpos do xidx_pos[i], xidx_val[i] = i, i end
  table.sort(xidx_pos, function(a, b) return xpos[a] < xpos[b] end)
  table.sort(xidx_val, function(a, b) return xval[a] > xval[b] end)

  local ypos, yval, yidx_pos, yidx_val = {}, {}, {}, {}
  for k, v in pairs(ys) do
    table.insert(ypos, k)
    table.insert(yval, v)
  end
  for i = 1, #ypos do yidx_pos[i], yidx_val[i] = i, i end
  table.sort(yidx_pos, function(a, b) return ypos[a] < ypos[b] end)
  table.sort(yidx_val, function(a, b) return yval[a] > yval[b] end)

  local xcosts = getmincosts(n - 1, xpos, xval, xidx_pos, xidx_val)
  local ycosts = getmincosts(n - 1, ypos, yval, yidx_pos, yidx_val)
  for ik = 0, n - 1 do
    cand = xcosts[0 + 1] + ycosts[ik + 1]
    for i = 1, ik do
      cand = mmi(cand, xcosts[i + 1] + ycosts[ik - i + 1])
    end
    if not retik[ik + 1] then
      retik[ik + 1] = cand
    elseif cand < retik[ik + 1] then
      retik[ik + 1] = cand
    end
  end
end

local function grayWalk(size, add_func, rm_func, work_func)
  local prv = 0
  local total = bls(1, size) - 1
  local bpos = {}
  for i = 1, size do
    bpos[bls(1, i - 1)] = i
  end
  work_func()
  for i = 1, total do
    local v = grayCode(i)
    if prv < v then
      prv, v = v, v - prv
      add_func(bpos[v])
    else
      prv, v = v, prv - v
      rm_func(bpos[v])
    end
    work_func()
  end
end

for j = 1, n do
  local xj = x[j]
  if not xs[xj] then
    xs[xj] = p[j]
  else
    xs[xj] = xs[xj] + p[j]
  end
end
for ik = 0, n do
  retik[ik + 1] = false
end

grayWalk(n, add_func, rm_func, work_func)
retik[n + 1] = 0
print(table.concat(retik, "\n"))
-- print(os.clock())