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 Haskell (GHC 8.8.3) (judge verdict: Time Limit Exceeded)

import Data.List
import Control.Monad

data Axis = X|Y deriving Show
type City = [(Int,Int,Int)]
type RoadMap = [(Axis, Int)]

fst3 (x,y,z) = x
snd3 (x,y,z) = y

x -: f = f x

getInt = (fmap read) getLine :: IO Int
getInts = ((fmap ((fmap read) . words)) getLine) :: IO [Int]

getFromLines :: Int -> IO a -> IO [a]
getFromLines 0 _ = return []
getFromLines n getA = do
  a <- getA
  l <- getFromLines (n-1) getA
  return (a:l)

main = do
  n <- getInt
  city' <- getFromLines n getInts
  let city = fmap (\[x,y,w]->(x,y,w)) city'
  putStrLn $ [0..(n-1)] -: (fmap (\i-> show $ minCost i city)) -: (intercalate "\n")
  print 0

minCost :: Int -> City -> Int
minCost nRoad city =
  let xRange = city -: (fmap fst3) in
  let yRange = city -: (fmap snd3) in
  (roadCands nRoad xRange yRange) -: (fmap (\roadMap-> calcCost city ((X,0):(Y,0):roadMap))) -: minimum

calcCost :: City -> RoadMap -> Int
calcCost city roadMap =
  city -: (fmap (cityMinCost roadMap)) -: sum
  where
  cityMinCost :: RoadMap -> (Int,Int,Int) -> Int
  cityMinCost roadMap (x,y,w) = w * (roadMap -: (fmap (distance (x,y))) -: minimum)
  distance:: (Int,Int) -> (Axis, Int) -> Int
  distance (x,y) (X, x') = abs (x-x')
  distance (x,y) (Y, y') = abs (y-y')

roadCands :: Int -> [Int] -> [Int] -> [RoadMap]
roadCands nRoad xRange yRange = chose
  ((xRange -: (fmap (\x->(X,x)))) ++ (yRange -: (fmap (\y->(Y,y)))) :: [(Axis, Int)])
  nRoad

chose :: [a] -> Int -> [[a]]
chose l 0 = [[]]
chose [] k = []
chose (x:xs) k = (chose xs k) ++ ((chose xs (k-1)) -: (fmap (x:)))