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 Rust (1.42.0) (judge verdict: Wrong Answer)

//use itertools::Itertools;
use std::cmp;
use std::collections::BTreeMap;
use std::collections::BTreeSet;
use std::collections::BinaryHeap;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::VecDeque;
use std::io::Read;
use std::usize::MAX;

macro_rules! input {(source = $s:expr, $($r:tt)*) => {let mut iter = $s.split_whitespace();let mut next = || { iter.next().unwrap() };input_inner!{next, $($r)*}};($($r:tt)*) => {let stdin = std::io::stdin();let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));let mut next = move || -> String{bytes.by_ref().map(|r|r.unwrap() as char).skip_while(|c|c.is_whitespace()).take_while(|c|!c.is_whitespace()).collect()};input_inner!{next, $($r)*}};}
macro_rules! input_inner {($next:expr) => {};($next:expr, ) => {};($next:expr, $var:ident : $t:tt $($r:tt)*) => {let $var = read_value!($next, $t);input_inner!{$next $($r)*}};}
macro_rules! read_value {($next:expr, ( $($t:tt),* )) => {( $(read_value!($next, $t)),* )};($next:expr, [ $t:tt ; $len:expr ]) => {(0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()};($next:expr, chars) => {read_value!($next, String).chars().collect::<Vec<char>>()};($next:expr, usize1) => {read_value!($next, usize) - 1};($next:expr, [ $t:tt ]) => {{let len = read_value!($next, usize);(0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()}};($next:expr, $t:ty) => {$next().parse::<$t>().expect("Parse error")};}

fn solve() {
    input! {
        n: usize,
        shuraku: [(i64,i64,usize);n]
    }

    // -のみ
    let mut only_x_walk = vec![vec![MAX; n]; 1 << n];
    let mut only_x_count = vec![0; 1 << n];

    // |のみ
    let mut only_y_walk = vec![vec![MAX; n]; 1 << n];
    let mut only_y_count = vec![0; 1 << n];

    // 大通りのみ
    for i in 0..n {
        only_x_walk[0][i] = shuraku[i].1.abs() as usize * shuraku[i].2;
        only_y_walk[0][i] = shuraku[i].0.abs() as usize * shuraku[i].2;
    }

    for i in 0..1 << n {
        let mut x_set = HashSet::new();
        let mut x_count = 0;
        let mut y_set = HashSet::new();
        let mut y_count = 0;
        // j = iにおいて有効になっている集落
        for j in 0..n {
            only_x_walk[i][j] = cmp::min(only_x_walk[i][j], only_x_walk[0][j]);
            only_y_walk[i][j] = cmp::min(only_y_walk[i][j], only_y_walk[0][j]);

            if i >> j & 1 == 1 {
                if x_set.get(&shuraku[j].1) == None {
                    x_set.insert(shuraku[j].1);
                    x_count += 1;
                }
                if y_set.get(&shuraku[j].0) == None {
                    y_set.insert(shuraku[j].0);
                    y_count += 1;
                }

                // k = jが有効になったことによる最短距離の更新対象となる集落
                for k in 0..n {
                    only_x_walk[i][k] = cmp::min(
                        only_x_walk[i][k],
                        (shuraku[k].1 - shuraku[j].1).abs() as usize * shuraku[k].2,
                    );
                    only_y_walk[i][k] = cmp::min(
                        only_y_walk[i][k],
                        (shuraku[k].0 - shuraku[j].0).abs() as usize * shuraku[k].2,
                    );
                }
            }
        }

        only_x_count[i] = x_count;
        only_y_count[i] = y_count;
    }

    let mut min_dists = vec![MAX; n + 1];

    let mut three_pow = 3usize.pow(n as u32);

    for i in 0..three_pow {
        let mut x_on = 0;
        let mut y_on = 0;

        let mut san_num = i;

        // 各集落
        for j in 0..n {
            let k = san_num % 3;

            if k == 1 {
                x_on += 1 << j;
            }

            if k == 2 {
                y_on += 1 << j;
            }

            san_num /= 3;
        }

        let active_senro = only_x_count[x_on] + only_y_count[y_on];

        let mut sum_dist = 0;
        for j in 0..n {
            sum_dist += cmp::min(only_x_walk[x_on][j], only_y_walk[y_on][j]);
        }

        //println!("{} {}", active_senro, sum_dist);

        min_dists[active_senro] = cmp::min(min_dists[active_senro], sum_dist);
    }

    for i in 0..n + 1 {
        println!("{}", min_dists[i]);
    }
}

fn main() {
    let thd = std::thread::Builder::new().stack_size(104_857_600);
    thd.spawn(|| solve()).unwrap().join().unwrap();

    /*
    // 入力を一括で読み込む場合
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut input = buf.split_whitespace();

    // inputに対しnext()で読み込んでいく
    let q: usize = input.next().unwrap().parse().unwrap();
    */
}

const LARGE_PRIME: u64 = 1_000_000_007;

// @〜でsnippet