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: Accepted)

#![allow(
  non_snake_case,
  unused_variables,
  unused_assignments,
  unused_mut,
  unused_imports,
  dead_code
)]
use proconio::fastout;
use proconio::input;
use proconio::marker::*;
use std::cmp::*;
use std::collections::*;
//use std::collections::VecDeque;

macro_rules! debug {
  ($($a:expr),* $(,)*) => {
      #[cfg(debug_assertions)]
      eprintln!(concat!($("| ", stringify!($a), "={:?} "),*, "|"), $(&$a),*);
  };
}

#[fastout]
fn main() {
  input! {
     N:usize,
     XYP:[(i64,i64,i64);N]
  }
  let N: usize = N;
  let mut XYP: Vec<(i64, i64, i64)> = XYP;
  let inf = std::i64::MAX;
  let mut xkyori = vec![vec![inf; N]; 2usize.pow(N as u32)];
  let mut ykyori = vec![vec![inf; N]; 2usize.pow(N as u32)];

  for pat_i in 0..2usize.pow(N as u32) {
    // 組み合わせの作成
    for i in 0..N {
      if pat_i >> i & 1 == 1 {
        // 都市 i に鉄道が引かれている場合の距離は0。
        xkyori[pat_i][i] = 0;
        ykyori[pat_i][i] = 0;
      } else {
        // そうでない場合、すべての路線への距離の最小値を求める。
        let (x, y, p) = XYP[i];
        xkyori[pat_i][i] = x.abs();
        ykyori[pat_i][i] = y.abs();
        for j in 0..N {
          if pat_i >> j & 1 == 1 {
            xkyori[pat_i][i] = min(xkyori[pat_i][i], (x - XYP[j].0).abs());
            ykyori[pat_i][i] = min(ykyori[pat_i][i], (y - XYP[j].1).abs());
          }
        }
        xkyori[pat_i][i] *= p;
        ykyori[pat_i][i] *= p;
      }
      debug!(pat_i, i, xkyori[pat_i]);
    }
  }
  debug!(xkyori);
  debug!(ykyori);
  // ここまでで前処理。xkyoriとykyoriが計算できた。
  let mut ans = vec![inf; N + 1];
  // 3進数を計算。0=線路なし,1=X路線, 2=Y路線
  for pat_i in 0..3usize.pow(N as u32) {
    // pat_xとpat_yの作成
    let mut k = 0;
    let mut pat = pat_i;
    let mut pat_x = 0;
    let mut pat_y = 0;
    for i in 0..N {
      match pat % 3 {
        1 => {
          pat_x += 2usize.pow(i as u32);
          k += 1;
        }
        2 => {
          pat_y += 2usize.pow(i as u32);
          k += 1;
        }
        _ => (),
      }
      pat = pat / 3;
    }
    let mut t_ans = 0;
    for i in 0..N {
      t_ans += min(xkyori[pat_x][i], ykyori[pat_y][i]);
    }
    ans[k] = min(ans[k], t_ans);
  }
  for a in ans {
    println!("{}", a);
  }
}