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: Time Limit Exceeded)

use std::collections::*;
use std::io::*;
use std::*;
// template {{{
macro_rules! get {
    ($($t:tt),*; $n:expr) => {
        {
            let stdin = std::io::stdin();
            let ret = std::io::BufRead::lines(stdin.lock()).take($n).map(|line| {
                let line = line.unwrap();
                let mut it = line.split_whitespace();
                _get!(it; $($t),*)
            }).collect::<Vec<_>>();
            ret
        }
    };
    ($($t:tt),*) => {
        {
            let mut line = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut it = line.split_whitespace();
            _get!(it; $($t),*)
        }
    };
}
macro_rules! _get {
    ($it:ident; [char]) => {
        _get!($it; String).chars().collect::<Vec<_>>()
    };
    ($it:ident; [u8]) => {
        _get!($it; String).bytes().collect::<Vec<_>>()
    };
    ($it:ident; usize1) => {
        $it.next().unwrap().parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1
    };
    ($it:ident; [usize1]) => {
        $it.map(|s| s.parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1).collect::<Vec<_>>()
    };
    ($it:ident; [$t:ty]) => {
        $it.map(|s| s.parse::<$t>().unwrap_or_else(|e| panic!("{}", e))).collect::<Vec<_>>()
    };
    ($it:ident; $t:ty) => {
        $it.next().unwrap().parse::<$t>().unwrap_or_else(|e| panic!("{}", e))
    };
    ($it:ident; $($t:tt),+) => {
        ($(_get!($it; $t)),*)
    };
}
// }}}

fn main() {
    let n = get!(usize);
    let mut vil = get!(i32,i32,i64;n);
    vil.push((0, 0, 0));
    let vil = vil;
    let mut ans = vec![1 << 60; n + 1];
    for i in 0..(1 << n) {
        let overs: Vec<_> = (0..n).filter(|x| i as i32 & (1 << x) != 0).collect();
        for j in 0..(1 << overs.len()) {
            let mut hori = Vec::new();
            let mut vert = Vec::new();
            for k in 0..overs.len() {
                if j & (1 << k) == 0 {
                    hori.push(overs[k]);
                } else {
                    vert.push(overs[k]);
                }
            }
            hori.push(n);
            vert.push(n);
            let horimin: Vec<_> = (0..n)
                .map(|x| hori.iter().map(|y| (vil[*y].1 - vil[x].1).abs()).min())
                .collect();
            let vertmin: Vec<_> = (0..n)
                .map(|x| vert.iter().map(|y| (vil[*y].0 - vil[x].0).abs()).min())
                .collect();
            let mut s = 0;
            for k in 0..n {
                s += horimin[k].min(vertmin[k]).unwrap() as i64 * vil[k].2;
            }
            ans[overs.len()] = ans[overs.len()].min(s);
        }
    }
    ans.iter().map(|x| println!("{}", x)).count();
}