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)

#[allow(unused_imports)]
use itertools::Itertools;
#[allow(unused_imports)]
use itertools_num::ItertoolsNum;
#[allow(unused_imports)]
use std::cmp;
#[allow(unused_imports)]
use std::iter;
#[allow(unused_imports)]
use superslice::*;

fn run() {
    let (r, w) = (std::io::stdin(), std::io::stdout());
    let mut sc = IO::new(r.lock(), w.lock());

    let n: usize = sc.read();
    let mut x = vec![];
    let mut y = vec![];
    let mut p = vec![];
    for _ in 0..n {
        x.push(sc.read::<i64>());
        y.push(sc.read::<i64>());
        p.push(sc.read::<i64>());
    }

    let mut xsel = vec![vec![0; n]; 1 << n];
    let mut ysel = vec![vec![0; n]; 1 << n];
    for i in 0..(1 << n) {
        for j in 0..n {
            xsel[i][j] = x[j].abs();
            ysel[i][j] = y[j].abs();
            for k in 0..n {
                if (i >> k) & 1 > 0 {
                    xsel[i][j] = cmp::min(xsel[i][j], (x[j] - x[k]).abs());
                    ysel[i][j] = cmp::min(ysel[i][j], (y[j] - y[k]).abs());
                }
            }
        }
    }

    let mut ans = vec![std::i64::MAX; n + 1];
    for i in 0..((1_usize) << n) {
        let count = i.count_ones() as usize;
        for j in (0..=i).rev() {
            let j = j & i;
            let mut cost = 0;
            for k in 0..n {
                if (i >> k) & 1 == 0 {
                    cost += cmp::min(xsel[j][k], ysel[i - j][k]) * p[k];
                }
            }
            ans[count] = cmp::min(ans[count], cost);
        }
    }

    for i in 0..=n {
        println!("{}", ans[i]);
    }
}

fn main() {
    std::thread::Builder::new()
        .name("run".into())
        .stack_size(256 * 1024 * 1024)
        .spawn(run)
        .unwrap()
        .join()
        .unwrap();
}

pub struct IO<R, W: std::io::Write>(R, std::io::BufWriter<W>);

impl<R: std::io::Read, W: std::io::Write> IO<R, W> {
    pub fn new(r: R, w: W) -> IO<R, W> {
        IO(r, std::io::BufWriter::new(w))
    }
    pub fn write<S: std::ops::Deref<Target = str>>(&mut self, s: S) {
        use std::io::Write;
        self.1.write(s.as_bytes()).unwrap();
    }
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .0
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n' || b == b'\r' || b == b'\t')
            .take_while(|&b| b != b' ' && b != b'\n' && b != b'\r' && b != b'\t')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn read_vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }
    pub fn read_pairs<T: std::str::FromStr>(&mut self, n: usize) -> Vec<(T, T)> {
        (0..n).map(|_| (self.read(), self.read())).collect()
    }
    pub fn read_pairs_1_indexed(&mut self, n: usize) -> Vec<(usize, usize)> {
        (0..n)
            .map(|_| (self.read::<usize>() - 1, self.read::<usize>() - 1))
            .collect()
    }
    pub fn read_chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
    pub fn read_char_grid(&mut self, n: usize) -> Vec<Vec<char>> {
        (0..n).map(|_| self.read_chars()).collect()
    }
    pub fn read_matrix<T: std::str::FromStr>(&mut self, n: usize, m: usize) -> Vec<Vec<T>> {
        (0..n)
            .map(|_| (0..m).map(|_| self.read()).collect())
            .collect()
    }
}