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(dead_code)]
fn main() {
    let stdin = stdin();
    let mut reader = StdinReader::new(stdin.lock());
    let n = reader.u();
    let xyp = reader.iv3(n);

    let mut ans = vec![std::i64::MAX; 16];

    let mut pattern = 1;
    for _ in 0..n {
        pattern *= 3;
    }

    //各集落に対し、縦、横、引かないの3通り
    for i in 0..pattern {
        //今回伸ばした路線の数
        let mut count = 0;

        let mut rosen = vec![vec![false; n]; 3];
        let mut i = i;
        for j in 0..n {
            rosen[i % 3][j] = true;
            if i % 3 != 0 {
                count += 1;
            }
            i /= 3;
        }

        let mut yoko = vec![0];
        let mut tate = vec![0];
        for i in 0..n {
            if rosen[1][i] {
                yoko.push(xyp[i].1);
            }
            if rosen[2][i] {
                tate.push(xyp[i].0);
            }
        }
        yoko.sort();
        tate.sort();

        let mut score = 0;
        for i in 0..n {
            let mut m = std::i64::MAX;
            for &tate in &tate {
                m = min((xyp[i].0 - tate).abs(), m);
            }
            for &yoko in &yoko {
                m = min((xyp[i].1 - yoko).abs(), m);
            }
            score += m * xyp[i].2;
        }

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

#[allow(unused_imports)]
use itertools::Itertools;
#[allow(unused_imports)]
use std::{cmp::*, collections::*, io::*, num::*, str::*};
#[allow(unused_imports)]
use stdin_reader::StdinReader;

#[allow(dead_code)]
mod stdin_reader {
    use std::fmt::Debug;
    use std::io::*;
    use std::str::*;

    pub struct StdinReader<R: BufRead> {
        reader: R,
        buf: Vec<u8>,
        // Should never be empty
        pos: usize, // Should never be out of bounds as long as the input ends with '\n'
    }

    impl<R: BufRead> StdinReader<R> {
        pub fn new(r: R) -> StdinReader<R> {
            StdinReader {
                reader: r,
                buf: Vec::new(),
                pos: 0,
            }
        }
        pub fn next<T: FromStr>(&mut self) -> T
        where
            T::Err: Debug,
        {
            if self.buf.is_empty() {
                self._read_next_line();
            }
            let mut start = None;
            loop {
                if self.pos == self.buf.len() {
                    break;
                }
                match (self.buf[self.pos], start.is_some()) {
                    (b' ', true) | (b'\n', true) => break,
                    (_, true) | (b' ', false) => self.pos += 1,
                    (b'\n', false) => self._read_next_line(),
                    (_, false) => start = Some(self.pos),
                }
            }
            let target = &self.buf[start.unwrap()..self.pos];
            from_utf8(target).unwrap().parse().unwrap()
        }

        fn _read_next_line(&mut self) {
            self.pos = 0;
            self.buf.clear();
            if self.reader.read_until(b'\n', &mut self.buf).unwrap() == 0 {
                panic!("Reached EOF");
            }
        }

        pub fn str(&mut self) -> String {
            self.next()
        }
        pub fn s(&mut self) -> Vec<char> {
            self.next::<String>().chars().collect()
        }
        pub fn i(&mut self) -> i64 {
            self.next()
        }
        pub fn i2(&mut self) -> (i64, i64) {
            (self.next(), self.next())
        }
        pub fn i3(&mut self) -> (i64, i64, i64) {
            (self.next(), self.next(), self.next())
        }
        pub fn u(&mut self) -> usize {
            self.next()
        }
        pub fn u2(&mut self) -> (usize, usize) {
            (self.next(), self.next())
        }
        pub fn u3(&mut self) -> (usize, usize, usize) {
            (self.next(), self.next(), self.next())
        }
        pub fn u4(&mut self) -> (usize, usize, usize, usize) {
            (self.next(), self.next(), self.next(), self.next())
        }
        pub fn u5(&mut self) -> (usize, usize, usize, usize, usize) {
            (
                self.next(),
                self.next(),
                self.next(),
                self.next(),
                self.next(),
            )
        }
        pub fn u6(&mut self) -> (usize, usize, usize, usize, usize, usize) {
            (
                self.next(),
                self.next(),
                self.next(),
                self.next(),
                self.next(),
                self.next(),
            )
        }
        pub fn f(&mut self) -> f64 {
            self.next()
        }
        pub fn f2(&mut self) -> (f64, f64) {
            (self.next(), self.next())
        }
        pub fn c(&mut self) -> char {
            self.next::<String>().pop().unwrap()
        }
        pub fn iv(&mut self, n: usize) -> Vec<i64> {
            (0..n).map(|_| self.i()).collect()
        }
        pub fn iv2(&mut self, n: usize) -> Vec<(i64, i64)> {
            (0..n).map(|_| self.i2()).collect()
        }
        pub fn iv3(&mut self, n: usize) -> Vec<(i64, i64, i64)> {
            (0..n).map(|_| self.i3()).collect()
        }
        pub fn uv(&mut self, n: usize) -> Vec<usize> {
            (0..n).map(|_| self.u()).collect()
        }
        pub fn uv2(&mut self, n: usize) -> Vec<(usize, usize)> {
            (0..n).map(|_| self.u2()).collect()
        }
        pub fn uv3(&mut self, n: usize) -> Vec<(usize, usize, usize)> {
            (0..n).map(|_| self.u3()).collect()
        }
        pub fn uv4(&mut self, n: usize) -> Vec<(usize, usize, usize, usize)> {
            (0..n).map(|_| self.u4()).collect()
        }
        pub fn fv(&mut self, n: usize) -> Vec<f64> {
            (0..n).map(|_| self.f()).collect()
        }
        pub fn cmap(&mut self, h: usize) -> Vec<Vec<char>> {
            (0..h).map(|_| self.s()).collect()
        }
    }
}