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)

use std::cmp::*;
use input_mcr::*;

fn main() {
    input! {
        n: usize,
        cs: [(i64,i64,i64); n],
    }
    
    /*
    let n = 15;
    let mut cs = vec![];
    for i in 0..n as i64 {
        cs.push((i,i,i));
    }
    */
    
    let inf = 1i64 << 60;
    let mut xs = vec![vec![]; n+1];
    for bits in 0usize..1<<n {
        let mut res = [inf; 16];
        for i in 0..n {
            res[i] = cs[i].0.abs() * cs[i].2;
        }
        for i in 0..n {
            if bits & (1 << i) != 0 {
                for k in 0..n {
                    res[k] = min(res[k], (cs[k].0 - cs[i].0).abs() * cs[k].2);
                }
            }
        }
        xs[bits.count_ones() as usize].push((res,bits));
    }
    eprintln!("xs end");
    let mut ys = vec![vec![]; n+1];
    for bits in 0usize..1<<n {
        let mut res = [inf; 16];
        for i in 0..n {
            res[i] = cs[i].1.abs() * cs[i].2;
        }
        for i in 0..n {
            if bits & (1 << i) != 0 {
                for k in 0..n {
                    res[k] = min(res[k], (cs[k].1 - cs[i].1).abs() * cs[k].2);
                }
            }
        }
        ys[bits.count_ones() as usize].push((res,bits));
    }
    eprintln!("ys end");
    let mut res = vec![inf; n+1];
    for i in 0..n+1 {
        for k in 0..n+1 {
            if i + k <= n {
                for xs in &xs[i] {
                    for ys in &ys[k] {
                        if xs.1 & ys.1 == 0 {
                            let mut t = 0;
                            for j in 0..n {
                                t += min(xs.0[j], ys.0[j]);
                            }
                            res[i+k] = min(res[i+k], t);
                        }
                    }
                }
            }
        }
    }
    eprintln!("all match end");
    for &r in &res {
        println!("{}", r);
    }
}
pub mod input_mcr {
    // ref: tanakh <https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8>
    #[macro_export(local_inner_macros)]
    macro_rules! input {
        (source = $s:expr, $($r:tt)*) => {
            let mut parser = Parser::from_str($s);
            input_inner!{parser, $($r)*}
        };
        (parser = $parser:ident, $($r:tt)*) => {
            input_inner!{$parser, $($r)*}
        };
        (new_stdin_parser = $parser:ident, $($r:tt)*) => {
            let stdin = std::io::stdin();
            let reader = std::io::BufReader::new(stdin.lock());
            let mut $parser = Parser::new(reader);
            input_inner!{$parser, $($r)*}
        };
        ($($r:tt)*) => {
            input!{new_stdin_parser = parser, $($r)*}
        };
    }
    
    #[macro_export(local_inner_macros)]
    macro_rules! input_inner {
        ($parser:ident) => {};
        ($parser:ident, ) => {};
        ($parser:ident, $var:ident : $t:tt $($r:tt)*) => {
            let $var = read_value!($parser, $t);
            input_inner!{$parser $($r)*}
        };
    }
    
    #[macro_export(local_inner_macros)]
    macro_rules! read_value {
        ($parser:ident, ( $($t:tt),* )) => {
            ( $(read_value!($parser, $t)),* )
        };
        ($parser:ident, [ $t:tt ; $len:expr ]) => {
            (0..$len).map(|_| read_value!($parser, $t)).collect::<Vec<_>>()
        };
        ($parser:ident, chars) => {
            read_value!($parser, String).chars().collect::<Vec<char>>()
        };
        ($parser:ident, char_) => {
            read_value!($parser, String).chars().collect::<Vec<char>>()[0]
        };
        ($parser:ident, usize1) => {
            read_value!($parser, usize) - 1
        };
        ($parser:ident, line) => {
            $parser.next_line()
        };
        ($parser:ident, line_) => {
            $parser.next_line().chars().collect::<Vec<char>>()
        };
        ($parser:ident, $t:ty) => {
            $parser.next::<$t>().expect("Parse error")
        };
    }
    
    use std::io;
    use std::io::BufRead;
    use std::str;
    use std::collections::VecDeque;
    
    pub struct Parser<R> {
        pub reader: R,
        buf: VecDeque<u8>,
        parse_buf: Vec<u8>,
    }
    
    impl Parser<io::Empty> {
        pub fn from_str(s: &str) -> Parser<io::Empty> {
            Parser {
                reader: io::empty(),
                buf: VecDeque::from(s.as_bytes().to_vec()),
                parse_buf: vec![],
            }
        }
    }
    
    impl<R: BufRead> Parser<R> {
        pub fn new(reader: R) -> Parser<R> {
            Parser {
                reader: reader,
                buf: VecDeque::new(),
                parse_buf: vec![],
            }
        }
        pub fn update_buf(&mut self) {
            loop {
                let (len, complete) = {
                    let buf2 = self.reader.fill_buf().unwrap();
                    self.buf.extend(buf2.iter());
                    let len = buf2.len();
                    (len, buf2.last() < Some(&0x20))
                };
                self.reader.consume(len);
                if complete {
                    break;
                }
            }
        }
        pub fn next<T: str::FromStr>(&mut self) -> Result<T, T::Err> {
            loop {
                while let Some(c) = self.buf.pop_front() {
                    if c > 0x20 {
                        self.buf.push_front(c);
                        break;
                    }
                }
                self.parse_buf.clear();
                while let Some(c) = self.buf.pop_front() {
                    if c <= 0x20 {
                        self.buf.push_front(c);
                        break;
                    } else {
                        self.parse_buf.push(c);
                    }
                }
                if self.parse_buf.is_empty() {
                    self.update_buf();
                } else {
                    return unsafe { str::from_utf8_unchecked(&self.parse_buf) }.parse::<T>();
                }
            }
        }
        pub fn next_line(&mut self) -> String {
            loop {
                while let Some(c) = self.buf.pop_front() {
                    if c >= 0x20 {
                        self.buf.push_front(c);
                        break;
                    }
                }
                self.parse_buf.clear();
                while let Some(c) = self.buf.pop_front() {
                    if c < 0x20 {
                        self.buf.push_front(c);
                        break;
                    } else {
                        self.parse_buf.push(c);
                    }
                }
                if self.parse_buf.is_empty() {
                    self.update_buf();
                } else {
                    return unsafe { str::from_utf8_unchecked(&self.parse_buf) }.to_string();
                }
            }
        }
    }
}