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)]
use {
    std::error::Error,
    std::fmt::Debug,
    std::str::FromStr,
    std::io::prelude::*,
    std::io::{stdin, stdout, BufReader, BufWriter},
    std::collections::HashSet,
    std::cmp::min,
};

type BoxResult<T> = Result<T, Box<dyn std::error::Error>>;

fn main() -> BoxResult<()> {
    let mut scanner = Scanner::new(stdin());
    
    let n = scanner.read::<usize>()?;
    let mut cities = vec![];
    for _ in 0..n {
        let (x, y, p) = scanner.read_three::<i64, i64, i64>()?;
        cities.push((x, y, p));
    }
    
    let ans = solve(n, &cities);

    for i in 0..(n + 1) {
        println!("{}", ans[i]);
    }

    Ok(())
}

fn solve(n: usize, cities: &[(i64, i64, i64)]) -> Vec<i64> {
    let mut dists = vec![std::i64::MAX; n + 1];
    let mut stack = vec![(0, vec![])];
    while stack.len() > 0 {
        let (index, coords) = stack.pop().unwrap();
        if index == cities.len() {
            let mut roads = HashSet::new();
            roads.insert((0, true));
            roads.insert((0, false));
            for i in 0..n {
                if coords[i] == 1 {
                    roads.insert((cities[i].0, true));
                } else if coords[i] == 2 {
                    roads.insert((cities[i].1, false));
                }
            }
            let roads = roads.into_iter().collect::<Vec<_>>();
            let d = calc(&cities, &roads);
            let k = roads.len() - 2;
            dists[k] = min(d, dists[k]);
            continue;
        }
        for i in 0..3 {
            let mut new_coords = coords.clone();
            new_coords.push(i);
            stack.push((index + 1, new_coords));
        }
    }
    dists
}

fn calc(cities: &[(i64, i64, i64)], roads: &[(i64, bool)]) -> i64 {
    let mut count = 0;
    for &(x, y, p) in cities {
        let mut d = std::i64::MAX;
        for &(k, vertical) in roads {
            if vertical {
                d = min((x - k).abs(), d);
            } else {
                d = min((y - k).abs(), d);
            }
        }
        count += d * p;
    }
    count
}

pub struct Scanner<R> {
    reader: BufReader<R>,
    buffer: String,
}
impl<R> Scanner<R>
where
    R: Read,
{
    pub fn new(inner: R) -> Scanner<R> {
        let reader = BufReader::new(inner);
        let buffer = String::new();
        Scanner {
            reader: reader,
            buffer: buffer,
        }
    }
    pub fn with_capacity(capacity: usize, inner: R) -> Scanner<R> {
        let reader = BufReader::with_capacity(capacity, inner);
        let buffer = String::new();
        Scanner {
            reader: reader,
            buffer: buffer,
        }
    }
    pub fn read_line(&mut self) -> BoxResult<String> {
        self.reader.read_line(&mut self.buffer)?;
        let cloned = self.buffer.trim().to_string();
        self.buffer.clear();
        Ok(cloned)
    }
 
    pub fn read<T>(&mut self) -> BoxResult<T>
    where
        T: FromStr,
        T::Err: Debug + Error + 'static,
    {
        self.reader.read_line(&mut self.buffer)?;
        let parsed = self.buffer.trim().parse()?;
        self.buffer.clear();
        Ok(parsed)
    }
    pub fn read_two<T, U>(&mut self) -> BoxResult<(T, U)>
    where
        T: FromStr + Copy,
        T::Err: Debug + Error + 'static,
        U: FromStr + Copy,
        U::Err: Debug + Error + 'static,
    {
        self.reader.read_line(&mut self.buffer)?;
        let (a, b) = {
            let params = self.buffer.trim().split_whitespace().collect::<Vec<_>>();
            let a = params[0].parse()?;
            let b = params[1].parse()?;
            (a, b)
        };
        self.buffer.clear();
        Ok((a, b))
    }
    pub fn read_three<T, U, V>(&mut self) -> BoxResult<(T, U, V)>
    where
        T: FromStr + Copy,
        T::Err: Debug + Error + 'static,
        U: FromStr + Copy,
        U::Err: Debug + Error + 'static,
        V: FromStr + Copy,
        V::Err: Debug + Error + 'static,
    {
        self.reader.read_line(&mut self.buffer)?;
        let (a, b, c) = {
            let params = self.buffer.trim().split_whitespace().collect::<Vec<_>>();
            let a = params[0].parse()?;
            let b = params[1].parse()?;
            let c = params[2].parse()?;
            (a, b, c)
        };
        self.buffer.clear();
        Ok((a, b, c))
    }
    pub fn read_four<T, U, V, W>(&mut self) -> BoxResult<(T, U, V, W)>
    where
        T: FromStr + Copy,
        T::Err: Debug + Error + 'static,
        U: FromStr + Copy,
        U::Err: Debug + Error + 'static,
        V: FromStr + Copy,
        V::Err: Debug + Error + 'static,
        W: FromStr + Copy,
        W::Err: Debug + Error + 'static,
    {
        self.reader.read_line(&mut self.buffer)?;
        let (a, b, c, d) = {
            let params = self.buffer.trim().split_whitespace().collect::<Vec<_>>();
            let a = params[0].parse()?;
            let b = params[1].parse()?;
            let c = params[2].parse()?;
            let d = params[3].parse()?;
            (a, b, c, d)
        };
        self.buffer.clear();
        Ok((a, b, c, d))
    }
    pub fn read_vector<T>(&mut self) -> BoxResult<Vec<T>>
    where
        T: FromStr + Copy,
        T::Err: Debug + Error + 'static,
    {
        self.reader.read_line(&mut self.buffer)?;
        let parsed_vec = self
            .buffer
            .trim()
            .split_whitespace()
            .map(|s| s.parse())
            .collect::<Result<Vec<_>, _>>()?;
        self.buffer.clear();
        Ok(parsed_vec)
    }
    pub fn read_matrix<T>(&mut self, lines: u64) -> BoxResult<Vec<Vec<T>>>
    where
        T: FromStr + Copy,
        T::Err: Debug + Error + 'static,
    {
        let mut parsed_matrix = vec![];
        for _ in 0..lines {
            self.reader.read_line(&mut self.buffer)?;
            parsed_matrix.push(
                self.buffer
                    .trim()
                    .split_whitespace()
                    .map(|s| s.parse())
                    .collect::<Result<Vec<_>, _>>()?,
            );
            self.buffer.clear();
        }
        Ok(parsed_matrix)
    }
    pub fn read_chars(&mut self) -> BoxResult<Vec<char>> {
        self.reader.read_line(&mut self.buffer)?;
        let chars = self.buffer.trim().chars().collect::<Vec<_>>();
        self.buffer.clear();
        Ok(chars)
    }
}