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::min;

const INF: i64 = 1 << 62;
fn main() {
    let (r, w) = (std::io::stdin(), std::io::stdout());
    let mut sc = IO::new(r.lock(), w.lock());
    let first_n: usize = sc.read();

    let mut pos = vec![];
    for _ in 0..first_n {
        let x: i64 = sc.read();
        let y: i64 = sc.read();
        let p: i64 = sc.read();
        if x == 0 || y == 0 {
            continue;
        }
        pos.push((x, y, p));
    }

    let n = pos.len();
    let mut xi = vec![];
    let mut yi = vec![];
    for i in 0..n {
        let (px, py, _) = pos[i];
        xi.push((px, i));
        yi.push((py, i));
    }
    xi.sort();
    yi.sort();

    let mut dp = vec![INF; n + 1];
    dfs(0, 0, 0, 0, &mut dp, &pos, &xi, &yi);

    for i in 0..=first_n {
        if dp.len() > i {
            sc.write(dp[i]);
        } else {
            sc.write("0");
        }
        sc.write("\n")
    }
}

fn dfs(
    lx: usize,
    ly: usize,
    depth: usize,
    count: usize,
    dp: &mut Vec<i64>,
    pos: &[(i64, i64, i64)],
    xi: &[(i64, usize)],
    yi: &[(i64, usize)],
) {
    if pos.len() == depth {
        dp[count] = min(calc(pos, xi, yi, lx, ly), dp[count]);
        return;
    }
    dfs(lx, ly, depth + 1, count, dp, pos, xi, yi);
    dfs(lx | (1 << depth), ly, depth + 1, count + 1, dp, pos, xi, yi);
    dfs(lx, ly | (1 << depth), depth + 1, count + 1, dp, pos, xi, yi);
}

fn calc(
    pos: &[(i64, i64, i64)],
    xi: &[(i64, usize)],
    yi: &[(i64, usize)],
    lx: usize,
    ly: usize,
) -> i64 {
    let n = pos.len();
    let mut min_cost = [INF; 15];
    for i in 0..n {
        let (px, py, _) = pos[i];
        min_cost[i] = min(px.abs(), py.abs());
    }

    let mut left_x = -INF;
    let mut left_y = -INF;
    for p in 0..n {
        let (x, i) = xi[p];
        if lx & (1 << i) != 0 {
            left_x = x;
        }
        let dx = (left_x - x).abs();
        min_cost[i] = min(min_cost[i], dx);

        let (y, i) = yi[p];
        if ly & (1 << i) != 0 {
            left_y = y;
        }
        let dy = (left_y - y).abs();
        min_cost[i] = min(min_cost[i], dy);
    }

    let mut right_x = INF;
    let mut right_y = INF;
    for p in (0..n).rev() {
        let (x, i) = xi[p];
        if lx & (1 << i) != 0 {
            right_x = x;
        }
        let dx = (right_x - x).abs();
        min_cost[i] = min(min_cost[i], dx);

        let (y, i) = yi[p];
        if ly & (1 << i) != 0 {
            right_y = y;
        }
        let dy = (right_y - y).abs();
        min_cost[i] = min(min_cost[i], dy);
    }

    let mut sum = 0;
    for i in 0..n {
        sum += min_cost[i] * pos[i].2;
    }
    sum
}

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) -> Self {
        Self(r, std::io::BufWriter::new(w))
    }
    pub fn write<S: ToString>(&mut self, s: S) {
        use std::io::Write;
        self.1.write_all(s.to_string().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 vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }
    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}