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::io;
use std::io::{stdout, BufWriter, Write};

#[allow(dead_code)]
fn read<T: std::str::FromStr>() -> T {
    let mut n = String::new();
    io::stdin().read_line(&mut n).unwrap();
    n.trim().parse().ok().unwrap()
}
#[allow(dead_code)]
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    let mut n = String::new();
    io::stdin().read_line(&mut n).unwrap();
    n.trim()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn main() {
    let out = stdout();
    let mut writer = BufWriter::new(out.lock());
    solve(&mut writer);
    writer.flush().unwrap();
}

fn solve<W: Write>(mut writer: W) {
    #[allow(unused_macros)]
    macro_rules! print {
        ($fmt:expr) => {
            write!(writer, $fmt).unwrap()
        };
        ($fmt:expr, $($arg:tt)*) => {
            write!(writer, $fmt, $($arg)*).unwrap()
        };
    }
    #[allow(unused_macros)]
    macro_rules! println {
        ($fmt:expr) => {
            writeln!(writer, $fmt).unwrap()
        };
        ($fmt:expr, $($arg:tt)*) => {
            writeln!(writer, $fmt, $($arg)*).unwrap()
        };
    }

    // Main Code
    let n: usize = read();
    let mut x = Vec::new();
    let mut y = Vec::new();
    let mut p = Vec::new();
    for i in 0..n {
        let data = read_vec::<i64>();
        x.push(data[0]);
        y.push(data[1]);
        p.push(data[2]);
    }
    let mut dist_x = vec![vec![0; 1 << n]; n];
    let mut dist_y = vec![vec![0; 1 << n]; n];
    for i in 0..1 << n {
        let mut b = vec![false; n];
        for j in 0..n {
            if ((1 << j) & i) != 0 {
                b[j] = true;
            }
        }
        for j in 0..n {
            let mut min = i64::abs(x[j]);
            for k in 0..n {
                if b[k] {
                    min = std::cmp::min(min, i64::abs(x[j]-x[k]));
                }
            }
            dist_x[j][i] = min * p[j];
        }
        for j in 0..n {
            let mut min = i64::abs(y[j]);
            for k in 0..n {
                if b[k] {
                    min = std::cmp::min(min, i64::abs(y[j]-y[k]));
                }
            }
            dist_y[j][i] = min * p[j];
        }
    }
    let mut pow = 1;
    for i in 0..n {
        pow *= 3;
    }
    let mut ans = vec![std::i64::MAX; n + 1];
    for i in 0..pow {
        let mut state = vec![0; n];
        let mut m = i;
        for j in 0..n {
            state[j] = m % 3;
            m /= 3;
        }
        let mut count = 0;
        let mut state_x = 0;
        let mut state_y = 0;
        for j in 0..n {
            if state[j] == 1 {
                count += 1;
                state_x += 1 << j;
            }
            else if state[j] == 2 {
                count += 1;
                state_y += 1 << j;
            }
        }
        let mut dist = 0;
        for j in 0..n {
            dist += std::cmp::min(dist_x[j][state_x], dist_y[j][state_y]);
        }
        ans[count] = std::cmp::min(ans[count], dist);
    }
    for i in 0..=n {
        println!("{}", ans[i]);
    }
}