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)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

fn main() {
    let n = read::<usize>();
    let mut points = vec![];
    for i in 0..n {
        let v = read_vec::<i64>();
        let (x, y, p) = (v[0], v[1], v[2]);
        points.push((x, y, p));
    }

    let mut answers = vec![std::i64::MAX; n + 1];
    for state in 0..1 << n {
        let mut xs = vec![];
        let mut ys = vec![];
        for j in 0..n {
            if (state >> j) & 1 == 1 {
                xs.push(j);
            } else {
                ys.push(j);
            }
        }

        let mut xs = xs
            .iter()
            .map(|&x| (points[x].0, points[x].2))
            .collect::<Vec<_>>();
        xs.sort();
        let mut ys = ys
            .iter()
            .map(|&x| (points[x].1, points[x].2))
            .collect::<Vec<_>>();
        ys.sort();

        let dp_x = solve_two(&xs);
        let dp_y = solve_two(&ys);

        for xi in 0..dp_x.len() {
            for yi in 0..dp_y.len() {
                if xi + yi > n {
                    break;
                }
                answers[xi + yi] = min(answers[xi + yi], dp_x[xi] + dp_y[yi]);
            }
        }
    }
    assert_eq!(answers[n], 0);
    for ans in answers {
        println!("{}", ans);
    }
}

fn solve_two(xs: &Vec<(i64, i64)>) -> Vec<i64> {
    let n = xs.len();
    let mut dp = vec![vec![std::i64::MAX; n + 1]; n + 1];
    dp[0][0] = 0;

    for l in 0..n {
        for r in l + 1..n + 1 {
            let mut min_cost = std::i64::MAX;
            for k in l..r {
                let cost = (l..r)
                    .map(|i| (xs[i].0 - xs[k].0).abs() * xs[i].1)
                    .sum::<i64>();
                min_cost = min(cost, min_cost);
            }
            for ki in 0..n {
                if dp[l][ki] == std::i64::MAX {
                    continue;
                }
                dp[r][ki + 1] = min(dp[r][ki + 1], dp[l][ki] + min_cost);
            }
            let no_use_cost = (l..r).map(|i| xs[i].0.abs() * xs[i].1).sum::<i64>();
            for ki in 0..n + 1 {
                if dp[l][ki] == std::i64::MAX {
                    continue;
                }
                dp[r][ki] = min(dp[r][ki], dp[l][ki] + no_use_cost);
            }
        }
    }
    dp[n].clone()
}
fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}