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: Wrong Answer)

#![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 xs = xs
            .iter()
            .map(|&x| (points[x].0, points[x].2))
            .collect::<Vec<_>>();

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

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

        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)>, n: usize) -> Vec<i64> {
    let mut x_positives = xs
        .iter()
        .filter(|&&x| x.0 > 0)
        .map(|&x| (x.0, x.1))
        .collect::<Vec<_>>();
    let mut x_negatives = xs
        .iter()
        .filter(|&&x| x.0 <= 0)
        .map(|&x| (-x.0, x.1))
        .collect::<Vec<_>>();
    x_positives.sort();
    x_negatives.sort();

    let dp_p = solve(&x_positives);
    let dp_n = solve(&x_negatives);

    let mut dp = vec![std::i64::MAX; xs.len() + 1];
    for pi in 0..dp_p.len() {
        for ni in 0..dp_n.len() {
            dp[pi + ni] = min(dp[pi + ni], dp_p[pi] + dp_n[ni]);
        }
    }
    while dp.len() < n {
        dp.push(0);
    }
    dp
}

fn solve(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 i in 0..n {
        dp[i + 1][0] = cost0(&xs, 0, i);
    }

    for i in 0..n {
        for ki in 0..n + 1 {
            for li in 0..i + 1 {
                if dp[li][ki] == std::i64::MAX {
                    continue;
                }
                dp[i + 1][ki + 1] = min(dp[i + 1][ki + 1], dp[li][ki] + cost(&xs, li, i));
            }
        }
    }

    let ret = (0..n + 1).map(|x| dp[n][x]).collect::<Vec<_>>();
    ret
}

fn cost0(xs: &Vec<(i64, i64)>, li: usize, ri: usize) -> i64 {
    (li..ri + 1).map(|x| xs[x].0.abs() * xs[x].1).sum::<i64>()
}

fn cost(xs: &Vec<(i64, i64)>, li: usize, ri: usize) -> i64 {
    let mut cands = vec![];
    let sum = (li..ri + 1).map(|x| xs[x].1).sum::<i64>();
    let averaged_sum = (li..ri + 1).map(|x| xs[x].0 * xs[x].1).sum::<i64>() / sum;
    cands.push(averaged_sum);
    cands.push(averaged_sum + 1);

    let mut min_cost = std::i64::MAX;
    for cand in cands {
        let mut cur_cost = 0;
        for i in li..ri + 1 {
            cur_cost += (xs[i].0 - cand).abs() * xs[i].1;
        }
        min_cost = min(cur_cost, min_cost);
    }
    min_cost
}

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()
}