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)

use std::io::Read;
use std::cmp::min;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();
    let cities: Vec<(isize, isize, isize)> = (0..n).map(|_| {
        let x: isize = iterator.next().unwrap().parse().unwrap();
        let y: isize = iterator.next().unwrap().parse().unwrap();
        let p: isize = iterator.next().unwrap().parse().unwrap();
        (x, y, p)
    }).collect();

    let lines: Vec<(bool, isize)> = vec![
        (true, 0), // x = 0
        (false, 0), // y = 0
    ];

    let mut results: Vec<isize> = vec![std::isize::MAX; n + 1];
    dfs(&cities, &lines, &mut results, 0);

    // // k = 0
    // let mut results: Vec<isize> = vec![std::isize::MAX; n + 1];
    // results.push(calc(&cities, &lines));
    //
    // for _ in 0..n {
    //     let mut ans = std::isize::MAX;
    //     let mut next_lines = Vec::new();
    //     for &(x, y, p) in cities.iter() {
    //         let mut lines1 = lines.clone();
    //         lines1.push((true, y));
    //         let s1 = calc(&cities, &lines1);
    //         // println!("s1: {}, lines: {:?}", s1, &lines1);
    //         if s1 < ans {
    //             ans = s1;
    //             next_lines = lines1;
    //         }
    //         let mut lines2 = lines.clone();
    //         lines2.push((false, x));
    //         let s2 = calc(&cities, &lines2);
    //         // println!("s2: {}, lines: {:?}", s2, lines2);
    //         if s2 < ans {
    //             ans = s2;
    //             next_lines = lines2;
    //         }
    //     }
    //     results.push(ans);
    //     lines = next_lines;
    // }

    let mut result = String::new();
    for r in results {
        result.push_str(&format!("{}\n", r));
    }
    result.trim().to_string()
}

fn dfs(
    cities: &Vec<(isize, isize, isize)>,
    lines: &Vec<(bool, isize)>,
    results: &mut Vec<isize>,
    depth: usize
) {
    if depth >= cities.len() {
        return;
    }

    results[lines.len() - 2] = min(results[lines.len() - 2], calc(cities, lines));
    dfs(cities, &lines.clone(), results, depth + 1);

    let (x, y, _) = cities[depth];

    let mut lines1 = lines.clone();
    lines1.push((true, y));
    results[lines1.len() - 2] = min(results[lines1.len() - 2], calc(cities, &lines1));
    dfs(cities, &lines1, results, depth + 1);

    let mut lines2 = lines.clone();
    lines2.push((false, x));
    results[lines2.len() - 2] = min(results[lines2.len() - 2], calc(cities, &lines2));
    dfs(cities, &lines2, results, depth + 1);
}

fn calc(cities: &Vec<(isize, isize, isize)>, lines: &Vec<(bool, isize)>) -> isize {
    let mut result: isize = 0;
    for &(x, y, p) in cities.iter() {
        let mut hosu: isize = std::isize::MAX;
        for &(horizontal, v) in lines.iter() {
            if horizontal {
                hosu = min(hosu, (y - v).abs());
            } else {
                hosu = min(hosu, (x - v).abs());
            }
        }
        result += hosu * p;
    }
    return result
}