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(unused_macros)]
use std::cmp::{max, min};
use std::collections::*;
use std::io::{stdin, Read};

#[allow(unused_macros)]
macro_rules! parse {
    ($it: ident ) => {};
    ($it: ident, ) => {};
    ($it: ident, $var:ident : $t:tt $($r:tt)*) => {
        let $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
    ($it: ident, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
    ($it: ident, $var:ident $($r:tt)*) => {
        let $var = parse_val!($it, usize);
        parse!($it $($r)*);
    };
}

#[allow(unused_macros)]
macro_rules! parse_val {
    ($it: ident, [$t:tt; $len:expr]) => {
        (0..$len).map(|_| parse_val!($it, $t)).collect::<Vec<_>>();
    };
    ($it: ident, ($($t: tt),*)) => {
        ($(parse_val!($it, $t)),*)
    };
    ($it: ident, u1) => {
        $it.next().unwrap().parse::<usize>().unwrap() -1
    };
    ($it: ident, $t: ty) => {
        $it.next().unwrap().parse::<$t>().unwrap()
    };
}

#[cfg(debug_assertions)]
macro_rules! debug {
    ($( $args:expr ),*) => { eprintln!( $( $args ),* ); }
}

#[cfg(not(debug_assertions))]
macro_rules! debug {
    ($( $args:expr ),*) => {
        ()
    };
}

fn solve(s: &str) {
    let mut it = s.split_whitespace();
    parse!(it, n: usize);
    let mut x = vec![];
    let mut y = vec![];
    let mut p = vec![];
    for _ in 0..n {
        parse!(it, _x: i64, _y: i64, _p: i64);
        x.push(_x);
        y.push(_y);
        p.push(_p);
    }
    let mut xs = vec![vec![std::i64::MAX; n]; 1 << n];
    let mut ys = vec![vec![std::i64::MAX; n]; 1 << n];
    for i in 0..(1 << n) {
        for j in 0..n {
            xs[i][j] = x[j].abs() * p[j];
            ys[i][j] = y[j].abs() * p[j];
            for k in 0..n {
                if (i >> k) & 1 == 1 {
                    for (xs, x) in [(&mut xs, &mut x), (&mut ys, &mut y)].iter_mut() {
                        xs[i][j] = min(xs[i][j], (x[j] - x[k]).abs() * p[j]);
                    }
                }
            }
        }
    }
    let mut ret = vec![std::i64::MAX; n + 1];
    for i in 0..(1 << n) {
        let mut xi: usize = i;
        let count = i.count_ones() as usize;
        loop {
            xi &= i;
            let yi = i - xi;
            let mut p = 0;
            for k in 0..n {
                p += min(xs[xi][k], ys[yi][k]);
            }
            ret[count] = min(ret[count], p);
            if xi == 0 {
                break;
            }
            xi -= 1;
        }
    }
    for i in 0..=n {
        println!("{}", ret[i]);
    }
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    solve(&s);
}

#[cfg(test)]
mod tests {
    use super::*;
    #[test]
    fn test_input() {
        let s = "
        ";
        solve(s);
    }
}