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)

// ---------- begin chmin, chmax ----------
trait ChangeMinMax {
    fn chmin(&mut self, x: Self) -> bool;
    fn chmax(&mut self, x: Self) -> bool;
}

impl<T: PartialOrd> ChangeMinMax for T {
    fn chmin(&mut self, x: Self) -> bool {
        *self > x && {
            *self = x;
            true
        }
    }
    fn chmax(&mut self, x: Self) -> bool {
        *self < x && {
            *self = x;
            true
        }
    }
}
// ---------- end chmin, chmax ----------

use proconio::*;
use std::cmp::*;

#[fastout]
fn run() {
    input! {
        n: usize,
        p: [(i64, i64, i64); n],
    }
    let mut cost = vec![0; 1 << n];
    for i in 1..(1 << n) {
        let mut a = vec![];
        let mut b = vec![];
        for (j, &(x, y, p)) in p.iter().enumerate() {
            if (i >> j) & 1 == 1 {
                a.push((x, p));
                b.push((y, p));
            }
        }
        let mut val = std::i64::MAX;
        for &(z, _) in a.iter() {
            let mut s = 0;
            for &(x, p) in a.iter() {
                s += (z - x).abs() * p;
            }
            val.chmin(s);
        }
        for &(z, _) in b.iter() {
            let mut s = 0;
            for &(x, p) in b.iter() {
                s += (z - x).abs() * p;
            }
            val.chmin(s);
        }
        cost[i] = val;
    }
    let mut ini = vec![0; 1 << n];
    for i in 1..(1 << n) {
        let mut s = 0;
        for (j, &(x, y, p)) in p.iter().enumerate() {
            if (i >> j) & 1 == 1 {
                s += min(x.abs(), y.abs()) * p;
            }
        }
        ini[i] = s;
    }
    let inf = std::i64::MAX / 100;
    let mut dp = vec![inf; 1 << n];
    dp[(1 << n) - 1] = 0;
    for _ in 0..n {
        let ans = dp.iter().zip(ini.iter()).map(|p| *p.0 + *p.1).min().unwrap();
        println!("{}", ans);
        let mut next = vec![inf; 1 << n];
        for (bit, v) in dp.iter().enumerate().filter(|p| *p.1 != inf && p.0 > 0) {
            let mut j = bit;
            let m = (bit + 1).next_power_of_two() >> 1;
            while j >= m {
                let x = bit ^ j;
                next[x].chmin(v + cost[j]);
                j = (j - 1) & bit;
            }
        }
        dp = next;
    }
    println!("0");
}

fn main() {
    run();
}