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)

// -*- coding:utf-8-unix -*-

use proconio::{fastout, input};

// Implementation according to Solution 1 in the editorial

#[fastout]
fn main() {
    input! {
        n: usize,
        xyp: [(isize, isize, usize); n],
    }
    macro_rules! pow2 (($i:expr) => (2 << $i));
    let mut pow3vec: Vec<usize> = Vec::with_capacity(n + 1);
    let mut tmp = 1;
    for _ in 0..=n {
        pow3vec.push(tmp);
        tmp *= 3;
    }
    let _ = tmp;
    let pow3vec = pow3vec;
    macro_rules! pow3 (($i:expr) => (pow3vec[$i]));
    macro_rules! has_railway (($bits:expr, $i:expr) => ($bits & pow2!($i-1) != 0));
    // Precompute the distances
    let mut tate_costs: Vec<Vec<usize>> = vec![vec![0; n]; pow2!(n)];
    for bits in 0..pow2!(n) {
        for i in 1..=n {
            let (x, y, p) = xyp[i - 1];
            let mut min_dist = x.abs().min(y.abs()) as usize;
            for j in 1..=n {
                if has_railway!(bits, j) {
                    let (xx, _yy, _pp) = xyp[j - 1];
                    let dist = (xx - x).abs() as usize;
                    min_dist = min_dist.min(dist);
                }
            }
            tate_costs[bits][i - 1] = min_dist * p;
        }
    }
    let mut yoko_costs: Vec<Vec<usize>> = vec![vec![0; n]; pow2!(n)];
    for bits in 0..pow2!(n) {
        for i in 1..=n {
            let (x, y, p) = xyp[i - 1];
            let mut min_dist = x.abs().min(y.abs()) as usize;
            for j in 1..=n {
                if has_railway!(bits, j) {
                    let (_xx, yy, _pp) = xyp[j - 1];
                    let dist = (yy - y).abs() as usize;
                    min_dist = min_dist.min(dist);
                }
            }
            yoko_costs[bits][i - 1] = min_dist * p;
        }
    }
    // Check 3^N patterns
    let mut min_costs: Vec<usize> = vec![10000 * 1000000 * 15; n + 1];
    for i in 0..pow3!(n) {
        // Compute corresponding tate and yoko bit sequences
        // and count the number of railways
        let mut num_railways = 0;
        let mut tate_railways = 0;
        let mut yoko_railways = 0;
        for j in 1..=n {
            let keta_3 = pow3!(j - 1);
            let keta_2 = pow2!(j - 1);
            let pat = (i / keta_3) % 3;
            match pat {
                0 => {
                    if tate_railways & keta_2 == 0 {
                        num_railways += 1;
                        tate_railways |= keta_2;
                    }
                }
                1 => {
                    if yoko_railways & keta_2 == 0 {
                        num_railways += 1;
                        yoko_railways |= keta_2;
                    }
                }
                2 => {}
                _ => {}
            }
        }
        // Compute the total cost
        let mut cost = 0;
        for j in 1..=n {
            cost += tate_costs[tate_railways][j - 1].min(yoko_costs[yoko_railways][j - 1]);
        }
        // Update
        min_costs[num_railways] = min_costs[num_railways].min(cost);
    }
    for c in min_costs {
        println!("{}", c);
    }
}