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)

use proconio::{input, fastout};
use std::cmp::min;

const INF: i64 = 9000000000000000000;
const INFS: i32 = 2000000000;

fn calc(a: &Vec<usize>, ss: usize, pc: &Vec<Vec<i32>>, p: &Vec<i32>) -> Vec<i64> {
    let n = a.len();
    let mut res = vec![INF; n+1];
    for s in 0..(1<<n) {
        let mut st = ss;
        let mut b = n;
        for i in 0..n {
            if s>>i&1 == 0 {
                st &= !(1<<a[i]);
                b -= 1;
            }
        }
        let mut now: i64 = 0;
        for i in 0..n {
            now += pc[st][a[i]] as i64 * p[a[i]] as i64;
        }
        res[b] = min(res[b], now);
    }
    res
}
#[fastout]
fn main() {
    input! {
        n: usize,
        xyp: [(i32, i32, i32); n],
    }
    let x: Vec<i32> = xyp.iter().map(|e| e.0).collect();
    let y: Vec<i32> = xyp.iter().map(|e| e.1).collect();
    let p: Vec<i32> = xyp.iter().map(|e| e.2).collect();

    // pre calc
    let mut pcx = vec![vec![INFS;n];1<<n];
    let mut pcy = vec![vec![INFS;n];1<<n];
    for s in 0..(1<<n) {
        let mut px = vec![0i32];
        let mut py = vec![0i32];
        for i in 0..n {
            if s>>i&1 != 0 {px.push(x[i])}
            else           {py.push(y[i])}
        }
        for i in 0..n {
            let mut m = INFS;
            for j in 0..(px.len()) {
                m = min(m, (px[j]-x[i]).abs())
            }
            pcx[s][i] = m;
            let mut m = INFS;
            for j in 0..(py.len()) {
                m = min(m, (py[j]-y[i]).abs())
            }
            pcy[((1<<n)-1)&!s][i] = m;
        }
    }
    //main
    let mut ans = vec![INF;n+1];
    for s in 0..(1<<n) {
        let mut ax: Vec<usize> = Vec::new();
        let mut ay: Vec<usize> = Vec::new();
        for i in 0..n {
            if s>>i&1 != 0 {ax.push(i);}
            else           {ay.push(i);}
        }
        let dx: Vec<i64> = calc(&ax, s,&pcx,&p);
        let dy: Vec<i64> = calc(&ay,((1<<n)-1)&!s,&pcy,&p);
        for i in 0..dx.len() {
            for j in 0..dy.len() {
                ans[i+j] = min(ans[i+j], dx[i]+dy[j]);
            }
        } 
    }
    for i in 0..(n+1) {
        println!("{}", ans[i]);
    }
}