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: Memory Limit Exceeded)

#![allow(
    non_snake_case,
    unused_variables,
    unused_assignments,
    unused_mut,
    unused_imports,
    dead_code
)]
//use proconio::fastout;
use proconio::input;
use proconio::marker::*;
use std::cmp::*;
use std::collections::*;
//use std::collections::VecDeque;

fn main() {
    input! {
       N:usize,
       XYP:[(i64,i64,i64);N]
    }
    // 各鉄道パターンについて、集落からの距離
    let inf = std::i64::MAX;
    let mut dtate = vec![vec![0i64; N]; 2i64.pow(N as u32) as usize];
    let mut dyoko = vec![vec![0i64; N]; 2i64.pow(N as u32) as usize];
    for i in 0..dtate.len() {
        for j in 0..N {
            dtate[i][j] = XYP[j].0.abs();
            dyoko[i][j] = XYP[j].1.abs();
        }
        //println!("i={},l={}", i, format!("{:0>15b}", i));
        let l: Vec<bool> = format!("{:0>15b}", i).chars().map(|c| c == '1').collect();
        let sum = l.iter().fold(0, |sum, &a| sum + if a { 1 } else { 0 });

        for j in 0..N {
            for k in 0..N {
                //println!("L[k]={}", l[14 - k]);
                if l[14 - k] {
                    dtate[i][j] = min(dtate[i][j], (XYP[j].0 - XYP[k].0).abs());
                    dyoko[i][j] = min(dyoko[i][j], (XYP[j].1 - XYP[k].1).abs());
                }
            }
            dtate[i][j] *= XYP[j].2;
            dyoko[i][j] *= XYP[j].2;
        }
    }
    //println!("tate:{:?}", dtate);
    //println!("yoko:{:?}", dyoko);
    let mut m = vec![vec![0i64; N]; 3i64.pow(N as u32) as usize];
    let mut ans = vec![std::i64::MAX; N + 1];
    for i in 0..m.len() {
        // i の j bit目が0なら、j番目の町に路線なし、1なら東西線、2なら南北線とする。
        // iから、dtate[x]とdyoko[y]のxとyを求める。
        let mut x = 0;
        let mut y = 0;
        let mut ii = i;
        let mut n = N;
        for j in 0..N {
            let tmp = ii % 3;
            // 順番がおかしくなるけど、全部合計するので気にしない。
            x *= 2;
            y *= 2;
            match tmp {
                1 => x += 1,
                2 => y += 1,
                _ => n -= 1,
            }
            ii /= 3;
        }
        let mut t = 0;
        for j in 0..N {
            t += min(dtate[x][j], dyoko[y][j]);
        }
        ans[n] = min(ans[n], t);
    }
    for i in 0..=N {
        println!("{}", ans[i]);
    }
}