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;
use std::cmp::*;

#[allow(unused_imports)]
use proconio::marker::{Bytes, Chars};

const MAX_SCORE: i64 = 10000*1000000i64*15+1;



#[allow(dead_code)]

//#[proconio::fastout]
fn main() {
    input! {
        n:usize,
        v:[(i32,i32,i64);n],
    }

    let mut distx = vec![vec![0;n];1<<n]; // distx[j][i] : ビットパターンj の線路が南北に引いてあるときに、集落iの人がx方向に移動して線路に到達するまでの距離
    let mut disty = vec![vec![0;n];1<<n];
    for i in 0..n {
        distx[0][i] = v[i].0.abs();
        disty[0][i] = v[i].1.abs();
    }

    for j in 1..1<<n {
        // ビットパターンjの、最も下の1になっているビットを探す
        let mut k = 1;
        let mut m = 0;
        while j & k == 0 {
            k <<= 1;
            m += 1;
        }

        for i in 0..n {
            distx[j][i] = min(distx[j^k][i], (v[m].0-v[i].0).abs());
        }
        for i in 0..n {
            disty[j][i] = min(disty[j^k][i], (v[m].1-v[i].1).abs());
        }
    }

    let mut score = vec![MAX_SCORE;n+1];
    let mut jx = 0;
    let mut jy = 0;
    let max_t = 1<<n;
    loop {
        let mut num_rail = 0;
        for i in 0..n {
            if (1<<i) & (jx|jy) != 0 {
                num_rail += 1;
            }
        }
        
        let mut s = 0;
        for i in 0..n {
            s += min(distx[jx][i], disty[jy][i]) as i64 * v[i].2;
        }
        score[num_rail] = min( score[num_rail], s );
        
        // 3進数(jx&jy=0)で1を足す
        //println!("{} {} {}", jx, jy, jx&jy );

        let b = (jy+1) ^ (jy & (jy+1)); // jyに1を足してどこまで繰り上がるか
        jy += 1;
        if b == max_t {
            break;
        }
        if jx & b == 0 {
            jx |= b;
            jy &= !b;
        } else {
            jx &= !b;
        }
    }

    for i in 0..=n {
        println!("{}", score[i] );
    }

}