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)

#![allow(non_snake_case,unused_imports,dead_code)]
use std::str::FromStr;
use std::collections::{HashMap,HashSet,VecDeque,BinaryHeap};
use std::cmp::{max,min,Reverse,Ordering};
use std::time::{Duration, Instant};
use rand::Rng;

use proconio::input;
use proconio::marker::{Chars,Usize1,Isize1};

fn main() {

    input!(
        N:usize,
        mut XYP:[(isize,isize,isize);N],
    );

    /*
    let N = 15usize;
    let XYP = (0..N).map(|n| ((n+1) as isize,(n+1) as isize,1000000isize)).collect::<Vec<_>>();
    */

    let INF = 1e15 as isize;
    let mut ans = vec![INF;N+1];
    ans[N] = 0;

    for i in 0..(1<<N){
        let mut X = vec![0];
        let mut J = Vec::new();
        let mut Y = Vec::new();
        for j in 0..N{
            if i>>j&1==1{
                X.push(XYP[j].0);
            } else {
                J.push(j);
                Y.push(XYP[j].1);
            }
        }
        let n = X.len()-1;
        for m in 0..N-n{
            let mut tmp = INF;
            for mut Y2 in Combinations::new(&Y,m){
                Y2.push(0);
                let mut tmp2 = 0;
                for &j in &J{
                    let (a,b,p) = XYP[j];
                    let mut cur = INF;
                    for &x in &X{
                        cur = min(cur,(a-x).abs());
                    }
                    for &y in &Y2{
                        cur = min(cur,(b-y).abs());
                    }
                    tmp2 += cur*p;
                }
                tmp = min(tmp,tmp2);
            }
            ans[n+m] = min(ans[n+m],tmp);
        }
    }

    for out in ans{
        println!("{}",out);
    }

}


// #region Combinations
struct Combinations<'a,T:Copy>{
    N:usize,
    R:usize,
    V:&'a Vec<T>,
    curr:Option<Vec<usize>>,
}
impl<T:Copy> Combinations<'_,T>{
    fn new(V:&Vec<T>,r:usize) -> Combinations<T>{
        let N = V.len();
        Combinations{
            N:N,
            R:r,
            V:V,
            curr:None
        }
    }
    fn next_combination(&mut self){
        if let Some(curr) = &mut self.curr{
            if self.R==0{
                self.curr = None;
                return;
            }
            let mut i = self.R-1;
            curr[i] += 1;
            while i>0 && curr[i]>=self.N-i{
                i -= 1;
                curr[i] += 1;
            }
            for j in i+1..self.R{
                curr[j] = curr[j-1]+1;
            }
            if curr[self.R-1] >= self.N{
                self.curr = None;
                return;
            }
        } else {
            self.curr = Some((0..self.R).collect::<Vec<_>>());
        }
    }
}
impl<T:Copy> Iterator for Combinations<'_,T>{
    type Item = Vec<T>;
    fn next(&mut self)->Option<Vec<T>>{
        self.next_combination();
        if let Some(curr) = &self.curr{
            let mut ret = Vec::new();
            for &i in curr {
                ret.push(self.V[i]);
            }
            Some(ret)
        } else {
            None
        }
    }
}
// #endregion