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)

#[allow(unused_imports)]
use itertools::Itertools;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;

pub struct BitPattern {
    i: u64,
    mask: u64,
    end: bool,
}
impl BitPattern {
    #[inline]
    pub fn new(n: usize) -> Self {
        Self {
            i: (1 << n) - 1,
            mask: (1 << n) - 1,
            end: n == 0,
        }
    }
    #[inline]
    pub fn with_mask(n: usize, mask: u64) -> Self {
        let mask = mask & ((1 << n) - 1);
        Self {
            i: mask,
            mask: mask,
            end: n == 0,
        }
    }
}

impl Iterator for BitPattern {
    type Item = u64;
    fn next(&mut self) -> Option<Self::Item> {
        if self.end {
            None
        } else {
            let x = self.i;
            if self.i == 0 {
                self.end = true;
            } else {
                self.i = (self.i - 1) & self.mask;
            }
            Some(!x & self.mask)
        }
    }
}

fn main() {
    input! {
        n: usize,
        xyp: [(i64, i64, i64); n],
    }

    let mut x_val = (0..n).collect_vec();
    let mut y_val = (0..n).collect_vec();
    x_val.sort_by_key(|&i| xyp[i].0);
    y_val.sort_by_key(|&i| xyp[i].1);
    let mut x_sorted = vec![0; n];
    let mut y_sorted = vec![0; n];
    for i in 0..n {
        x_sorted[x_val[i]] = i;
        y_sorted[y_val[i]] = i;
    }

    let mut res = vec![std::i64::MAX; n + 1];

    for i in 0..1 << n {
        for j in BitPattern::with_mask(n, i) {
            let mut x_s = 0u32;
            let mut y_s = 0u32;
            for k in 0..n {
                if i >> k & 1 == 0 {
                    continue;
                }
                if j >> k & 1 == 0 {
                    x_s |= 1 << x_sorted[k];
                } else {
                    y_s |= 1 << y_sorted[k];
                }
            }
            let mut x_line = vec![];
            let mut xz = false;
            for k in 0..n {
                if x_s >> k & 1 == 0 {
                    continue;
                }
                let k = x_val[k];
                if !xz && xyp[k].0 > 0 {
                    x_line.push(0);
                    xz = true;
                }
                if x_line.last() != Some(&xyp[k].0) {
                    x_line.push(xyp[k].0);
                }
            }
            if x_line.is_empty() || matches!(x_line.last(), Some(&x) if x < 0) {
                x_line.push(0);
                xz = true;
            }
            let mut y_line = vec![];
            let mut yz = false;
            for k in 0..n {
                if y_s >> k & 1 == 0 {
                    continue;
                }
                let k = y_val[k];
                if !yz && xyp[k].1 > 0 {
                    y_line.push(0);
                    yz = true;
                }
                if y_line.last() != Some(&xyp[k].1) {
                    y_line.push(xyp[k].1);
                }
            }
            if y_line.is_empty() || matches!(y_line.last(), Some(&y) if y < 0) {
                y_line.push(0);
                yz = true;
            }

            use std::cmp::min;
            let mut mn = vec![std::i64::MAX; n];
            let mut idx = 0;
            for k in 0..n {
                let k = x_val[k];
                if i >> k & 1 == 1 {
                    continue;
                }
                while idx + 1 < x_line.len() && x_line[idx + 1] <= xyp[k].0 {
                    idx += 1;
                }
                let l = (x_line[idx] - xyp[k].0).abs();
                let r = if idx + 1 < x_line.len() {
                    (x_line[idx + 1] - xyp[k].0).abs()
                } else {
                    std::i64::MAX
                };
                mn[k] = min(mn[k], min(l, r));
            }
            let mut idx = 0;
            for k in 0..n {
                let k = y_val[k];
                if i >> k & 1 == 1 {
                    continue;
                }
                while idx + 1 < y_line.len() && y_line[idx + 1] <= xyp[k].1 {
                    idx += 1;
                }
                let l = (y_line[idx] - xyp[k].1).abs();
                let r = if idx + 1 < y_line.len() {
                    (y_line[idx + 1] - xyp[k].1).abs()
                } else {
                    std::i64::MAX
                };
                mn[k] = min(mn[k], min(l, r));
            }

            let mut s = 0;
            for k in 0..n {
                if i >> k & 1 == 1 {
                    continue;
                }
                s += mn[k] * xyp[k].2;
            }
            let mut i = x_line.len() + y_line.len();
            if xz {
                i -= 1;
            }
            if yz {
                i -= 1;
            }
            if n + 1 == i {
                println!(
                    "aa {} {} {} {} {} {}",
                    x_s.count_ones(),
                    y_s.count_ones(),
                    x_line.len(),
                    y_line.len(),
                    xz,
                    yz
                );
                println!("bb {:?} {:?}", x_line, y_line);
            }
            let r = &mut res[i];
            *r = min(*r, s);
        }
    }

    let mut min = res[0];
    for res in res {
        min = std::cmp::min(min, res);
        println!("{}", min);
    }
}