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_macros, dead_code)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros, dead_code)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros, dead_code)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[allow(dead_code)]
struct UnionFind {
    parent: Vec<usize>,
    rank: Vec<usize>,
    size: Vec<usize>,
}

#[allow(dead_code)]
impl UnionFind {
    fn new(n: usize) -> UnionFind {
        let mut p = vec![0; n];
        for i in 0..n {
            p[i] = i;
        }
        return UnionFind {
            parent: p,
            rank: vec![0; n],
            size: vec![1; n],
        };
    }

    fn find(&mut self, x: usize) -> usize {
        if x == self.parent[x] {
            x
        } else {
            let p = self.parent[x];
            let pr = self.find(p);
            self.parent[x] = pr;
            pr
        }
    }

    fn same(&mut self, a: usize, b: usize) -> bool {
        self.find(a) == self.find(b)
    }

    fn unite(&mut self, a: usize, b: usize) {
        let a_root = self.find(a);
        let b_root = self.find(b);
        if a_root == b_root {
            return;
        }
        if self.rank[a_root] > self.rank[b_root] {
            self.parent[b_root] = a_root;
            self.size[a_root] += self.size[b_root];
        } else {
            self.parent[a_root] = b_root;
            self.size[b_root] += self.size[a_root];
            if self.rank[a_root] == self.rank[b_root] {
                self.rank[b_root] += 1;
            }
        }
    }

    fn get_size(&mut self, x: usize) -> usize {
        let root = self.find(x);
        self.size[root]
    }
}

const MOD_P: usize = 1000000007;

#[allow(dead_code)]
// fact(n) = n! mod p
fn fact(n: usize) -> usize {
    let mut acc = 1;
    for i in 1..n + 1 {
        acc = acc * i % MOD_P;
    }
    acc
}

#[allow(dead_code)]
fn mod_pow(b: usize, mut e: usize) -> usize {
    let mut base = b;
    let mut acc = 1;

    while e > 1 {
        if e % 2 == 1 {
            acc = acc * base % MOD_P;
        }
        e /= 2;
        base = base * base % MOD_P;
    }

    if e == 1 {
        acc = acc * base % MOD_P;
    }

    acc
}

#[allow(dead_code)]
fn comb(n: usize, r: usize) -> usize {
    // nCr = n! / (r! (n-r)!) = n! (r!)^(p-2) ((n-r)!)^(p-2)
    let x = ((n - r + 1)..(n + 1)).fold(1, |p, x| p * x % MOD_P);
    let y = mod_pow(fact(r), MOD_P - 2);
    x * y % MOD_P
}

#[derive(Clone, Copy, Debug)]
struct GF(usize);

impl std::ops::Add for GF {
    type Output = GF;
    fn add(self, rhs: GF) -> Self::Output {
        let mut d = self.0 + rhs.0;
        if d >= MOD_P {
            d -= MOD_P;
        }
        GF(d)
    }
}

impl std::ops::AddAssign for GF {
    fn add_assign(&mut self, rhs: GF) {
        *self = *self + rhs;
    }
}

impl std::ops::Sub for GF {
    type Output = GF;
    fn sub(self, rhs: GF) -> Self::Output {
        let mut d = self.0 + MOD_P - rhs.0;
        if d >= MOD_P {
            d -= MOD_P;
        }
        GF(d)
    }
}

impl std::ops::SubAssign for GF {
    fn sub_assign(&mut self, rhs: GF) {
        *self = *self - rhs;
    }
}

impl std::ops::Mul for GF {
    type Output = GF;
    fn mul(self, rhs: GF) -> Self::Output {
        let mut d = self.0 * rhs.0;
        d %= MOD_P;
        GF(d)
    }
}

impl std::ops::MulAssign for GF {
    fn mul_assign(&mut self, rhs: GF) {
        *self = *self * rhs;
    }
}

impl std::fmt::Display for GF {
    fn fmt<'a>(&self, f: &mut std::fmt::Formatter<'a>) -> std::fmt::Result {
        write!(f, "{}", self.0)
    }
}

#[allow(dead_code)]
impl GF {
    pub fn new(n: usize) -> GF {
        GF(n % MOD_P)
    }
    pub fn zero() -> GF {
        GF(0)
    }
    pub fn one() -> GF {
        GF(1)
    }
    pub fn pow(self, mut e: usize) -> GF {
        let mut acc = GF::one();
        let mut b = self;
        while e > 1 {
            if e % 2 == 1 {
                acc *= b;
            }
            b *= b;
            e /= 2;
        }
        if e == 1 {
            acc *= b;
        }
        acc
    }
    pub fn fact(self) -> GF {
        let mut acc = GF::one();
        for i in 1..=self.0 {
            acc *= GF::new(i);
        }
        acc
    }
    pub fn inv(self) -> GF {
        self.pow(MOD_P - 2)
    }
    pub fn comb(n: GF, r: GF) -> GF {
        // nCr = n! / (r! (n-r)!) = n! (r!)^(p-2) ((n-r)!)^(p-2)
        let x = ((n.0 - r.0 + 1)..=n.0).fold(GF::one(), |p, x| p * GF(x));
        let y = r.fact().inv();
        x * y
    }
}

#[allow(dead_code)]
#[derive(Debug)]
struct MemComb {
    inv: Vec<GF>,
    fact: Vec<GF>,
    factinv: Vec<GF>,
}

#[allow(dead_code)]
impl MemComb {
    pub fn new(n: usize) -> MemComb {
        let mut inv = vec![GF::one(); n + 1];
        let mut fact = vec![GF::one(); n + 1];
        let mut factinv = vec![GF::one(); n + 1];

        for i in 2..=n {
            fact[i] = fact[i - 1] * GF(i);
        }

        factinv[n] = fact[n].inv();
        for i in (1..n).rev() {
            factinv[i] = factinv[i + 1] * GF(i + 1); // 1/n! = 1/(n+1)! * (n+1)
            inv[i] = factinv[i] * fact[i - 1]; // 1/n = 1/n! * (n-1)!
        }
        inv[n] = factinv[n] * fact[n - 1];

        MemComb { inv, fact, factinv }
    }
    pub fn comb(&self, n: usize, r: usize) -> GF {
        if n >= r {
            self.fact[n] * self.factinv[r] * self.factinv[n - r]
        } else {
            GF::zero()
        }
    }
}

#[allow(dead_code)]
fn gcd(a: usize, b: usize) -> usize {
    if b > a {
        gcd(b, a)
    } else if a % b == 0 {
        b
    } else {
        gcd(b, a % b)
    }
}

#[allow(dead_code)]
fn getline() -> String {
    let mut __ret = String::new();
    std::io::stdin().read_line(&mut __ret).ok();
    return __ret;
}

#[allow(unused_imports)]
use std::cmp::{max, min};

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

    // distmat(x|y)[i][j]: 路線i を引いたときの集落j の歩行距離
    let mut distmatx = vec![vec![std::isize::MAX; n]; 1 << n];
    let mut distmaty = vec![vec![std::isize::MAX; n]; 1 << n];
    for i in 0..1 << n {
        for j in 0..n {
            let (x, y, _) = xyp[j];
            distmatx[i][j] = x.abs();
            distmaty[i][j] = y.abs();
            for k in 0..n {
                if i & (1 << k) != 0 {
                    distmatx[i][j] = min(distmatx[i][j], (x - xyp[k].0).abs());
                    distmaty[i][j] = min(distmaty[i][j], (y - xyp[k].1).abs());
                }
            }
        }
    }

    let mut ansv = vec![std::isize::MAX; n + 1];
    for i in 0..1 << n {
        let cnt = (i as u64).count_ones() as usize;

        for j in 0..=i {
            let j = j & i;
            let mut cost = 0;
            for k in 0..n {
                if i & (1 << k) == 0 {
                    cost += min(distmatx[j][k], distmaty[i - j][k]) * xyp[k].2;
                }
            }
            ansv[cnt] = min(ansv[cnt], cost);
        }
    }
    for i in 0..=n {
        println!("{}", ansv[i]);
    }
}