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 {
   proconio::{fastout, input, marker::*},
   std::cmp::*,
   std::collections::*,
};

trait Monoid {
   fn unit() -> Self;
}
#[derive(Debug, Copy, Clone, PartialEq, Eq, PartialOrd, Ord)]
enum MinInt {
   Val(i64),
   Maximal,
}
impl MinInt {
   fn unwrap(self) -> i64 {
      if let Self::Val(x) = self {
         x
      } else {
         panic!();
      }
   }
}
impl std::ops::Add for MinInt {
   type Output = Self;
   fn add(self, other: Self) -> Self {
      match (self, other) {
         (Val(x), Val(y)) => Val(x + y),
         _ => self::MinInt::Maximal,
      }
   }
}
impl std::ops::Mul for MinInt {
   type Output = Self;
   fn mul(self, other: Self) -> Self {
      if self < other {
         self
      } else {
         other
      }
   }
}
impl Monoid for MinInt {
   fn unit() -> Self { MinInt::Maximal }
}
use MinInt::*;

fn set_cost(cost: &mut Vec<MinInt>, n: usize, xyp: &Vec<(i64, i64, i64)>) {
   let num = 1 << n;
   for s in 0..num {
      let mut idx = vec![];
      let mut tot = 0;
      for i in 0..n {
         if (s >> i) & 1 == 0 {
            continue;
         }
         idx.push(i);
         tot += xyp[i].2;
      }
      let mut cx: i64 = 0;
      let mut cy: i64 = 0;
      if idx.len() > 1 {
         idx.sort_by_key(|&i| xyp[i].0);
         let mut m = 0;
         let mut i = 0;
         while m < tot / 2 {
            m += xyp[idx[i]].2;
            i += 1;
         }
         let mx = xyp[idx[i - 1]].0;
         for (x, _, p) in idx.iter().map(|&i| xyp[i]) {
            cx += p * (x - mx).abs();
         }
         idx.sort_by_key(|&i| xyp[i].1);
         let mut m = 0;
         let mut i = 0;
         while m < tot / 2 {
            m += xyp[idx[i]].2;
            i += 1;
         }
         let my = xyp[idx[i - 1]].1;
         for (_, y, p) in idx.iter().map(|&i| xyp[i]) {
            cy += p * (y - my).abs();
         }
      }
      cost[s] = Val(min(cx, cy));
   }
}

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

   let num = 1 << n;
   let mut cost = vec![Maximal; num];
   set_cost(&mut cost, n, &xyp);

   let mut dp = vec![vec![Maximal; n + 1]; num];
   for s in 0..num {
      let mut c = 0i64;
      for i in 0..n {
         if (s >> i) & 1 == 1 {
            let (x, y, p) = xyp[i];
            c += p * min(x.abs(), y.abs());
         }
      }
      dp[s][0] = Val(c);
   }

   for i in 0..n {
      for t in 0..num {
         for s in t..num {
            dp[s][i + 1] = dp[s][i + 1] * (dp[s - t][i] + cost[t]);
         }
      }
   }
   for i in 0..=n {
      println!("{}", dp[num - 1][i].unwrap());
   }
}