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 C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using Int = long long;  // clang-format off
#define REP_(i, a_, b_, a, b, ...) for (Int i = (a), lim##i = (b); i < lim##i; i++)
#define REP(i, ...) REP_(i, __VA_ARGS__, __VA_ARGS__, 0, __VA_ARGS__)
#define ALL(v) std::begin(v), std::end(v)
struct SetupIO { SetupIO() { std::cin.tie(nullptr), std::ios::sync_with_stdio(false), std::cout << std::fixed << std::setprecision(13); } } setup_io;
#ifndef dump
#define dump(...)
#endif  // clang-format on

/**
 *    author:  knshnb
 *    created: Sat Jul 25 22:14:29 JST 2020
 **/

template <class T> bool chmin(T& a, const T& b) { return a > b ? a = b, true : false; }
template <class T> bool chmax(T& a, const T& b) { return a < b ? a = b, true : false; }

signed main() {
    const Int INF = 1e8;
    Int n;
    std::cin >> n;
    std::vector<Int> x(n), y(n), p(n);
    REP(i, n) std::cin >> x[i] >> y[i] >> p[i];
    std::vector<Int> ans(n + 1, 1e18);
    REP(S, 1 << n) {
        // Sの部分集合の列挙
        for (int T = S;; T = (T - 1) & S) {
            // 操作
            std::vector<Int> xs, ys;
            REP(i, n) {
                if (S >> i & 1) {
                    if (T >> i & 1) {
                        ys.push_back(y[i]);
                    } else {
                        xs.push_back(x[i]);
                    }
                }
            }
            // std::sort(ALL(xs));
            // std::sort(ALL(ys));
            Int acc = 0;
            REP(i, n) {
                if (S >> i & 1) continue;
                Int mi = std::min(std::abs(x[i]), std::abs(y[i]));
                for (Int a : xs) {
                    chmin(mi, std::abs(x[i] - a));
                }
                for (Int b : ys) {
                    chmin(mi, std::abs(y[i] - b));
                }
                // auto it1 = std::lower_bound(ALL(xs), x[i]);
                // auto it2 = std::lower_bound(ALL(ys), y[i]);
                // acc += p[i] * std::min({*it1 - x[i], x[i] - *std::prev(it1), *it2 - y[i], y[i] - *std::prev(it2)});
                acc += p[i] * mi;
            }
            chmin(ans[__builtin_popcount(S)], acc);
            if (T == 0) break;
        }
    }
    for (Int z : ans) std::cout << z << "\n";
}