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: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
#define REP(i,n) for (ll i = 0; i < (n); ++i)
#define RREP(i, n) for (ll i = (n) - 1; i >= 0; --i)
#define ALL(v) (v).begin(), (v).end()
template<class T>
using reverse_priority_queue = priority_queue<T, vector<T>, greater<T>>;


int main(){
    ll n;
    cin >> n;
    vector<ll> x(n), y(n), p(n);
    REP(i, n){
        cin >> x[i] >> y[i] >> p[i];
    }

    vector<vector<ll>> x2d(1LL << n, vector<ll>(n)), y2d(1LL << n, vector<ll>(n));
    REP(i, (1LL << n)){
        REP(j, n){
            ll dx = abs(x.at(j)), dy = abs(y.at(j));
            REP(k, n){
                if((i >> k) & 1){
                    dx = min(dx, abs(x.at(j) - x.at(k)));
                    dy = min(dy, abs(y.at(j) - y.at(k)));
                }
            }
            x2d.at(i).at(j) = dx;
            y2d.at(i).at(j) = dy;
        }
    }

    ll ub = 1;
    REP(i, n){ ub *= 3; }

    vector<ll> ans(n + 1, numeric_limits<ll>::max());

    REP(i, ub){
        ll ix = 0, iy = 0, k = 0;
        {
            ll m = i;
            REP(j, n){
                const ll l = m % 3;
                if(l == 1){
                    ix += (1LL << j);
                    k++;
                }else if(l == 2){
                    iy += (1LL << j);
                    k++;
                }
                m /= 3;
            }
        }

        ll s = 0;
        const auto& dx = x2d.at(ix), dy = y2d.at(iy);
        REP(j, n){
            s += p.at(j) * min(dx.at(j), dy.at(j));
        }

        ans.at(k) = min(ans.at(k), s);
    }

    for(auto ai : ans){
        cout << ai << endl;
    }
    return 0;
}