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: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (int)(n); i ++)
#define irep(i,n) for (int i = (int)(n) - 1;i >= 0;--i)
using namespace std;
using ll = long long;
using PL = pair<ll,ll>;
using P = pair<int,int>;
constexpr int INF = 1000000000;
constexpr long long HINF = 1000000000000000;
constexpr long long MOD = 1000000007;// = 998244353;
constexpr double EPS = 1e-4;
constexpr double PI = 3.14159265358979;

int main() {
    int N; cin >> N;
    vector<PL> XY(N);
    vector<ll> X,Y;
    vector<ll> P(N);
    rep(i,N) {
        cin >> XY[i].first >> XY[i].second >> P[i];
        X.push_back(XY[i].first);
        Y.push_back(XY[i].second);
    }
    sort(X.begin(),X.end());
    sort(Y.begin(),Y.end());
    X.erase(unique(X.begin(),X.end()),X.end());
    Y.erase(unique(Y.begin(),Y.end()),Y.end());

    int x = X.size(), y = Y.size();
    vector<ll> ans(N + 1,HINF);
    rep(i,N) {
        rep(bit1,1<<x) {
            vector<int> chose_x;
            rep(j,x) {
                if ((bit1>>j)&1) chose_x.push_back(j);
            }
            int ny = min((int)N - (int)chose_x.size(),(int)y);
            rep(bit2,1<<ny) {
                vector<int> chose_y;
                rep(j,ny) {
                    if ((bit2>>j)&1) chose_y.push_back(j);
                }
                if ((int)(chose_x.size() + chose_y.size()) > N) continue;
                int now = chose_x.size() + chose_y.size();
                ll ret = 0;
                rep(k,N) {
                    ll diff = min(abs(XY[k].first),abs(XY[k].second));
                    for (int x:chose_x) {
                        diff = min(diff,abs(X[x] - XY[k].first));
                    }
                    for (int y:chose_y) {
                        diff = min(diff,abs(Y[y] - XY[k].second));
                    }
                    ret += diff * P[k];
                }
                ans[now] = min(ans[now],ret);
            }
        }
    }
    rep(i,N + 1) cout << ans[i] << '\n';
    return 0;
}