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>
using namespace std;

void solve(int N, vector<int> X, vector<int> Y, vector<int> P)
{
    map<int, int> mpx, mpy;
    for (int i = 0; i < N; i++)
    {
        mpx[X.at(i)] += P.at(i);
        mpy[Y.at(i)] += P.at(i);
    }
    vector<pair<int, int>> px, py;
    for (auto x : mpx)
    {
        px.push_back(make_pair(x.second, x.first));
    }
    sort(px.rbegin(), px.rend());
    for (auto y : mpy)
    {
        py.push_back(make_pair(y.second, y.first));
    }
    sort(py.rbegin(), py.rend());

    for (int n = 0; n <= N; n++)
    {
        int64_t res = INT64_MAX;
        for (int p = 0; p < (1 << n); p++)
        {
            vector<int> xs, ys;
            int idx_px = 0;
            int idx_py = 0;
            for (int i = 0; i < n; i++)
            {
                if ((p >> i) & 1)
                {
                    if (idx_px < (int)px.size())
                    {
                        xs.push_back(px.at(idx_px).second);
                        idx_px++;
                    }
                }
                else
                {
                    if (idx_py < (int)py.size())
                    {
                        ys.push_back(py.at(idx_py).second);
                        idx_py++;
                    }
                }
            }
            int64_t cost = 0;
            for (int i = 0; i < N; i++)
            {
                int cost_i = min(abs(X.at(i)), abs(Y.at(i)));
                for (auto x : xs)
                {
                    cost_i = min(cost_i, abs(X.at(i) - x));
                }
                for (auto y : ys)
                {
                    cost_i = min(cost_i, abs(Y.at(i) - y));
                }
                cost += (int64_t)cost_i * P.at(i);
            }
            res = min(res, cost);
        }
        cout << res << endl;
    }
}

int main()
{
    int N;
    cin >> N;
    vector<int> X(N);
    vector<int> Y(N);
    vector<int> P(N);
    for (int i = 0; i < N; i++)
    {
        cin >> X.at(i);
        cin >> Y.at(i);
        cin >> P.at(i);
    }
    solve(N, move(X), move(Y), move(P));
    return 0;
}