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 <iostream>
#include <cstdio>
#include <vector>
#include <iomanip>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <map>
#include <set>
#include <cmath>
using namespace std;

using VLL = vector <long long>;
using VPILL = vector < pair<long long, int> >;
int N;
VLL X;
VLL Y;
VLL P;
VLL ans;
VPILL PX;
VPILL PY;
int xcho;
int ycho;

const long long INF = 1000000000LL * 1000000000LL;

void solve(int now, int chosen) {
    if (now == N) {
        VLL minDist(N, INF);

        long long res = 0;
        long long lastX = -INF;
        long long lastY = -INF;

        for (int i = 0; i < N; ++i) {
            // default
            minDist[i] = min(minDist[i], min(abs(X[i]), abs(Y[i])));

            // x axis
            if (xcho & (1 << PX[i].second)) {
                lastX = PX[i].first;
            }

            minDist[PX[i].second] = min(minDist[PX[i].second], PX[i].first-lastX);

            // y axis
            if (ycho & (1 << PY[i].second)) {
                lastY = PY[i].first;
            }

            minDist[PY[i].second] = min(minDist[PY[i].second], PY[i].first-lastY);
        }

        // backward
        lastX = INF;
        lastY = INF;

        for (int i = N-1; i >= 0; --i) {
            // x axis
            if (xcho & (1 << PX[i].second)) {
                lastX = PX[i].first;
            }

            minDist[PX[i].second] = min(minDist[PX[i].second], lastX-PX[i].first);

            // y axis
            if (ycho & (1 << PY[i].second)) {
                lastY = PY[i].first;
            }

            minDist[PY[i].second] = min(minDist[PY[i].second], lastY - PY[i].first);
        }

        for (int i = 0; i < N; ++i) {
            res += P[i] * minDist[i];
        }

        ans[chosen] = min(ans[chosen], res);
        return ;
    }

    solve(now+1, chosen);

    xcho |= (1<<now);
    solve(now+1, chosen+1);
    xcho &= ~(1<<now);

    ycho |= (1<<now);
    solve(now+1, chosen+1);
    ycho &= ~(1<<now);
}

int main() {
    cin >> N;

    X = VLL(N);
    Y = VLL(N);
    P = VLL(N);

    PX = VPILL(N);
    PY = VPILL(N);

    for (int i = 0; i < N; ++i) {
        cin >> X[i] >> Y[i] >> P[i];
        PX[i] = make_pair(X[i], i);
        PY[i] = make_pair(Y[i], i);
    }

    ans = VLL(N+1, INF);
    xcho = 0;
    ycho = 0;

    sort(PX.begin(), PX.end());
    sort(PY.begin(), PY.end());
    solve(0, 0);

    for (int k = 0; k <= N; ++k) {
        cout << ans[k] << endl;
    }

    return 0;
}