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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cassert>

using namespace std;

#define print(x) cout << x << endl
#define input(x) cin >> x

typedef long long llint;

const llint INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int INF = 0x3f3f3f3f;
const int N = 18;

struct Point {
    int x, y, p;
};

llint dp1[N][N];
llint dp2[N][N];

llint calc(const vector<Point>& ps, const vector<int>& disx, int l, int r) {
    if (l == r) {
        return 0;
    }
    int n = ps.size();
    if (dp2[l][r] < INFLL) {
        return dp2[l][r];
    }
    llint tot = 0;
    for (int i = l + 1; i <= r; i++) {
        if (i <= 0 || i > n) {
            continue;
        }
        int ll = l > 0?  ps[l - 1].y: -INF;
        int rr = r <= n? ps[r - 1].y: INF;

        int mini = min(ps[i - 1].y - ll, rr - ps[i - 1].y);
        mini = min(mini, disx[i - 1]);
        mini = min(mini, abs(ps[i - 1].y));

        tot += 1LL * mini * ps[i - 1].p;
    }

    return dp2[l][r] = tot;
}

void solve(const vector<Point>& ps, vector<int>& dis, vector<llint>& ans, int status) {
    int n = ps.size();
    int cntx = 0;
    for (int i = 0; i < n; i++) {
        if (status & (1 << i)) {
            cntx++;
        }
    }

    {
        llint tot = 0;
        for (int j = 0; j < n; j++) {
            tot += 1LL * dis[j] * ps[j].p;
        }
        ans[cntx] = min(ans[cntx], tot);
    }

    int cnty = n - cntx;

    memset(dp1, 0xFF & INFLL, sizeof(dp1));
    memset(dp2, 0xFF & INFLL, sizeof(dp2));
    dp1[0][0] = 0;

    /*
    for (int i = 1; i <= n; i++) {
        dp1[i][0] = dp1[i - 1][0] + 1LL * dis[i - 1] * ps[i - 1].p;
    }
    */

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= cnty; j++) {
            for (int k = 0; k <= i - 1; k++) {
                dp1[i][j] = min(dp1[i][j], dp1[k][j - 1] + calc(ps, dis, k, i));
            }
        }
    }
    /*
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= i; j++) {
            printf("%lld ", dp1[i][j]);
        }
        puts("");
    }
    puts("--");
    */

    // print(status);
    for (int i = 1; i <= cnty; i++) {
        for (int j = i; j <= n; j++) {
            ans[i + cntx] = min(ans[i + cntx], dp1[j][i] + calc(ps, dis, j, n + 1));
            // printf("%d %d, %lld %lld\n", j, i, dp1[j][i], calc(ps, dis, j, n + 1));
        }
    }
    // puts("--");
}

int main() {
    int n;
    vector<Point> ps;

    input(n);

    int x, y, p;
    for (int i = 0; i < n; i++) {
        scanf("%d%d%d", &x, &y, &p);
        ps.push_back({x, y, p});
    }

    vector<llint> ans(n + 1, INFLL);

    sort(ps.begin(), ps.end(), [](const Point& pa, const Point& pb) {
        return pa.y < pb.y;
    });

    for (int i = 0; i < (1 << n); i++) {
        vector<int> dis(n);

        for (int j = 0; j < n; j++) {
            dis[j] = min(abs(ps[j].y), abs(ps[j].x));
        }

        for (int j = 0; j < n; j++) {
            if ((i & (1 << j)) == 0) {
                continue;
            }
            for (int k = 0; k < n; k++) {
                dis[k] = min(dis[k], abs(ps[j].x - ps[k].x));
            }
        }

        solve(ps, dis, ans, i);
    }

    for (auto num: ans) {
        printf("%lld\n", num);
    }

    return 0;
}