Problem: 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.

Examples

Input

3
1 2 300
3 3 600
1 4 800


Output

2900
900
0
0


Input

5
3 5 400
5 3 700
5 5 1000
5 7 700
7 5 400


Output

13800
1600
0
0
0
0


Input

6
2 5 1000
5 2 1100
5 5 1700
-2 -5 900
-5 -2 600
-5 -5 2200


Output

26700
13900
3200
1200
0
0
0


Input

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


Output

2576709
1569381
868031
605676
366338
141903
0
0
0

Solution (in CPP):

Solution:
#ifdef _MSC_VER
#include <__msvc_all_public_headers.hpp>
#undef min
#undef max
#else
#define NDEBUG
#include <bits/stdc++.h>
#endif

using namespace std;
using large = long long;

struct Ville {
    large x, y;
    large p;

    large near_x, near_y;
};

int nx;
int xs[20];
int ny;
int ys[20];

static large solve(Ville* const v, const int n, const int start, const int k) {
    if (k == 0) {
        auto sum = 0LL;
        for (int i = 0; i < n; ++i) {
            auto xmin = LLONG_MAX;
            auto ymin = LLONG_MAX;
            for (int ix = 0; ix < nx; ++ix)
                xmin = min(xmin, abs(v[i].x - xs[ix]));
            for (int iy = 0; iy < ny; ++iy)
                ymin = min(ymin, abs(v[i].y - ys[iy]));
            sum += v[i].p * min(xmin, ymin);
        }
        return sum;
    }

    auto score = LLONG_MAX;
    for (int i = start; i + k <= n; ++i) {
        // horizontal
        xs[nx++] = v[i].x;
        auto case1 = solve(v, n, i + 1, k - 1);
        nx--;
        // vertical
        ys[ny++] = v[i].y;
        auto case2 = solve(v, n, i + 1, k - 1);
        ny--;

        score = min({score, case1, case2});
    }
    return score;
}

int main()
{
    int n;
    cin >> n;
    auto ville = vector<Ville>();
    for (auto i = 0; i < n; ++i) {
        auto v = Ville{};
        cin >> v.x >> v.y >> v.p;
        ville.push_back(v);
    }


    xs[nx++] = 0;
    ys[ny++] = 0;
    for (int k = 0; k < n; ++k) {
        cout << solve(ville.data(), n, 0, k) << "\n";
    }

    for (int k = n; k <= n; ++k) {
        cout << "0\n";
    }

    return 0;
}