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:
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <array>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
using ll = long long;

template <class T>
void chmin(T &a, T b) {
    if (b < a) a = b;
}

struct P {
    int x[2], z;
};

int main() {
    int n;
    cin >> n;

    P p[15];
    for (int i = 0; i < n; i++) {
        int x, y, z;
        cin >> x >> y >> z;

        p[i] = { x, y, z };
    }

    vector<array<array<int, 2>, 15>> v(1 << n);
    for (int i = 0; i < n; i++) {
        for (int h = 0; h < 2; h++) {
            v[0][i][h] = min(abs(p[i].x[0]), abs(p[i].x[1]));
        }
    }
    for (int k = 1; k < 1 << n; k++) {
        int i = __builtin_ctz(k);
        auto u = v[k ^ 1 << i];
        for (int j = 0; j < n; j++) {
            for (int h = 0; h < 2; h++) {
                chmin(u[j][h], abs(p[i].x[h] - p[j].x[h]));
            }
        }
        v[k] = u;
    }

    ll r[15 + 1];
    for (int i = 0; i <= n; i++) {
        r[i] = 1LL << 60;
    }

    for (int k = 0; k < 1 << n; k++) {
        for (int l = k; l >= 0; l--) {
            l &= k;
            int m = k ^ l;

            ll s = 0;
            for (int i = 0; i < n; i++) {
                s += (ll)min(v[l][i][0], v[m][i][1]) * p[i].z;
            }
            chmin(r[__builtin_popcount(k)], s);
        }
    }

    for (int i = 0; i <= n; i++) {
        cout << r[i] << endl;
    }

    return 0;
}