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<bits/stdc++.h>
using namespace std;
using ll = long long;
const int nax = 16;
int x[nax], y[nax], p[nax];
int n;
ll d[nax][2][1 << nax];

ll res[nax];
/// mimimum distance distance (x or y) to nearest line
/// denote that line pass through point in mask

inline void upd_min (ll &a, ll b) {
    a = min(a, b);
}

int main () {
    scanf("%d", &n);

    for (int i = 0 ; i < n ; ++ i)
        scanf("%d %d %d", x + i, y + i, p + i);

    for (int i = 0 ; i < n ; ++ i) {
        ll base = 1LL * p[i] * min(abs(x[i]) , abs(y[i]));
        for (int msk = 0 ; msk < (1 << n); ++ msk) {
            d[i][0][msk] = d[i][1][msk] = base;
            for (int j = 0 ; j < n ; ++ j) {
                if (msk >> j & 1) {
                    upd_min(d[i][0][msk], 1LL * p[i] * abs(x[j] - x[i]));
                    upd_min(d[i][1][msk], 1LL * p[i] * abs(y[j] - y[i]));
                }
            }
        }
    }

    fill(res, res + nax, LLONG_MAX);

    for (int m = 0 ; m < (1 << n) ; ++ m) {
        ll cur = LLONG_MAX;
        for (int sub = m; ; sub = (sub - 1) & m) {
            ll sum = 0;
            for (int k = 0 ; k < n ; ++ k)
                sum += min(d[k][0][sub], d[k][1][m ^ sub]);
            cur = min(cur, sum);
            if (sub == 0) break;
        }
        int pcnt = __builtin_popcount(m);
        res[pcnt] = min(res[pcnt], cur);
    }


    for (int i = 0 ; i <= n ; ++ i)
        printf("%lld\n", res[i]);


}
/*
    Good Luck
        -Lucina
*/