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 "algorithm"
#include "cstring"

using namespace std;
using ll = long long;
int n,x[16],y[16],d[16],D[16],p[16];
ll ans[16];
int main() {
#ifdef LOCAL
    freopen("in1.txt", "r", stdin);
    freopen("out1.txt", "w", stdout);
#endif



    memset(ans , 0x3f , sizeof(ans));
    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++){
        D[i] = min(abs(x[i]) , abs(y[i]));
    }
    ans[0] = 0;
    for(int i=0;i<n;i++){
        ans[0] += (ll)D[i]*p[i];
    }
    for(register int i=1;i<(1<<n);i++){
        int j = i;
        do{
            memcpy(d,D,sizeof(D));
            for(register int k=0;k<n;k++){
                if(i>>k&1){
                    if(j>>k&1){
                        for(register int q=0;q<n;q++){
                            d[q] = min(d[q] , abs(x[q] - x[k]));
                        }
                    }
                    else{
                        for(register int q=0;q<n;q++){
                            d[q] = min(d[q] , abs(y[q] - y[k]));
                        }
                    }
                }
            }
            ll tmp = 0;
            for(register int i=0;i<n;i++) tmp += (ll)d[i] * p[i];
            ans[__builtin_popcount(i)] = min(ans[__builtin_popcount(i)] , tmp);
            j = (j-1) & i;
        }while(j != i);
    }
    for(register int i=0;i<=n;i++) printf("%lld\n",ans[i]);
    return 0;
}