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<cstdio>
#include<algorithm>
#include<cmath>
typedef long long ll;
const int M = 17;
const ll inf = 0x3f3f3f3f3f3f3f;
int m;
int x[M],y[M],p[M];
ll ans[M];
int q[M],dis[M];
int popc[1 << M];
template<typename T>
inline void read(T &x){
    x = 0; char ch = getchar(); int f = 1;
    for(;ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') f = -1;
    for(;ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
    x *= f;
}
int main(){
    read(m);
    for(int i = 1;i < (1 << m); i++) popc[i] = popc[i >> 1] + (i & 1);
    for(int i = 1;i <= m; i++) read(x[i]), read(y[i]), read(p[i]), ans[i] = inf;
    ans[0] = inf;
    for(int i = 0,cnt;i < (1 << m); i++){
        cnt = 0;
        for(int j = 0;j < m; j++) if((i >> j) & 1) q[++cnt] = j + 1;
        for(int j = 0;j < (1 << cnt); j++){
            for(int k = 1;k <= m; k++) dis[k] = std::min(std::abs(x[k]),std::abs(y[k]));
            for(int k = 0;k < cnt; k++)
                if((j >> k) & 1)
                    for(int l = 1;l <= m; l++) dis[l] = std::min(dis[l],std::abs(x[l] - x[q[k + 1]]));
                else
                    for(int l = 1;l <= m; l++) dis[l] = std::min(dis[l],std::abs(y[l] - y[q[k + 1]]));
            ll sum = 0;
            for(int k = 1;k <= m; k++) sum += 1ll * dis[k] * p[k];
            ans[cnt] = std::min(ans[cnt],sum);
        }
    }
    for(int i = 0;i <= m; i++) printf("%lld\n",ans[i]);
    return 0;
}
/*
3
1 2 300
3 3 600
1 4 800

2900
900
0
0
*/