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;
#define SZ(a) ((int)(a).size())
#define PB push_back
#define MP make_pair
const int maxn = 200009;
const int MOD = 1e9 + 7;
int64_t X[1<<15][15],Y[1<<15][15];
int64_t AX[15],AY[15],AP[15],ans[16];
int main(){
#ifdef LOCAL
  freopen(".a.in", "r", stdin);
#endif
  ios_base::sync_with_stdio(false),cout.tie(0),cin.tie(0);
  int n;cin>>n;
  for(int i=0;i<n;++i)cin>>AX[i]>>AY[i]>>AP[i];
  for(int i=0;i<(1<<n);++i){
    for(int j=0;j<n;++j){
      X[i][j]=abs(AX[j]);
      for(int k=0;k<n;++k)if(i>>k&1)X[i][j]=min(X[i][j],abs(AX[k]-AX[j]));
      Y[i][j]=abs(AY[j]);
      for(int k=0;k<n;++k)if(i>>k&1)Y[i][j]=min(Y[i][j],abs(AY[k]-AY[j]));
    }
  }
  for(int i=0;i<=n;++i)ans[i]=1e18;
  for(int i=0;i<(1<<n);++i){
    int c=__builtin_popcount(i);
    for(int j=i;;j=(j-1)&i){
      int64_t tmp=0;
      for(int a=0;a<n;++a){
        tmp+=min(X[j][a],Y[j^i][a])*AP[a];
      }
      ans[c]=min(ans[c],tmp);
      if(j==0)break;
    }
  }
  for(int i=0;i<=n;++i)cout<<ans[i]<<'\n';
  
  return 0;
}
//2020.07.25 21:54:09