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;
using ull=unsigned long long;
using pii=pair<int,int>;

#define INF LONG_MAX
#define MOD 1000000007
#define rng(a) a.begin(),a.end()
#define rrng(a) a.end(),a.begin()
#define endl "\n"
#define int ll

signed main(){
  ios::sync_with_stdio(false);
  cin.tie(0);

  int N;
  cin>>N;
  vector<int>X(N),Y(N),P(N);
  for(int i=0;i<N;i++)cin>>X[i]>>Y[i]>>P[i];
  int bit2=pow(2,N);
  vector<vector<int>>tate(bit2,vector<int>(N));
  vector<vector<int>>yoko(bit2,vector<int>(N));
  for(int b=0;b<bit2;b++){
    for(int i=0;i<N;i++){
      int res=abs(X[i]);
      for(int j=0;j<N;j++){
        if(((b>>j)%2))res=min(res,abs(X[i]-X[j]));
      }
      tate[b][i]=res*P[i];
    }
    for(int i=0;i<N;i++){
      int res=abs(Y[i]);
      for(int j=0;j<N;j++){
        if(((b>>j)%2))res=min(res,abs(Y[i]-Y[j]));
      }
      yoko[b][i]=res*P[i];
    }
  }
  int bit=pow(3,N);
  vector<int>ANS(N+1,LONG_MAX);
  for(int b=0;b<bit;b++){//30000
    int s=0;
    int bbb=b;
    for(int j=0;j<N;j++){
      s+=(bbb%3!=0);
      bbb/=3;
    }
    int bb=b;
    int t=0,y=0,two=1;;
    for(int j=0;j<N;j++){//10
      if(bb%3==1)t+=two;
      if(bb%3==2)y+=two;
      bb/=3;
      two*=2;
    }
    int ans=0;
    for(int i=0;i<N;i++)ans+=min(tate[t][i],yoko[y][i]);
    ANS[s]=min(ans,ANS[s]);
  }
  for(auto it:ANS)cout<<it<<endl;
  return 0;
}