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>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
vector<int> f(int x,int n){
  vector<int> r(n);
  for(int i=0;i<n;i++){
    r[i]=x%3;
    x=x/3;
  }
  return r;
}

int main(){
  int N;
  scanf("%d",&N);
  vector<vector<long long>> L(N,vector<long long>(3));
  for(int i=0;i<N;i++){
    scanf("%lld%lld%lld",&L[i][0],&L[i][1],&L[i][2]);
  }
  long long INF=10000000000000000;
  vector<int> D(21);
  vector<int> T(21);
  D[0]=1;
  T[0]=1;
  for(int i=0;i<20;i++){
    D[i+1]=D[i]*2;
    T[i+1]=T[i]*3;
  }
  vector<vector<long long>> X(D[N],vector<long long>(N,INF));
  vector<vector<long long>> Y(D[N],vector<long long>(N,INF));
  for(int i=0;i<D[N];i++){
    for(int j=0;j<N;j++){
      X[i][j]=min(abs(L[j][0]),abs(L[j][1]));
      Y[i][j]=X[i][j];
      for(int k=0;k<N;k++){
        if(i&D[k]){
          X[i][j]=min(X[i][j],abs(L[j][0]-L[k][0]));
          Y[i][j]=min(Y[i][j],abs(L[j][1]-L[k][1]));
        }
      }
      X[i][j]*=L[j][2];
      Y[i][j]*=L[j][2];
    }
  }
  vector<int> s;
  int t;
  long long u;
  vector<long long> P(N+1,INF);
  int x,y;
  for(int i=0;i<T[N];i++){
    s=f(i,N);
    x=0;
    y=0;
    for(int j=0;j<N;j++){
      if(s[j]==1){
        x+=D[j];
      }else if(s[j]==2){
        y+=D[j];
      }
    }
    t=N-count(s.begin(),s.end(),0);
    u=0;
    for(int j=0;j<N;j++){
      u+=min(X[x][j],Y[y][j]);
    }
    P[t]=min(P[t],u);
  }
  for(int i=0;i<=N;i++){
    printf("%lld\n",P[i]);
  }
}