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 <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#include <set>
#include <tuple>
#define mkp make_pair
#define mkt make_tuple
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define all(v) v.begin(),v.end()
using namespace std;
typedef long long ll;
const ll MOD=1e9+7;
template<class T> void chmin(T &a,const T &b){if(a>b) a=b;}
template<class T> void chmax(T &a,const T &b){if(a<b) a=b;}

const ll INF=1e18;

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

  int N;
  cin>>N;
  vector<ll> X(N),Y(N),P(N);
  rep(i,N) cin>>X[i]>>Y[i]>>P[i];

  vector<vector<ll>> xmi((1<<N),vector<ll>(N,INF));
  rep(bit,(1<<N)){
      rep(i,N){
          rep(j,N){
              if(bit&(1<<j)) chmin(xmi[bit][i],abs(X[i]-X[j]));
          }
      }
  }
  vector<vector<ll>> ymi((1<<N),vector<ll>(N,INF));
  rep(bit,(1<<N)){
      rep(i,N){
          rep(j,N){
              if(bit&(1<<j)) chmin(ymi[bit][i],abs(Y[i]-Y[j]));
          }
      }
  }

  vector<ll> ans(31,INF);
  for(int xbit=0;xbit<(1<<N);xbit++){
    vector<ll> mires(N,INF);
    rep(i,N) mires[i]=min(abs(X[i]),abs(Y[i]));
    rep(i,N) chmin(mires[i],xmi[xbit][i]);

    int szx=__builtin_popcount(xbit);
    int T=0;
    rep(i,N){
        if(xbit&(1<<i));
        else T+=(1<<i);
    } 
    for(int S=T;;S=(S-1)&T){
      vector<ll> tres=mires;
      rep(i,N) chmin(tres[i],ymi[S][i]);
      int szy=__builtin_popcount(S);
      ll res=0;
      rep(i,N) res+=P[i]*tres[i];
      chmin(ans[szx+szy],res);
      if(S==0) break;
    }
  }

  rep(i,N+1) cout<<ans[i]<<endl;

  return 0;
}