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>
#define rep(i,n) for (ll i = 0; i < n; ++i)
using namespace std;
typedef long long ll;
typedef pair<ll,int> P;
const ll INF=1e18;

int main() {
  ll n;
  cin>>n;
  vector<ll> x(n);
  vector<ll> y(n);
  vector<ll> p(n);
  rep(i,n){
    ll x1,y1,p1;
    cin>>x1>>y1>>p1;
    x[i]=x1;y[i]=y1;p[i]=p1;
  }

  vector<ll> ans(n+1);
  rep(i,n+1)ans[i]=INF;
  vector<vector<ll>> xmin((1<<n), vector<ll>(n));
  vector<vector<ll>> ymin((1<<n), vector<ll>(n));
  rep(i,(1<<n)){
    rep(k,n){//各てん
      ll xnow=abs(x[k]); ll ynow=abs(y[k]);

      rep(j,n){

        if (!((i>>j)&1))continue;
      //  cout<<"j"<<j<<endl;

        xnow=min(xnow,abs(x[k]-x[j]));
        ynow=min(ynow,abs(y[k]-y[j]));

      }
  //    cout<<"i"<<i<<"k"<<k<<endl;
    //  cout<<"xnow"<<xnow<<endl;
    //  cout<<"ynow"<<ynow<<endl;

      xmin[i][k]=xnow;
      ymin[i][k]=ynow;
    }
  }


  rep(i,(1<<n)){
  //  cout<<"i"<<i<<endl;
    ll cnt=0;
    rep(j,n){
      if ((i>>j)&1) cnt++;
    }
  //  cout<<"cnt"<<cnt<<endl;
    for(ll j=i;j>=0;j--){
      j&=i;
  //    cout<<"j"<<j<<endl;
      ll a=0;
      rep(k,n){
        ll temp=min(xmin[i-j][k],ymin[j][k]);
        a+=temp*p[k];
      }

      ans[cnt]=min(ans[cnt],a);
  //    cout<<ans[cnt]<<endl;

    }
  }

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


}