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 REP(i,a,n) for(int i=(a); i<(int)(n); i++)
#define rep(i,n) REP(i,0,n)
#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin(); it!=(c).end(); ++it)
#define ALLOF(c) (c).begin(), (c).end()
typedef long long ll;
typedef unsigned long long ull;

static const ll INF = (ll)(1e18);

vector<ll> xx[(1<<15)+10];
vector<ll> yy[(1<<15)+10];

struct ST {
  ll x, y;
  ll p;
};

void prepare(int S, const vector<ST>& v){
  int N = v.size();
  {
    rep(i,N) xx[S].push_back(abs(v[i].x));
    rep(i,N){
      rep(j,N){
        if((S>>j)&1){
          xx[S][i] = min(xx[S][i], abs(v[i].x - v[j].x));
        }
      }
    }
  }
  {
    rep(i,N) yy[S].push_back(abs(v[i].y));
    rep(i,N){
      rep(j,N){
        if((S>>j)&1){
          yy[S][i] = min(yy[S][i], abs(v[i].y - v[j].y));
        }
      }
    }
  }
}

ll calc(const vector<ST>& v, int T, int R){
  int N = v.size();
  ll ret = 0;
  rep(i,N){
    ret += min(xx[T][i], yy[R][i]) * v[i].p;
  }
  return ret;
}


int main(){
  int N;
  cin >> N;

  vector<ST> v;
  rep(i,N){
    ST st;
    cin >> st.x >> st.y >> st.p;
    v.push_back(st);
  }

  for(int S=0; S<(1<<N); S++){
    prepare(S, v);
  }
  
  vector<ll> ret(N+1, INF);

  for(int S=0; S<(1<<N); S++){
    int cnt = 0;
    rep(i,N) if((S>>i)&1) cnt++;
    for(int T=S; T>=0; T--){
      T &= S;
      int R = S^T;
      ret[cnt] = min(ret[cnt], calc(v, T, R));
    }
  }

  rep(i,ret.size()){
    cout << ret[i] << endl;
  }

  return 0;
}