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:
#pragma GCC optimize("Ofast")

#include <algorithm>
#include <iostream>
#include <vector>
#include <set>
using namespace std;

int main(){
  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];
  }
  const int INF = 1e9+1;
  int t = (1<<N)-1;
  vector<long long> ans(N+1,1e18);

  vector<vector<long long>> DX(1<<N,vector<long long>(N,INF)), DY = DX;
  for(int i = 0; i < (1<<N); ++i){
    vector<long long> XX = {-INF,0,INF};
    for(int l = 0; l < N; ++l){
      if(i>>l&1){
        XX.emplace_back(X[l]);
      }
    }
    sort(XX.begin(), XX.end());
    for(int l = 0; l < N; ++l){
      auto xlb = lower_bound(XX.begin(), XX.end(), X[l]);
      long long dx = (*xlb)-X[l];
      dx = min(dx,X[l]-(*prev(xlb)));
      DX[i][l] = dx;
    }
    
    vector<long long> YY = {-INF,0,INF};
    for(int l = 0; l < N; ++l){
      if(i>>l&1){
        YY.emplace_back(Y[l]);
      }
    }
    sort(YY.begin(), YY.end());
    long long s = 0;
    for(int l = 0; l < N; ++l){
      auto ylb = lower_bound(YY.begin(), YY.end(), Y[l]);
      long long dy = (*ylb)-Y[l];
      dy = min(dy,Y[l]-(*prev(ylb)));
      DY[i][l] = dy;
    }
  }
  
  for(int i = 0; i < (1<<N); ++i){
    int u = t^i;
    for(int k = u; ; k = (k-1)&u){

      int p = __builtin_popcount(i|k);
      long long s = 0;
      for(int l = 0; l < N; ++l){
        s += min(DX[i][l],DY[k][l])*P[l];
        if(s >= ans[p]) break;
      }
      ans[p] = min(ans[p],s);
      if(!k) break;
    }
  }
  for(auto e : ans)
    cout << e << "\n";
}