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 <bits/stdc++.h>
#define ll long long
using namespace std;

int main(){
  int n;
  cin>>n;
  ll vec[n][3];
  ll minabs[n];
  for(int i=0;i<n;i++){
    cin>>vec[i][0]>>vec[i][1]>>vec[i][2];
    minabs[i]=min(abs(vec[i][0]),abs(vec[i][1]));
  }
  ll ans[n+1];
  for(int i=0;i<=n;i++){
    ans[i]=1e13;
  }
  int vecb[n];
  int vecc[n];
  ll cost[n];
  ll sum,x;
  for(int bit=0;bit<(1<<n);bit++){
    int t=0;
    for(int i=0;i<n;i++){
      if(bit&(1<<i)){
        vecb[i]=1;
        vecc[t]=i;
        t++;
      }
      else{
        vecb[i]=0;
      }
    }
    if(t==n){
      ans[t]=0;
      continue;
    }
    for(int i=0;i<n;i++){
      cost[i]=minabs[i];
    }
    /*x=0;
    for(int i=0;i<n;i++){
      if(vecb[i]==1){
        vecc[x]=i;
        x++;
      }
    }*/
    for(int bitb=0;bitb<(1<<t);bitb++){
      for(int i=0;i<t;i++){
        if(bitb&(1<<i)){ //縦に引く
          for(int j=0;j<n;j++){
            cost[j]=min(cost[j],abs(vec[j][0]-vec[vecc[i]][0]));
          }
        }
        else{ //横に引く
          for(int j=0;j<n;j++){
            cost[j]=min(cost[j],abs(vec[j][1]-vec[vecc[i]][1]));
          }
        }
      }
      sum=0;
      for(int i=0;i<n;i++){
        sum+=cost[i]*vec[i][2];
        cost[i]=minabs[i];
      }
      ans[t]=min(ans[t],sum);
    }
  }
  
  for(int i=0;i<n+1;i++){
    cout<<ans[i]<<endl;
  }
}