// I SELL YOU...! 
#include<iostream>
#include<vector>
#include<algorithm>
#include<functional>
#include<queue>
#include<chrono>
#include<iomanip>
#include<map>
#include<set>
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
using TP = tuple<ll,ll,ll>;
void init_io(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(18);
}
ll n;
const ll MAX =17;
const ll INF = 1e18;
vector<ll> ans(MAX,INF);
vector<ll> x(MAX),y(MAX),p(MAX);
void solve(ll t,ll c,vector<ll> &xv,vector<ll> &yv){
  if(t==n){
    ll sum = 0;
    for(int i=0;i<n;i++){
      ll tmp = min(abs(x[i]),abs(y[i]));
      for(auto lx: xv){
        tmp = min(tmp,abs(x[i]-lx));
      }
      for(auto ly: yv){
        tmp = min(tmp,abs(y[i]-ly));
      }
      sum += tmp*p[i];
    }
    ans[c] = min(ans[c],sum);
    return;
  }
  xv.push_back(x[t]);
  solve(t+1,c+1,xv,yv);
  xv.pop_back();

  yv.push_back(y[t]);
  solve(t+1,c+1,xv,yv);
  yv.pop_back();

  solve(t+1,c,xv,yv);
}
signed main(){
  init_io();
  cin >> n;
  for(int i=0;i<n;i++){
    cin >> x[i] >> y[i] >> p[i];
  }
  vector<ll> tx,ty;
  solve(0,0,tx,ty);
  for(int i=0;i<=n;i++){
    cout << ans[i]<<endl;
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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

### Response

```cpp
// I SELL YOU...! 
#include<iostream>
#include<vector>
#include<algorithm>
#include<functional>
#include<queue>
#include<chrono>
#include<iomanip>
#include<map>
#include<set>
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
using TP = tuple<ll,ll,ll>;
void init_io(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(18);
}
ll n;
const ll MAX =17;
const ll INF = 1e18;
vector<ll> ans(MAX,INF);
vector<ll> x(MAX),y(MAX),p(MAX);
void solve(ll t,ll c,vector<ll> &xv,vector<ll> &yv){
  if(t==n){
    ll sum = 0;
    for(int i=0;i<n;i++){
      ll tmp = min(abs(x[i]),abs(y[i]));
      for(auto lx: xv){
        tmp = min(tmp,abs(x[i]-lx));
      }
      for(auto ly: yv){
        tmp = min(tmp,abs(y[i]-ly));
      }
      sum += tmp*p[i];
    }
    ans[c] = min(ans[c],sum);
    return;
  }
  xv.push_back(x[t]);
  solve(t+1,c+1,xv,yv);
  xv.pop_back();

  yv.push_back(y[t]);
  solve(t+1,c+1,xv,yv);
  yv.pop_back();

  solve(t+1,c,xv,yv);
}
signed main(){
  init_io();
  cin >> n;
  for(int i=0;i<n;i++){
    cin >> x[i] >> y[i] >> p[i];
  }
  vector<ll> tx,ty;
  solve(0,0,tx,ty);
  for(int i=0;i<=n;i++){
    cout << ans[i]<<endl;
  }
}

```