#include <bits/stdc++.h>
#define rep(i,n)for(long long i=0;i<(long long)(n);++i)
#define all(a) a.begin(), a.end()
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
 
const ll MOD=1e9+7;
const ll INF=1e18;
const int MAX=510000;
const double pi=acos(-1);
int dx[8] = {1,0,-1,0,1,1,-1,-1};
int dy[8] = {0,1,0,-1,-1,1,1,-1};
// int d[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false); 
  
  int n;
  cin >> n;
  vector<ll>x(n),y(n),p(n); 
  rep(i,n) cin >> x[i] >> y[i] >> p[i]; 
  vector<ll>ans(n+1,1e18);
  
  //前計算パート．
  //縦の路線の選び方をbitで持ったときの集落iの最短移動距離をdistx[bit][i]とする．
  vector<vector<ll>> distx(1<<n,vector<ll>(n));
  vector<vector<ll>> disty(1<<n,vector<ll>(n));
  for(int bit=0;bit<(1<<n);bit++){
    rep(i,n){
      //大通りとの距離
      distx[bit][i]=abs(x[i]);
      disty[bit][i]=abs(y[i]);
      rep(j,n){
        if(bit&(1<<j)){
          distx[bit][i]=min(distx[bit][i],abs(x[i]-x[j]));
          disty[bit][i]=min(disty[bit][i],abs(y[i]-y[j]));
        }
      }
    }
  }
  for(int bit=0;bit<(1<<n);++bit){
    int cnt = __builtin_popcount(bit);
    for(int i=bit;i>=0;--i){
      i&=bit;//bitで表せる選び方の部分集合i
      ll cost = 0;
      rep(j,n){
        if(!(bit&(1<<j))){
          // bitに含まれる場合は距離0
          cost += min(distx[i][j],disty[bit-i][j])*p[j];
        }
      }
      ans[cnt]=min(ans[cnt],cost);
    }
  }
  rep(i,n+1) cout << ans[i] << endl;
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
#define rep(i,n)for(long long i=0;i<(long long)(n);++i)
#define all(a) a.begin(), a.end()
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
 
const ll MOD=1e9+7;
const ll INF=1e18;
const int MAX=510000;
const double pi=acos(-1);
int dx[8] = {1,0,-1,0,1,1,-1,-1};
int dy[8] = {0,1,0,-1,-1,1,1,-1};
// int d[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false); 
  
  int n;
  cin >> n;
  vector<ll>x(n),y(n),p(n); 
  rep(i,n) cin >> x[i] >> y[i] >> p[i]; 
  vector<ll>ans(n+1,1e18);
  
  //前計算パート．
  //縦の路線の選び方をbitで持ったときの集落iの最短移動距離をdistx[bit][i]とする．
  vector<vector<ll>> distx(1<<n,vector<ll>(n));
  vector<vector<ll>> disty(1<<n,vector<ll>(n));
  for(int bit=0;bit<(1<<n);bit++){
    rep(i,n){
      //大通りとの距離
      distx[bit][i]=abs(x[i]);
      disty[bit][i]=abs(y[i]);
      rep(j,n){
        if(bit&(1<<j)){
          distx[bit][i]=min(distx[bit][i],abs(x[i]-x[j]));
          disty[bit][i]=min(disty[bit][i],abs(y[i]-y[j]));
        }
      }
    }
  }
  for(int bit=0;bit<(1<<n);++bit){
    int cnt = __builtin_popcount(bit);
    for(int i=bit;i>=0;--i){
      i&=bit;//bitで表せる選び方の部分集合i
      ll cost = 0;
      rep(j,n){
        if(!(bit&(1<<j))){
          // bitに含まれる場合は距離0
          cost += min(distx[i][j],disty[bit-i][j])*p[j];
        }
      }
      ans[cnt]=min(ans[cnt],cost);
    }
  }
  rep(i,n+1) cout << ans[i] << endl;
  return 0;
}

```