#include<bits/stdc++.h>
#define int long long
using namespace std;
struct T{int X,Y,P;};
T A[20];
set<pair<int,int>> S;
int N,ans[100],h[1<<20][20],w[1<<20][20];
void a(int n,int x,int y,int k){
  if(n==N){
    int cnt=0;
    for(int i=0;i<N;i++)
      cnt+=min(h[x][i],w[y][i])*A[i].P;
    ans[S.size()-2]=min(ans[S.size()-2],cnt);
    return;
  }
  int s=S.size();
  S.insert({0,A[n].X});
  a(n+1,x+k,y,k*2);
  if(s!=S.size())S.erase({0,A[n].X});
  s=S.size();
  S.insert({1,A[n].Y});
  a(n+1,x,y+k,k*2);
  if(s!=S.size())S.erase({1,A[n].Y});
  a(n+1,x,y,k*2);
}
signed main(){
  S.insert({0,0});
  S.insert({1,0});
  cin>>N;
  for(int i=0;i<N;i++)
    cin>>A[i].X>>A[i].Y>>A[i].P;
  for(int bit=0;bit<(1<<N);bit++){
    for(int i=0;i<N;i++)h[bit][i]=abs(A[i].X),w[bit][i]=abs(A[i].Y);
    bitset<20> B(bit);
    for(int i=0;i<N;i++)
      for(int j=0;j<N;j++)
        if(B.test(j)){
          h[bit][i]=min(h[bit][i],abs(A[i].X-A[j].X));
          w[bit][i]=min(w[bit][i],abs(A[i].Y-A[j].Y));
        }
  }
  for(int i=0;i<=N;i++)ans[i]=1000000000000000000;
  a(0,0,0,1);
  int n=1000000000000000000;
  for(int i=0;i<=N;i++){
    n=min(n,ans[i]);
    cout<<n<<endl;
  }
}
### Prompt

Your challenge is to write 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 int long long
using namespace std;
struct T{int X,Y,P;};
T A[20];
set<pair<int,int>> S;
int N,ans[100],h[1<<20][20],w[1<<20][20];
void a(int n,int x,int y,int k){
  if(n==N){
    int cnt=0;
    for(int i=0;i<N;i++)
      cnt+=min(h[x][i],w[y][i])*A[i].P;
    ans[S.size()-2]=min(ans[S.size()-2],cnt);
    return;
  }
  int s=S.size();
  S.insert({0,A[n].X});
  a(n+1,x+k,y,k*2);
  if(s!=S.size())S.erase({0,A[n].X});
  s=S.size();
  S.insert({1,A[n].Y});
  a(n+1,x,y+k,k*2);
  if(s!=S.size())S.erase({1,A[n].Y});
  a(n+1,x,y,k*2);
}
signed main(){
  S.insert({0,0});
  S.insert({1,0});
  cin>>N;
  for(int i=0;i<N;i++)
    cin>>A[i].X>>A[i].Y>>A[i].P;
  for(int bit=0;bit<(1<<N);bit++){
    for(int i=0;i<N;i++)h[bit][i]=abs(A[i].X),w[bit][i]=abs(A[i].Y);
    bitset<20> B(bit);
    for(int i=0;i<N;i++)
      for(int j=0;j<N;j++)
        if(B.test(j)){
          h[bit][i]=min(h[bit][i],abs(A[i].X-A[j].X));
          w[bit][i]=min(w[bit][i],abs(A[i].Y-A[j].Y));
        }
  }
  for(int i=0;i<=N;i++)ans[i]=1000000000000000000;
  a(0,0,0,1);
  int n=1000000000000000000;
  for(int i=0;i<=N;i++){
    n=min(n,ans[i]);
    cout<<n<<endl;
  }
}
```