#include <bits/stdc++.h>
#include <stdlib.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<ll,ll> llP;

ll mod(ll a, ll b){
  ll ret=a%b;
  if(ret<0) ret+=b;
  return ret;
}

ll modpow(ll a,ll b,ll c){
  ll res=1;
  while(b>0){
    if(b&1) res=mod(res*a,c);
    a=mod(a*a,c);
    b>>=1;
  }
  return res;
}

int n;
ll ans[16];
int cur[16];
int x[16],y[16],p[16];
int minx[40000][16],miny[40000][16];

void dfs(int m,int rx,int ry,int kisuu,int kazu){
  if(m==n){
    ll temp=0;
    for(int i=0;i<n;i++){
      temp+=1ll*p[i]*min(minx[rx][i],miny[ry][i]);
    }
    ans[kazu]=min(ans[kazu],temp);
  }
  else{
    cur[m]=1;
    dfs(m+1,rx+kisuu,ry,2*kisuu,kazu+1);
    cur[m]=0;
    dfs(m+1,rx,ry,2*kisuu,kazu);
    cur[m]=-1;
    dfs(m+1,rx,ry+kisuu,2*kisuu,kazu+1);
  }
  return;
}

int main() {
  cin>>n;
  for(int i=0;i<n;i++){
    cin>>x[i]>>y[i]>>p[i];
  }
  for(int i=0;i<n+1;i++){
    ans[i]=1e18;
  }
  for(int i=0;i<(1<<n);i++){
    bool ch[n];
    for(int j=0;j<n;j++){
      ch[j]=false;
    }
    for(int j=0;j<n;j++){
      if((i>>j)&1) ch[j]=true;
    }
    for(int j=0;j<n;j++){
      if(ch[j]){
        minx[i][j]=0;
        miny[i][j]=0;
      }
      else{
        int tempx=min(abs(x[j]),abs(y[j]));
        int tempy=tempx;
        for(int k=0;k<n;k++){
          if(ch[k]){
            tempx=min(tempx,abs(x[j]-x[k]));
            tempy=min(tempy,abs(y[j]-y[k]));
          } 
        }
        minx[i][j]=tempx;
        miny[i][j]=tempy;
      }
    }
  }
  dfs(0,0,0,1,0);
  for(int i=0;i<=n;i++){
    if(ans[i]==1e18) cout<<0<<endl;
    else cout<<ans[i]<<endl;
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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
#include <bits/stdc++.h>
#include <stdlib.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<ll,ll> llP;

ll mod(ll a, ll b){
  ll ret=a%b;
  if(ret<0) ret+=b;
  return ret;
}

ll modpow(ll a,ll b,ll c){
  ll res=1;
  while(b>0){
    if(b&1) res=mod(res*a,c);
    a=mod(a*a,c);
    b>>=1;
  }
  return res;
}

int n;
ll ans[16];
int cur[16];
int x[16],y[16],p[16];
int minx[40000][16],miny[40000][16];

void dfs(int m,int rx,int ry,int kisuu,int kazu){
  if(m==n){
    ll temp=0;
    for(int i=0;i<n;i++){
      temp+=1ll*p[i]*min(minx[rx][i],miny[ry][i]);
    }
    ans[kazu]=min(ans[kazu],temp);
  }
  else{
    cur[m]=1;
    dfs(m+1,rx+kisuu,ry,2*kisuu,kazu+1);
    cur[m]=0;
    dfs(m+1,rx,ry,2*kisuu,kazu);
    cur[m]=-1;
    dfs(m+1,rx,ry+kisuu,2*kisuu,kazu+1);
  }
  return;
}

int main() {
  cin>>n;
  for(int i=0;i<n;i++){
    cin>>x[i]>>y[i]>>p[i];
  }
  for(int i=0;i<n+1;i++){
    ans[i]=1e18;
  }
  for(int i=0;i<(1<<n);i++){
    bool ch[n];
    for(int j=0;j<n;j++){
      ch[j]=false;
    }
    for(int j=0;j<n;j++){
      if((i>>j)&1) ch[j]=true;
    }
    for(int j=0;j<n;j++){
      if(ch[j]){
        minx[i][j]=0;
        miny[i][j]=0;
      }
      else{
        int tempx=min(abs(x[j]),abs(y[j]));
        int tempy=tempx;
        for(int k=0;k<n;k++){
          if(ch[k]){
            tempx=min(tempx,abs(x[j]-x[k]));
            tempy=min(tempy,abs(y[j]-y[k]));
          } 
        }
        minx[i][j]=tempx;
        miny[i][j]=tempy;
      }
    }
  }
  dfs(0,0,0,1,0);
  for(int i=0;i<=n;i++){
    if(ans[i]==1e18) cout<<0<<endl;
    else cout<<ans[i]<<endl;
  }
  return 0;
}

```