Problem:
Score: 500 points

Problem Statement

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.

Sample Input 1

3
1 2 300
3 3 600
1 4 800

Sample Output 1

2900
900
0
0

When K = 0, the residents of Area 1, 2, and 3 have to walk the distances of 1, 3, and 1, respectively, to reach a railroad.

Thus, the sum of the walking distances of all citizens, S, is 1 \times 300 + 3 \times 600 + 1 \times 800 = 2900.

When K = 1, if we build a railroad along the street corresponding to the line y = 4 in the coordinate plane, the walking distances of the citizens of Area 1, 2, and 3 become 1, 1, and 0, respectively.

Then, S = 1 \times 300 + 1 \times 600 + 0 \times 800 = 900.

We have many other options for where we build the railroad, but none of them makes S less than 900.

When K = 2, if we build a railroad along the street corresponding to the lines x = 1 and x = 3 in the coordinate plane, all citizens can reach a railroad with the walking distance of 0, so S = 0. We can also have S = 0 when K = 3.

The figure below shows the optimal way to build railroads for the cases K = 0, 1, 2.

The street painted blue represents the roads along which we build railroads.

Sample Input 2

5
3 5 400
5 3 700
5 5 1000
5 7 700
7 5 400

Sample Output 2

13800
1600
0
0
0
0

The figure below shows the optimal way to build railroads for the cases K = 1, 2.

Sample Input 3

6
2 5 1000
5 2 1100
5 5 1700
-2 -5 900
-5 -2 600
-5 -5 2200

Sample Output 3

26700
13900
3200
1200
0
0
0

The figure below shows the optimal way to build railroads for the case K = 3.

Sample Input 4

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

Sample Output 4

2576709
1569381
868031
605676
366338
141903
0
0
0

The figure below shows the optimal way to build railroads for the case K = 4.

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include "bits/stdc++.h"
using namespace std;
#define REP(i, n) for(ll i = 0;i < n;i++)
#define ll long long
#define MOD 1000000007LL
#define llMAX 9223372036854775807LL
#define llMIN -9223372036854775808LL
using vi = vector<ll>; // intの1次元の型に vi という別名をつける
using vvi = vector<vi>; // intの2次元の型に vvi という別名をつける

ll n;










//bit全探索
vi bt1(16);
vi bt2(16);
ll w1=0;
ll w2=0;






bool chk(ll j,ll i)
{
  ll jj=j;
  ll ret=0;
  for(ll k=0;k<n;k++){
    ret+=jj%2;
    jj/=2;
    ret+=jj%2;
    jj/=2;
  }
  if (ret!=i)return false;

  w1=0;
  w2=0;
  jj=j;
  for(ll k=0;k<n;k++){
    if (jj%2==1){
      bt1[w1]=k;
      w1++;
    }
    jj/=2;
    if (jj%2==1){
      bt2[w2]=k;
      w2++;
    }
    jj/=2;
  }
  return true;
}








int main(){
  ll ans=0;
  cin >> n ;
  
  vi x(n),y(n),p(n);
  REP(i,n){
    cin>>x[i]>>y[i]>>p[i];
  }


  ll nn=1;
  for(ll j=0;j<n;j++)
    nn*=4;


  for(ll i=0;i<=n;i++){
    if (i==0){
      ans=0;
      for(ll k=0;k<n;k++){
        ans+=min(abs(x[k]),abs(y[k]))*p[k];
      }
      cout<<ans<<endl;
      continue;
    }
    if (i==n)
    {
      cout<<0<<endl;
      continue;
    }


    //4bit全探索
    ans=llMAX;
    ll tmpans=0;
    for(ll j=0;j<nn;j++){
      if (chk(j,i))
      {
        tmpans=0;
        //cout<<"j"<<j<<endl;
        for(ll ii=0;ii<n;ii++){//全部の集落
          ll leng=1000000000000;
          for(ll kk=0;kk<w1;kk++){
            ll k=bt1[kk];
            leng=min(leng,abs(x[k]-x[ii]));
            //cout<<"?"<<k<<"?";
          }
          leng=min(leng,abs(x[ii]));
         
          for(ll kk=0;kk<w2;kk++){
            ll k=bt2[kk];
            leng=min(leng,abs(y[k]-y[ii]));
            //cout<<"?"<<k<<"?";
          }
          leng=min(leng,abs(y[ii]));

          tmpans+=p[ii]*leng;
        }//tmpansがきまる
        //cout<<"tmpans"<<tmpans<<endl;
        
        ans=min(ans,tmpans);
      }
    }//4bit全探索のおわり
    cout<<ans<<endl;

  }


  //cout<<ans<<endl;
  return 0;
}