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: Wrong Answer)

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define REP(a,b) for(int a=0;a<(b);++a)
#define REP1(i,n) for(int i=1;i<=(n);++i)
#define debug(x) cerr<<#x<<": "<<x<<'\n'
#define all(x) (x).begin(),(x).end()
#define YES() printf("YES\n")
#define NO() printf("NO\n")
#define isYES(x) printf("%s\n",(x) ? "YES" : "NO")
#define Yes() printf("Yes\n")
#define No() printf("No\n")
#define isYes(x) printf("%s\n",(x) ? "Yes" : "No")
#define isPossible(x) printf("%s\n",(x) ? "Possible" : "Impossible")
#define SZ(x) ((int)(x).size())
#define pb push_back
#define mp make_pair
#define INF (1<<29)

// const long long INF = 1LL<<60;

#define Sp(p) cout<<setprecision(25)<< fixed<<p<<endl
#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
#define pie 3.14159265358979323846
using namespace std;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> PP;
typedef pair<ll, LP> LPP;
template<class T=int>
T in(){T x;cin>>x;return (x);}
template<class T>
void print(T& x){cout<<x<<'\n';}

const int MOD =(int)1e9+7;
// const int mod =(int)998244353;
const int mod =(int)1e9+7;

const int MAX =510000;
ll fac[MAX],finv[MAX],inv[MAX];
void COMint(){
    fac[0]=fac[1]=1;
    finv[0]=finv[1]=1;
    inv[1]=1;
    for(int i=2;i<MAX;i++){
        fac[i]=fac[i-1]*i%MOD;
        inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;
        finv[i]=finv[i-1]*inv[i]%MOD;
    }
}
ll COM(int n,int k){
    if(n<k) return 0;
    if(n<0||k<0)return 0;
    return fac[n]*(finv[k]*finv[n-k]%MOD)%MOD;
}
ll gcd(ll a,ll b){
    if(a<0)a=-a;
    if(b<0)b=-b;
    if(b==0)return a;
    if(a>b){
        swap(a,b);
    }
    return gcd(a,b%a);
}
ll lcm(ll a,ll b){
  if(a<0)a=-a;
  if(b<0)b=-b;
    ll g;g=gcd(a,b);
    return b/g*a;
}
bool compare_by_b(pair<int, int> a, pair<int, int> b) {
    if(a.second != b.second){
        return a.second < b.second;
    }else{
        return a.first < b.first;
    }
}
bool compare_by_a(pair<int, int> a, pair<int, int> b) {
    if(a.first != b.first){
        return a.first < b.first;
    }else{
        return a.second < b.second;
    }
}
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
ll RS(ll N,ll P){
    if(P==0){
        return 1;
    }else{
        if(P%2==0){
            ll t=RS(N,P/2);
            return t*t;
        }else{
            return N*RS(N,P-1);
        }
    }
}
 struct UnionFind{
     vi d;
     UnionFind(int n):d(n,-1){};
     int deb(int x){
         return d[x];
     }
     int root(int x){
         if(d[x]<0)return x;
         return d[x]=root(d[x]);
     }
     int unite(int x,int y){
         x=root(x);y=root(y);
         if(d[x]>d[y])swap(x,y);
         d[x]+=d[y];
         d[y]=x;
         return true;
     }
     bool same(int x,int y){return root(x)==root(y);}
     int size(int x){return -d[root(x)];}
 };
struct mint {
   ll x; // typedef long long ll;
   mint(ll x=0):x((x%mod+mod)%mod){}
   mint& operator+=(const mint a) {
     if ((x += a.x) >= mod) x -= mod;
     return *this;
   }
   mint& operator-=(const mint a) {
     if ((x += mod-a.x) >= mod) x -= mod;
     return *this;
   }
   mint& operator*=(const mint a) {
     (x *= a.x) %= mod;
     return *this;
   }
   mint operator+(const mint a) const {
     mint res(*this);
     return res+=a;
   }
   mint operator-(const mint a) const {
     mint res(*this);
     return res-=a;
   }
   mint operator*(const mint a) const {
     mint res(*this);
     return res*=a;
   }
   mint pow(ll t) const {
     if (!t) return 1;
     mint a = pow(t>>1);
     a *= a;
     if (t&1) a *= *this;
     return a;
   }
   mint inv() const {
     return pow(mod-2);
   }
   mint& operator/=(const mint a) {
     return (*this) *= a.inv();
   }
   mint operator/(const mint a) const {
     mint res(*this);
     return res/=a;
   }
 };
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
struct combination {
vector<mint> fact, ifact;
   combination(int n):fact(n+1),ifact(n+1) {
       assert(n < mod);
       fact[0] = 1;
       for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
       ifact[n] = fact[n].inv();
       for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
     }
     mint operator()(int n, int k) {
       if (k < 0 || k > n) return 0;
       return fact[n]*ifact[k]*ifact[n-k];
     }
}c(2000005);
int N;
int main() {
  ios::sync_with_stdio(false);
  cin>>N;
  vector<int>X(N),Y(N),P(N);
  REP(i,N){
    cin>>X[i]>>Y[i]>>P[i];
  }
  vector<vector<ll>>x((1<<N),vector<ll>(N)),y((1<<N),vector<ll>(N));
  REP(i,(1<<N)){
    REP(j,N){
      int temp=min(abs(X[j]),abs(Y[j]));
      REP(k,N){
        if((i>>k)&1){
          chmin(temp,abs(Y[j]-Y[k]));
        }
      }
      x[i][j]=temp*P[j];
    }
  }
  REP(i,(1<<N)){
    REP(j,N){
      int temp=min(abs(X[j]),abs(Y[j]));
      REP(k,N){
        if((i>>k)&1){
          chmin(temp,abs(X[j]-X[k]));
        }
      }
      y[i][j]=temp*P[j];
    }
  }
  vector<ll>ans(N,INF);
  int n=RS(3,N);
  REP(i,n){
    int nx=0,ny=0;
    int temp=i;
    int cnt=0;
    REP(j,N){
      if(temp%3==1){
        nx+=RS(2,j);
        cnt++;
      }
      if(temp%3==2){
        ny+=RS(2,j);
        cnt++;
      }
      temp/=3;
    }
    if(cnt==N)continue;
    ll t=0;
    REP(j,N){
      t+=min(x[nx][j],y[ny][j]);
    }
    ans[cnt]=min(ans[cnt],t);
  }  
  REP(i,N){
    cout<<ans[i]<<endl;
  }
  cout<<0<<endl;
  return 0;
}