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: Accepted)

#include <bits/stdc++.h>
#include <stdlib.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef vector<double> Vec;
typedef vector<Vec> Mat;
typedef pair<ll,ll> P;
typedef pair<double,ll> Pd;
typedef pair<double,double> PD;
typedef priority_queue<P,vector<P>,greater<P> > P_queue;
typedef priority_queue<Pd,vector<Pd>,greater<Pd> > Pd_queue;

const ll MOD=998244353;
const ll mod=1000000007;
const ll INF=1e15;
const double DEL=1e-6;


#define REP(i,a,b) for(int i=(int)a;i<(int)b;i++)
#define rep(i,n) REP(i,0,n)
#define pb push_back
#define mp make_pair
#define ALL(a) a.begin(),a.end()
#define SORT(a) sort(ALL(a))
#define U_ERASE(V) V.erase(unique(ALL(V)), V.end());
#define ADD(a,b) a=(a+b)%mod
#define MUL(a,b) a=(a*b)%mod

ll N;
ll d[16][3];
ll mx[40000][16], my[40000][16];

ll MX(ll a, ll k){
    if(mx[a][k]>=0) return mx[a][k];
    if(!a) return mx[a][k]=abs(d[k][0])*d[k][2];
    vec f;
    rep(i,N) if((1<<i) & a) {
        f.pb(i);
        if(i==k) return mx[a][k]=0;
    }
    ll P=MX(a-(1<<f[0]),k);
    return mx[a][k]=min(P,abs(d[k][0]-d[f[0]][0])*d[k][2]);
}

ll MY(ll a, ll k){
    if(my[a][k]>=0) return my[a][k];
    if(!a) return my[a][k]=abs(d[k][1])*d[k][2];
    vec f;
    rep(i,N) if((1<<i) & a) {
        f.pb(i);
        if(i==k) return my[a][k]=0;
    }
    ll P=MY(a-(1<<f[0]),k);
    return my[a][k]=min(P,abs(d[k][1]-d[f[0]][1])*d[k][2]);
}
ll ans[16];

void Solve(ll a){
    ll F=N;
    ll A=a;
    ll xx=0,yy=0;
    rep(i,N){
        if(A%3==1) xx+=(1<<i);
        if(A%3==2) yy+=(1<<i);
        if(A%3==0) F--;
        A/=3;
    }
    ll ret=0;
    rep(i,N) ret+=min(MX(xx,i),MY(yy,i));
    //cout<<a<<' '<<ret<<endl;
    ans[F]=min(ret,ans[F]);
    return;
}

ll ruijou(ll n, ll k){
    if(!k) return 1;
    ll a=ruijou(n,k/2);
    a=(a*a);
    return k%2 ? (a*n): a;
}


int main(){
    rep(i,40000) rep(j,16){
        mx[i][j]=-INF; my[i][j]=-INF;
    }
    rep(j,16) ans[j]=INF;
    cin>>N;
    rep(i,N) rep(j,3) cin>>d[i][j];
    //cout<<"ARRIVED"<<endl;
    rep(i,ruijou(3,N)) Solve(i);
    rep(i,N+1) cout<<ans[i]<<endl;


}