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

Solution (in CPP):

Solution:
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <numeric>
#include <cassert>
#include <vector>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<28;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=998244353;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

int n;
vl a,x,y,p;

int main(){
	cin>>n;
	a=vl(n+1,INF);
	x=y=p=vl(n);
	for(int i=0;i<n;i++) cin>>x[i]>>y[i]>>p[i];
	int N=1;
	for(int i=0;i<n;i++) N*=3;
	for(int i=0;i<N;i++){
		vi b(n);
		int t=i;
		for(int j=0;j<n;j++){
			b[j]=t%3;
			t/=3;
		}
		t=0;
		for(int j=0;j<n;j++) if(b[j]) t++;
		ll sum=0;
		for(int j=0;j<n;j++) if(!b[j]){
			ll tmp=min(abs(x[j]),abs(y[j]));
			for(int k=0;k<n;k++){
				if(b[k]==1) tmp=min(tmp,abs(x[j]-x[k]));
				if(b[k]==2) tmp=min(tmp,abs(y[j]-y[k]));
			}
			sum+=tmp*p[j];
			if(sum>=a[t]) break;
		}
		a[t]=min(a[t],sum);
	}
	for(auto i:a) cout<<i<<endl;
}