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<iostream>
#include<vector>
#include<algorithm>
using namespace std;
vector<long>calc(vector<pair<int,long> >X)
{
	sort(X.begin(),X.end());
	long pre[16][16],zr[16][16];
	const long INF=3e18;
	for(int i=0;i<X.size();i++)for(int j=i+1;j<=X.size();j++)
	{
		pre[i][j]=INF;
		long R=0,mR=0;
		for(int k=i;k<j;k++)
		{
			R+=X[k].first*X[k].second;
			mR+=X[k].second;
		}
		long L=0,mL=0;
		for(int k=i;k<j;k++)
		{
			R-=X[k].first*X[k].second;
			mR-=X[k].second;
			L+=X[k].first*X[k].second;
			mL+=X[k].second;
			pre[i][j]=min(pre[i][j],R-mR*X[k].first+mL*X[k].first-L);
		}
		zr[i][j]=0;
		for(int k=i;k<j;k++)
		{
			zr[i][j]+=abs(X[k].first)*X[k].second;
		}
	}
	long dp[17][16];
	for(int i=0;i<=15;i++)for(int j=0;j<=15;j++)dp[i][j]=INF;
	dp[0][0]=0;
	for(int i=0;i<X.size();i++)for(int j=0;j<=i;j++)
	{
		for(int k=i+1;k<=X.size();k++)
		{
			dp[j+1][k]=min(dp[j+1][k],dp[j][i]+pre[i][k]);
			dp[j][k]=min(dp[j][k],dp[j][i]+zr[i][k]);
		}
	}
	vector<long>ret(16);
	for(int i=0;i<=15;i++)ret[i]=dp[i][X.size()];
	return ret;
}
long ans[16];
int main()
{
	int N;
	cin>>N;
	int X[15],Y[15];
	long P[15];
	for(int i=0;i<N;i++)
	{
		cin>>X[i]>>Y[i]>>P[i];
	}
	for(int i=0;i<=N;i++)ans[i]=9e18;
	for(int i=0;i<1<<N;i++)
	{
		vector<pair<int,long> >x,y;
		for(int j=0;j<N;j++)
		{
			if(i>>j&1)x.push_back(make_pair(X[j],P[j]));
			else y.push_back(make_pair(Y[j],P[j]));
		}
		vector<long>Ax=calc(x);
		vector<long>Ay=calc(y);
		for(int j=0;j<=N;j++)for(int k=0;j+k<=N;k++)
		{
			ans[j+k]=min(ans[j+k],Ax[j]+Ay[k]);
		}
	}
	for(int i=0;i<=N;i++)cout<<ans[i]<<endl;
}