#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------


int N;
int X[15],Y[15],P[15];
ll ret[16];

ll xmin[1<<15][15];
ll ymin[1<<15][15];

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N;
	FOR(i,N) {
		cin>>X[i]>>Y[i]>>P[i];
	}
	FOR(i,N+1) ret[i]=1LL<<60;
	
	FOR(i,1<<N) {
		FOR(j,N) {
			xmin[i][j]=abs(X[j])*1LL*P[j];
			ymin[i][j]=abs(Y[j])*1LL*P[j];
			FOR(x,N) if(i&(1<<x)) {
				xmin[i][j]=min(xmin[i][j],abs(X[j]-X[x])*1LL*P[j]);
				ymin[i][j]=min(ymin[i][j],abs(Y[j]-Y[x])*1LL*P[j]);
			}
		}
	}
	
	
	for(int xmask=0;xmask<1<<N;xmask++) {
		int lef=((1<<N)-1)^xmask;
		for(int ymask=lef; ymask>=0; ymask--) {
			ymask&=lef;
			ll sum=0;
			FOR(i,N) sum+=min(xmin[xmask][i],ymin[ymask][i]);
			y=__builtin_popcount(xmask|ymask);
			ret[y]=min(ret[y],sum);
		}
	}
	
	FOR(i,N) ret[i+1]=min(ret[i],ret[i+1]);
	FOR(i,N+1) cout<<ret[i]<<endl;
	
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}

### Prompt

Generate a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------


int N;
int X[15],Y[15],P[15];
ll ret[16];

ll xmin[1<<15][15];
ll ymin[1<<15][15];

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N;
	FOR(i,N) {
		cin>>X[i]>>Y[i]>>P[i];
	}
	FOR(i,N+1) ret[i]=1LL<<60;
	
	FOR(i,1<<N) {
		FOR(j,N) {
			xmin[i][j]=abs(X[j])*1LL*P[j];
			ymin[i][j]=abs(Y[j])*1LL*P[j];
			FOR(x,N) if(i&(1<<x)) {
				xmin[i][j]=min(xmin[i][j],abs(X[j]-X[x])*1LL*P[j]);
				ymin[i][j]=min(ymin[i][j],abs(Y[j]-Y[x])*1LL*P[j]);
			}
		}
	}
	
	
	for(int xmask=0;xmask<1<<N;xmask++) {
		int lef=((1<<N)-1)^xmask;
		for(int ymask=lef; ymask>=0; ymask--) {
			ymask&=lef;
			ll sum=0;
			FOR(i,N) sum+=min(xmin[xmask][i],ymin[ymask][i]);
			y=__builtin_popcount(xmask|ymask);
			ret[y]=min(ret[y],sum);
		}
	}
	
	FOR(i,N) ret[i+1]=min(ret[i],ret[i+1]);
	FOR(i,N+1) cout<<ret[i]<<endl;
	
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}

```