#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=15;
int x[N],y[N],p[N],mix[1<<N][N],miy[1<<N][N];
LL ans[N+5];
int main(){
	int n,s;scanf("%d",&n),s=(1<<n);
	for(int i=0;i<n;i++)scanf("%d%d%d",x+i,y+i,p+i);
	for(int i=0;i<s;i++){
		for(int j=0;j<n;j++){
			mix[i][j]=abs(x[j]),miy[i][j]=abs(y[j]);
			for(int k=0;k<n;k++)if(i&(1<<k))
				mix[i][j]=min(mix[i][j],abs(x[j]-x[k])),miy[i][j]=min(miy[i][j],abs(y[j]-y[k]));
		}
	}
	LL w,w1;memset(ans,0x1f,sizeof(ans));
	for(int i=0,j=0;i<s;i++,j=i){
		do{
			w1=w=0;
			for(int k=0;k<n;k++)if(i&(1<<k))w++;
				else w1+=1ll*min(mix[j][k],miy[i^j][k])*p[k];
			ans[w]=min(ans[w],w1);
			j=(j-1)&i;
		}while(i!=j);
	}
	for(int i=0;i<=n;i++)printf("%lld\n",ans[i]);
	return 0;
} 
### Prompt

Please formulate 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 long long LL;
const int N=15;
int x[N],y[N],p[N],mix[1<<N][N],miy[1<<N][N];
LL ans[N+5];
int main(){
	int n,s;scanf("%d",&n),s=(1<<n);
	for(int i=0;i<n;i++)scanf("%d%d%d",x+i,y+i,p+i);
	for(int i=0;i<s;i++){
		for(int j=0;j<n;j++){
			mix[i][j]=abs(x[j]),miy[i][j]=abs(y[j]);
			for(int k=0;k<n;k++)if(i&(1<<k))
				mix[i][j]=min(mix[i][j],abs(x[j]-x[k])),miy[i][j]=min(miy[i][j],abs(y[j]-y[k]));
		}
	}
	LL w,w1;memset(ans,0x1f,sizeof(ans));
	for(int i=0,j=0;i<s;i++,j=i){
		do{
			w1=w=0;
			for(int k=0;k<n;k++)if(i&(1<<k))w++;
				else w1+=1ll*min(mix[j][k],miy[i^j][k])*p[k];
			ans[w]=min(ans[w],w1);
			j=(j-1)&i;
		}while(i!=j);
	}
	for(int i=0;i<=n;i++)printf("%lld\n",ans[i]);
	return 0;
} 
```