#include<bits/stdc++.h>
using namespace std;
int x[20],y[20],p[20];
int xd[20][40010],yd[20][40010],u[40010],v[40010];
int X[20],Y[20],tot;
long long ans[30];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++)scanf("%d%d%d",&x[i],&y[i],&p[i]);
	int k=1;
	for(int i=0;i<n;i++)k*=3;
	for(int i=0;i<=n;i++)ans[i]=1e18;
	int K=1<<n;
	for(int i=0;i<n;i++){
		for(int j=0;j<K;j++)xd[i][j]=abs(x[i]),yd[i][j]=abs(y[i]);
		for(int j=0;j<K;j++){
			for(int l=0;l<n;l++){
				if(j&(1<<l)){
					xd[i][j]=min(xd[i][j],abs(x[i]-x[l]));
					yd[i][j]=min(yd[i][j],abs(y[i]-y[l]));
				}
			}
		}
	}
	for(int i=0;i<K;i++){
		tot=0;
		for(int j=0;j<n;j++){	
			if(i&(1<<j)){
				X[tot]=x[j];
				Y[tot++]=y[j];
			}
		}
		if(tot){
			sort(X,X+tot);
			u[i]=unique(X,X+tot)-X;
			sort(Y,Y+tot);
			v[i]=unique(Y,Y+tot)-Y;
		}
	}
	for(int i=0;i<k;i++){
		int a=0,b=0;
		for(int j=0,m=1;j<n;j++,m*=3){
			int c=i/m%3;
			if(c==1)a|=(1<<j);
			if(c==2)b|=(1<<j);
		}
		long long tmp=0;
		for(int j=0;j<n;j++)tmp+=(long long)p[j]*min(xd[j][a],yd[j][b]);
		if(ans[u[a]+v[b]]>tmp)ans[u[a]+v[b]]=tmp;
	}
	for(int i=0;i<=n;i++){
		if(i)ans[i]=min(ans[i],ans[i-1]);
		printf("%lld\n",ans[i]);
	}
} 
### Prompt

Your task is to create 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;
int x[20],y[20],p[20];
int xd[20][40010],yd[20][40010],u[40010],v[40010];
int X[20],Y[20],tot;
long long ans[30];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++)scanf("%d%d%d",&x[i],&y[i],&p[i]);
	int k=1;
	for(int i=0;i<n;i++)k*=3;
	for(int i=0;i<=n;i++)ans[i]=1e18;
	int K=1<<n;
	for(int i=0;i<n;i++){
		for(int j=0;j<K;j++)xd[i][j]=abs(x[i]),yd[i][j]=abs(y[i]);
		for(int j=0;j<K;j++){
			for(int l=0;l<n;l++){
				if(j&(1<<l)){
					xd[i][j]=min(xd[i][j],abs(x[i]-x[l]));
					yd[i][j]=min(yd[i][j],abs(y[i]-y[l]));
				}
			}
		}
	}
	for(int i=0;i<K;i++){
		tot=0;
		for(int j=0;j<n;j++){	
			if(i&(1<<j)){
				X[tot]=x[j];
				Y[tot++]=y[j];
			}
		}
		if(tot){
			sort(X,X+tot);
			u[i]=unique(X,X+tot)-X;
			sort(Y,Y+tot);
			v[i]=unique(Y,Y+tot)-Y;
		}
	}
	for(int i=0;i<k;i++){
		int a=0,b=0;
		for(int j=0,m=1;j<n;j++,m*=3){
			int c=i/m%3;
			if(c==1)a|=(1<<j);
			if(c==2)b|=(1<<j);
		}
		long long tmp=0;
		for(int j=0;j<n;j++)tmp+=(long long)p[j]*min(xd[j][a],yd[j][b]);
		if(ans[u[a]+v[b]]>tmp)ans[u[a]+v[b]]=tmp;
	}
	for(int i=0;i<=n;i++){
		if(i)ans[i]=min(ans[i],ans[i-1]);
		printf("%lld\n",ans[i]);
	}
} 
```