#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
ll n;
ll x[20],y[20],p[20];
ll ret[20];
ll D[20];

int a[20];
int cnt;
void solve(int k){
	if(k==0){
		for(int i=0;i<n;i++)D[i]=min(abs(x[i]),abs(y[i]));
	}
	if(k==n-2){
		for(int i=0;i<n;i++){
			if(i<k&&a[i]!=0){
				D[i]=0;
			}
			else {
				D[i]=min(abs(x[i]),abs(y[i]));
				for(int j=0;j<k;j++){
					if(a[j]==1)D[i]=min(D[i],abs(x[i]-x[j]));
					if(a[j]==2)D[i]=min(D[i],abs(y[i]-y[j]));
				}
			}
		}
	}
	if(k==n){
		ll sum=0;
		for(int i=0;i<n;i++){
			if(a[i]!=0)continue;
			ll d=D[i];
			for(int j=max((int)n-2,0);j<n;j++){
				if(a[j]==1)d=min(d,abs(x[i]-x[j]));
				if(a[j]==2)d=min(d,abs(y[i]-y[j]));
			}
			sum+=d*p[i];
		}
		ret[cnt]=min(ret[cnt],sum);
		return;
	}
	for(int j=0;j<3;j++){
		a[k]=j;
		if(j>0)cnt++;
		solve(k+1);
		if(j>0)cnt--;
	}
}

int main(){
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>x[i]>>y[i]>>p[i];
	}
	for(int k=0;k<=n;k++){
		ret[k]=1000000000000000000;
	}
	solve(0);
	for(int k=0;k<=n;k++){
		cout<<ret[k]<<endl;
	}
}
### 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;

typedef long long ll;
ll n;
ll x[20],y[20],p[20];
ll ret[20];
ll D[20];

int a[20];
int cnt;
void solve(int k){
	if(k==0){
		for(int i=0;i<n;i++)D[i]=min(abs(x[i]),abs(y[i]));
	}
	if(k==n-2){
		for(int i=0;i<n;i++){
			if(i<k&&a[i]!=0){
				D[i]=0;
			}
			else {
				D[i]=min(abs(x[i]),abs(y[i]));
				for(int j=0;j<k;j++){
					if(a[j]==1)D[i]=min(D[i],abs(x[i]-x[j]));
					if(a[j]==2)D[i]=min(D[i],abs(y[i]-y[j]));
				}
			}
		}
	}
	if(k==n){
		ll sum=0;
		for(int i=0;i<n;i++){
			if(a[i]!=0)continue;
			ll d=D[i];
			for(int j=max((int)n-2,0);j<n;j++){
				if(a[j]==1)d=min(d,abs(x[i]-x[j]));
				if(a[j]==2)d=min(d,abs(y[i]-y[j]));
			}
			sum+=d*p[i];
		}
		ret[cnt]=min(ret[cnt],sum);
		return;
	}
	for(int j=0;j<3;j++){
		a[k]=j;
		if(j>0)cnt++;
		solve(k+1);
		if(j>0)cnt--;
	}
}

int main(){
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>x[i]>>y[i]>>p[i];
	}
	for(int k=0;k<=n;k++){
		ret[k]=1000000000000000000;
	}
	solve(0);
	for(int k=0;k<=n;k++){
		cout<<ret[k]<<endl;
	}
}
```