#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,x[25],y[25],p[25],tot;
struct Line{
	bool f;//f=0:x=k f=1:y=k
	int k;
}l[25];
ll ans=0;
ll Calc(){
	ll r=0;
	for(int i=1;i<=n;i++){
		int minn=1e9;
		for(int j=1;j<=tot;j++){
			if(l[j].f)minn=min(minn,abs(l[j].k-y[i]));
			else minn=min(minn,abs(l[j].k-x[i]));
		}
		r+=1ll*minn*p[i];
		if(r>=ans)return r;
	}
	return r;
}
void dfs(int now,int nd){
	if(n-now+1<nd)return ;
	if(!nd)return ans=min(ans,Calc()),void();
	dfs(now+1,nd);
	l[++tot]={0,x[now]};
	dfs(now+1,nd-1);
	l[tot]={1,y[now]};
	dfs(now+1,nd-1);
	tot--;
}
int main() {
	//for(int i=1;i<=15;i++)cout<<rand()%15<<' '<<rand()%15<<' '<<rand()%15<<'\n';
	cin>>n;
	for(int i=1;i<=n;i++)cin>>x[i]>>y[i]>>p[i];
	for(int i=1;i<=n;i++)ans+=1ll*min(abs(x[i]),abs(y[i]))*p[i];
	cout<<ans<<endl;
	l[1]={0,0},l[1]={1,0};
	for(int i=1,F=0;i<=n;i++){
		tot=2;
		if(!F)dfs(1,i);
		cout<<ans<<endl;
		if(!ans)F=1;
	}
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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;
int n,x[25],y[25],p[25],tot;
struct Line{
	bool f;//f=0:x=k f=1:y=k
	int k;
}l[25];
ll ans=0;
ll Calc(){
	ll r=0;
	for(int i=1;i<=n;i++){
		int minn=1e9;
		for(int j=1;j<=tot;j++){
			if(l[j].f)minn=min(minn,abs(l[j].k-y[i]));
			else minn=min(minn,abs(l[j].k-x[i]));
		}
		r+=1ll*minn*p[i];
		if(r>=ans)return r;
	}
	return r;
}
void dfs(int now,int nd){
	if(n-now+1<nd)return ;
	if(!nd)return ans=min(ans,Calc()),void();
	dfs(now+1,nd);
	l[++tot]={0,x[now]};
	dfs(now+1,nd-1);
	l[tot]={1,y[now]};
	dfs(now+1,nd-1);
	tot--;
}
int main() {
	//for(int i=1;i<=15;i++)cout<<rand()%15<<' '<<rand()%15<<' '<<rand()%15<<'\n';
	cin>>n;
	for(int i=1;i<=n;i++)cin>>x[i]>>y[i]>>p[i];
	for(int i=1;i<=n;i++)ans+=1ll*min(abs(x[i]),abs(y[i]))*p[i];
	cout<<ans<<endl;
	l[1]={0,0},l[1]={1,0};
	for(int i=1,F=0;i<=n;i++){
		tot=2;
		if(!F)dfs(1,i);
		cout<<ans<<endl;
		if(!ans)F=1;
	}
}
```