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

#define rep(i, n) for(int i=0; i<n; ++i)
#define repit(it, li) for(auto it=li.begin(); it!=li.end(); it++)
#define ll long long int

int k, n;
int ci[15], cf[15];
int x[15], y[15], p[15], d[15];
bool used[15]={0};
ll mini;

void tan(int d, int s){
	if(d==k){
		//count
		ll sm=0;
		rep(i, n) if(!used[i]){
			int mn=min(abs(x[i]), abs(y[i]));
			rep(j, k){
				if(cf[j]) mn=min(mn, abs(x[i]-x[ci[j]]));
				else mn=min(mn, abs(y[i]-y[ci[j]]));
			}
			sm+=(ll)p[i]*mn;
		}
		mini=min(mini, sm);
		return ;
	}
	for(int i=s; i<n-(k-1-d); i++){
		ci[d]=i;
		used[i]=true;
		rep(j, 2){
			cf[d]=j;
			tan(d+1, i+1);
		}
		used[i]=false;
	}
}

int main(){
	cin>>n;
	rep(i, n) cin>>x[i]>>y[i]>>p[i];

	for(k=0; k<n+1; k++){
		mini=LLONG_MAX;
		tan(0, 0);
		cout<<mini<<endl;
	}

	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;

#define rep(i, n) for(int i=0; i<n; ++i)
#define repit(it, li) for(auto it=li.begin(); it!=li.end(); it++)
#define ll long long int

int k, n;
int ci[15], cf[15];
int x[15], y[15], p[15], d[15];
bool used[15]={0};
ll mini;

void tan(int d, int s){
	if(d==k){
		//count
		ll sm=0;
		rep(i, n) if(!used[i]){
			int mn=min(abs(x[i]), abs(y[i]));
			rep(j, k){
				if(cf[j]) mn=min(mn, abs(x[i]-x[ci[j]]));
				else mn=min(mn, abs(y[i]-y[ci[j]]));
			}
			sm+=(ll)p[i]*mn;
		}
		mini=min(mini, sm);
		return ;
	}
	for(int i=s; i<n-(k-1-d); i++){
		ci[d]=i;
		used[i]=true;
		rep(j, 2){
			cf[d]=j;
			tan(d+1, i+1);
		}
		used[i]=false;
	}
}

int main(){
	cin>>n;
	rep(i, n) cin>>x[i]>>y[i]>>p[i];

	for(k=0; k<n+1; k++){
		mini=LLONG_MAX;
		tan(0, 0);
		cout<<mini<<endl;
	}

	return 0;
}

```