#include <bits/stdc++.h>
using namespace std;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 1000000000000000002


int main(){
	int N;
	cin>>N;
	
	vector<long long> X(N),Y(N),P(N);
	for(int i=0;i<N;i++)cin>>X[i]>>Y[i]>>P[i];
	
	vector<long long> ans(N+1,Inf);
	
	for(int i=0;i<(1<<N);i++){
		vector<int> ind0,ind1;
		for(int j=0;j<N;j++){
			if((i>>j)&1)ind0.push_back(j);
			else ind1.push_back(j);
		}
		
		for(int j=0;j<(1<<ind0.size());j++){
			long long t = 0LL;
			for(int k=0;k<ind1.size();k++){
				long long xx = min(abs(Y[ind1[k]]),abs(X[ind1[k]]))*P[ind1[k]];;;
				for(int l=0;l<ind0.size();l++){
					if((j>>l)&1){
						xx = min(xx,abs(Y[ind1[k]]-Y[ind0[l]])*P[ind1[k]]);
					}
					else{
						xx = min(xx,abs(X[ind1[k]]-X[ind0[l]])*P[ind1[k]]);
					}
				}
				t+=xx;
				if(t>=ans[ind0.size()])break;
			}
			
			ans[ind0.size()] = min(ans[ind0.size()],t);
		}
	}
	
	for(int i=0;i<=N;i++)cout<<ans[i]<<endl;
			
	
	return 0;
}
### Prompt

Please create a solution in cpp 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 modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 1000000000000000002


int main(){
	int N;
	cin>>N;
	
	vector<long long> X(N),Y(N),P(N);
	for(int i=0;i<N;i++)cin>>X[i]>>Y[i]>>P[i];
	
	vector<long long> ans(N+1,Inf);
	
	for(int i=0;i<(1<<N);i++){
		vector<int> ind0,ind1;
		for(int j=0;j<N;j++){
			if((i>>j)&1)ind0.push_back(j);
			else ind1.push_back(j);
		}
		
		for(int j=0;j<(1<<ind0.size());j++){
			long long t = 0LL;
			for(int k=0;k<ind1.size();k++){
				long long xx = min(abs(Y[ind1[k]]),abs(X[ind1[k]]))*P[ind1[k]];;;
				for(int l=0;l<ind0.size();l++){
					if((j>>l)&1){
						xx = min(xx,abs(Y[ind1[k]]-Y[ind0[l]])*P[ind1[k]]);
					}
					else{
						xx = min(xx,abs(X[ind1[k]]-X[ind0[l]])*P[ind1[k]]);
					}
				}
				t+=xx;
				if(t>=ans[ind0.size()])break;
			}
			
			ans[ind0.size()] = min(ans[ind0.size()],t);
		}
	}
	
	for(int i=0;i<=N;i++)cout<<ans[i]<<endl;
			
	
	return 0;
}
```