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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#pragma optimize("-O3")

#define f first
#define s second
#define pb push_back

int main() {
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int n;
	cin>>n;
	int x[n];
	int y[n];
	long long int p[n];
	int mini[n];
	for(int i=0; i<n; i++){
		cin>>x[i]>>y[i]>>p[i];
		mini[i]=min(abs(x[i]), abs(y[i]));
	}
	
	long long int ans[n+1];
	for(int i=0; i<=n; i++){
		ans[i]=1e18;
	}
	for(int i=0; i<(1<<n); i++){
		int num=__builtin_popcount(i);
		int mask=i;
		vector<int>vec;
		vector<int>vecl;
		for(int j=0; j<n; j++){
			if((i&(1<<j))){
				vec.pb(j);
			}
			else{
				vecl.pb(j);
			}
		}
	
		while(mask>0){
			long long int tot=0;
			for(auto k:vecl){
				int xk=x[k];
				int yk=y[k];
				
				
				int val=mini[k];
				
				for(auto l:vec){
						if((mask&(1<<l))){
							val=min(val, abs(xk-x[l]));
						}
						else{
							val=min(val, abs(yk-y[l]));
						}
					
				}
				tot+=(val*(p[k]));
			}
			ans[num]=min(ans[num], tot);
			mask=(mask-1)&i;
		}
		long long int tot=0;
	
			for(auto k:vecl){
				int xk=x[k];
				int yk=y[k];
			
				int val=mini[k];
				
				for(auto l:vec){
					
						if((mask&(1<<l))){
							val=min(val, abs(xk-x[l]));
						}
						else{
							val=min(val, abs(yk-y[l]));
						}
					
				}
				tot+=(val*(p[k]));
			}
			ans[num]=min(ans[num], tot);
	}
	for(int i=0; i<=n; i++){
		cout<<ans[i]<<endl;
	}
	return 0;
}