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>
#define abs(x) ((x)>0?(x):(-(x)))
using namespace std;
const int maxn=19;

int n;
struct Point{
	int x,y,z;
	void input(){
		scanf("%d%d%d",&x,&y,&z);
	}
}p[maxn];

long long ans[maxn];

int count(int s){
	int ret=0;
	while(s){
		ret+=s&1;s>>=1;
	}
	return ret;
}

int q[maxn][2];

int d0[maxn][1<<15];
int d1[maxn][1<<15];

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)p[i].input();
	for(int i=1;i<=n;++i){
		d0[i][0]=abs(p[i].y);
		d1[i][0]=abs(p[i].x);
	}
	for(int s=0;s<(1<<n);++s){
		for(int i=1;i<=n;++i){
			d0[i][s]=abs(p[i].y);
			for(int j=1;j<=n;++j){
				if((s>>(j-1))&1){
					d0[i][s]=min(d0[i][s],abs(p[i].y-p[j].y));
				}
			}
			d1[i][s]=abs(p[i].x);
			for(int j=1;j<=n;++j){
				if((s>>(j-1))&1){
					d1[i][s]=min(d1[i][s],abs(p[i].x-p[j].x));
				}
			}
		}
	}
	for(int i=0;i<=n;++i)ans[i]=1e15;
	for(int s=0;s<(1<<n);++s){
		for(int t=s;;t=(t-1)&s){
			long long tmp=0;//int nn=0;
//			for(int i=1;i<=n;++i){
//				if((s>>(i-1))&1){
//					if((t>>(i-1))&1){
//						q[++nn][0]=p[i].x;
//						q[nn][1]=1;
//					}else{
//						q[++nn][0]=p[i].y;
//						q[nn][1]=0;
//					}
//				}
//			}
			for(int i=1;i<=n;++i){
				int d=min(d0[i][s^t],d1[i][t]);
//				for(int j=1;j<=nn;++j){
//					if(q[j][1]==0)d=min(d,abs(p[i].y-q[j][0]));
//					else d=min(d,abs(p[i].x-q[j][0]));
//				}
				tmp+=1LL*d*p[i].z;
			}
			int cnt=count(s);
			ans[cnt]=min(ans[cnt],tmp);
			if(t==0)break;
		}
	}
	for(int i=0;i<=n;++i)printf("%lld\n",ans[i]);
	return 0;
}