Problem:
Score: 500 points

Problem Statement

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.

Sample Input 1

3
1 2 300
3 3 600
1 4 800

Sample Output 1

2900
900
0
0

When K = 0, the residents of Area 1, 2, and 3 have to walk the distances of 1, 3, and 1, respectively, to reach a railroad.

Thus, the sum of the walking distances of all citizens, S, is 1 \times 300 + 3 \times 600 + 1 \times 800 = 2900.

When K = 1, if we build a railroad along the street corresponding to the line y = 4 in the coordinate plane, the walking distances of the citizens of Area 1, 2, and 3 become 1, 1, and 0, respectively.

Then, S = 1 \times 300 + 1 \times 600 + 0 \times 800 = 900.

We have many other options for where we build the railroad, but none of them makes S less than 900.

When K = 2, if we build a railroad along the street corresponding to the lines x = 1 and x = 3 in the coordinate plane, all citizens can reach a railroad with the walking distance of 0, so S = 0. We can also have S = 0 when K = 3.

The figure below shows the optimal way to build railroads for the cases K = 0, 1, 2.

The street painted blue represents the roads along which we build railroads.

Sample Input 2

5
3 5 400
5 3 700
5 5 1000
5 7 700
7 5 400

Sample Output 2

13800
1600
0
0
0
0

The figure below shows the optimal way to build railroads for the cases K = 1, 2.

Sample Input 3

6
2 5 1000
5 2 1100
5 5 1700
-2 -5 900
-5 -2 600
-5 -5 2200

Sample Output 3

26700
13900
3200
1200
0
0
0

The figure below shows the optimal way to build railroads for the case K = 3.

Sample Input 4

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

Sample Output 4

2576709
1569381
868031
605676
366338
141903
0
0
0

The figure below shows the optimal way to build railroads for the case K = 4.

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<queue>
#include<vector>
#include<cstring>
#include<map>
#include<set>
#include<cstdlib>
#include<bitset>
using namespace std;
typedef int ll;
struct node{
	short x=0,y=0;
	int p=0;
}a[20]; 
long long ans[20];
short mi[20],m1[20];
short px[20][20],py[20][20];
int main(void)
{
 
	short n;
	scanf("%hd",&n);
	for(short i=0;i<n;i++) scanf("%hd%hd%d",&a[i].x,&a[i].y,&a[i].p);
	int ma=1;
	for(short i=0;i<=n;i++) ans[i]=1000000000000000;
	for(short i=0;i<n;i++) ma*=3;
	for(short i=0;i<n;i++) m1[i]=min(abs(a[i].x),abs(a[i].y));
	for(short i=0;i<n;i++)
	{
		for(short j=0;j<n;j++)
		{
			px[i][j]=(short)abs(a[i].x-a[j].x);
			py[i][j]=(short)abs(a[i].y-a[j].y);
		}
	}
	for(int zhu=0;zhu<ma;zhu++)
	{
		int ppk=zhu;
		short cnt=0;
		int st=ma/3;
		for(short i=0;i<n;i++) mi[i]=m1[i];
		for(short i=0;i<n;i++,st/=3)
		{
			int now=ppk/st;
			if(now==2)
			{
				cnt++;
				for(ll j=0;j<n;j++) if(mi[j]>px[i][j]) mi[j]=px[i][j];
			}
			else if(now==1)
			{
				cnt++;
				for(ll j=0;j<n;j++) if(mi[j]>py[i][j]) mi[j]=py[i][j];
			}
			ppk-=st*now;
		}
		long long sum=0;
		for(short i=0;i<n;i++) sum+=(long long)mi[i]*a[i].p;
		ans[cnt]=min(ans[cnt],sum);
	}
	for(ll i=0;i<=n;i++) printf("%lld\n",ans[i]);
}