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: Wrong Answer)

#include<stdio.h>
int main()
{
int n,i,j,l;
scanf("%d",&n);
int x[n],y[n],sw[n],mind[n][n+1],dx[n],dy[n],absx[i],absy[i],cc;
  //一個めは町の番号、2個めは何本めの道について考えているか、
 
long long int p[n],ans,minans,ans1;
  ans=0;
  for(i=0;i<n;++i)
  {
  scanf("%d %d %lld",&x[i],&y[i],&p[i]);
    if(x[i]<0) absx[i]=-x[i];
    else absx[i]=x[i];
    if(y[i]<0) absy[i]=-y[i];
    else absy[i]=y[i];
  if(absx[i]>absy[i]) mind[i][0]=absy[i];
  else mind[i][0]=absx[i];
  sw[i]=0;
  
  ans=ans+p[i]*mind[i][0];
  }
  
  printf("%lld\n",ans);

  ans1=1500000000000;
  minans=0;
  for(i=0;i<n;++i) //i＋１は何本めの道を考えてるか
  {
   for(j=0;j<2*n;++j)//j(<n)はj番のx座標を考え、j(>=n)は(j-n)番のy座標を考える。
   {
   if(sw[j]==0 && j<n) 
    {
     for(l=0;l<n;++l)//lはl番めの町について
     {
       if(x[j]<x[l]) dx[l]=x[l]-x[j];
      else dx[l]=x[j]-x[l];
      if(dx[l]<mind[l][i]) mind[l][i+1]=dx[l];
      else mind[l][i+1]=mind[l][i];
      minans=minans+p[l]*mind[l][i+1];
     }
     if(minans<ans1) {ans1=minans;
                     cc=j;
                     }
    }
     
   else if(sw[j%n]==0 && j>=n)
    {
     for(l=0;l<n;++l)//lはl番めの町について
      {
       if(y[j%n]<y[l]) dy[l]=y[l]-y[j%n];
      else dy[l]=y[j%n]-y[l];
      if(dy[l]<mind[l][i]) mind[l][i+1]=dy[l];
      else mind[l][i+1]=mind[l][i];
      minans=minans+p[l]*mind[l][i+1];
      }
     if(minans<ans1) {ans1=minans;
                     cc=j%n;
                     }
     }
     sw[cc]=1;
     minans=0;
    }
    printf("%lld\n",ans1);
  ans1=1500000000000;
   }
return 0;}