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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <math.h>
#include <cassert>
#define rep(i,n) for(int i = 0; i < n; ++i )
using namespace std;
using ll = long long;
using P = pair<int,int>;

const ll inf = 1e9;
int main() {
  int n;
  cin >> n;
  vector<ll> x(n),y(n),p(n);
  rep(i,n) cin >> x[i] >> y[i] >> p[i];
  vector<vector<int>> v(1<<n,vector<int>(n)),h(1<<n,vector<int>(n));
  rep(b,1<<n){
    vector<int> vi={-inf,0,inf},hi={-inf,0,inf};
    rep(i,n) if(b>>i&1) vi.push_back(x[i]), hi.push_back(y[i]);
    sort(vi.begin(),vi.end());
    sort(hi.begin(),hi.end());
    rep(i,n){
      if(b>>i&1) continue;
      int xi = lower_bound(vi.begin(),vi.end(),x[i])-vi.begin();
      int yi = lower_bound(hi.begin(),hi.end(),y[i])-hi.begin();
      v[b][i] = min(vi[xi]-x[i],x[i]-vi[xi-1]);
      h[b][i] = min(hi[yi]-y[i],y[i]-hi[yi-1]);
    }
  }

  int t = 1;
  rep(_,n) t *= 3;
  vector<ll> ans(n+1,1e18L);
  rep(b,t){
    int k=0;
    int x=0,y=0,bi=b;
    rep(i,n){
      int t = bi%3;
      if(t!=0) ++k;
      if(t==1) x |= 1<<i;
      if(t==2) y |= 1<<i;
      bi /= 3;
    }
    ll res = 0;
    rep(i,n) res += p[i] * min(v[x][i],h[y][i]);
    ans[k] = min(ans[k],res);
  }
  rep(k,n+1) cout << ans[k] << endl;
}