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 rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;
#define chmin(x,y) x=min(x,y)

const ll INF = 1e18;

ll dp[20][20];

vector<ll> calc(vector<P>& a) {
  int n = a.size();
  sort(a.begin(), a.end());
  rep(i,n+1) rep(j,n+1) dp[i][j] = INF;
  dp[0][0] = 0;
  rep(l,n) {
    for(int r = l+1;r <= n;++r){
      {
        ll now = INF;
        for(int k = l;k < r;++k){
          ll s = 0;
          for(int i = l;i < r;++i){
            s += abs(a[i].first-a[k].first)*ll(a[i].second);
          }
          chmin(now, s);
        }
        rep(j,n) chmin(dp[r][j+1], dp[l][j]+now);
      }
      {
        ll now = 0;
        for(int i = l;i < r;++i){
          now += abs(a[i].first)*ll(a[i].second);
        }
        rep(j,n+1) chmin(dp[r][j],dp[l][j]+now);
      }
    }
  }
  vector<ll> res(n+1,INF);
  rep(j,n+1) res[j] = dp[n][j];
  return res;
}

int main() {
  int n;
  cin>>n;
  vector<int> x(n),y(n),p(n);
  rep(i,n)cin >> x[i] >> y[i] >> p[i];
  vector<ll> ans(n+1, INF);
  rep(s,1<<n){
    vector<P> ax,ay;
    rep(i,n){
      if(s>>i&1)ax.emplace_back(x[i], p[i]);
      else ay.emplace_back(y[i], p[i]);
    }
    vector<ll> dx = calc(ax);
    vector<ll> dy = calc(ay);
    rep(i,dx.size())rep(j,dy.size())
    {


      ans[i+j] = min(ans[i+j], dx[i]+dy[j]);
    }
  }

  rep(i,n+1){
    cout << ans[i] <<endl;
  }
  return 0;
}