#include<bits/stdc++.h>

using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++)

typedef long long ll;
typedef pair<int, int> P;
typedef pair<int, P> PP;

#define MAX 1000000020
#define INF 1e18

int main(int, char**)
{
  int n;
  cin >> n;
  vector<ll> x(n), y(n), p(n);
  rep(i,n) cin >> x[i] >> y[i] >> p[i];

  vector< vector<ll> > costx(n, vector<ll>(1<<n));
  vector< vector<ll> > costy(n, vector<ll>(1<<n));
  rep(i,n) {
    rep(j,1<<n) {
      costx[i][j] = abs(x[i])*p[i];
      costy[i][j] = abs(y[i])*p[i];
      rep(k,n) {
        if (j&1<<k) {
          costx[i][j] = min(costx[i][j], abs(x[i] - x[k])*p[i]);
          costy[i][j] = min(costy[i][j], abs(y[i] - y[k])*p[i]);
        }
      }
    }
  }

  vector<ll> ans(n+1,INF);
  rep(i,pow(3,n)) {
    int xx = 0, yy = 0;
    int tmp = i;
    int k = 0;
    int cnt = 0;
    while (tmp > 0) {
      if (tmp%3==1) xx |= 1<<cnt;
      if (tmp%3==2) yy |= 1<<cnt;
      if (tmp%3!=0) k++;
      cnt++;
      tmp /= 3;
    }
    ll dis = 0;
    rep(j,n) {
      dis += min(costx[j][xx],costy[j][yy]);
    }
    ans[k] = min(ans[k],dis);  
  }
  rep(i,n+1) cout << ans[i] << endl;

  return 0;
}

### Prompt

Please formulate a cpp solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>

using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++)

typedef long long ll;
typedef pair<int, int> P;
typedef pair<int, P> PP;

#define MAX 1000000020
#define INF 1e18

int main(int, char**)
{
  int n;
  cin >> n;
  vector<ll> x(n), y(n), p(n);
  rep(i,n) cin >> x[i] >> y[i] >> p[i];

  vector< vector<ll> > costx(n, vector<ll>(1<<n));
  vector< vector<ll> > costy(n, vector<ll>(1<<n));
  rep(i,n) {
    rep(j,1<<n) {
      costx[i][j] = abs(x[i])*p[i];
      costy[i][j] = abs(y[i])*p[i];
      rep(k,n) {
        if (j&1<<k) {
          costx[i][j] = min(costx[i][j], abs(x[i] - x[k])*p[i]);
          costy[i][j] = min(costy[i][j], abs(y[i] - y[k])*p[i]);
        }
      }
    }
  }

  vector<ll> ans(n+1,INF);
  rep(i,pow(3,n)) {
    int xx = 0, yy = 0;
    int tmp = i;
    int k = 0;
    int cnt = 0;
    while (tmp > 0) {
      if (tmp%3==1) xx |= 1<<cnt;
      if (tmp%3==2) yy |= 1<<cnt;
      if (tmp%3!=0) k++;
      cnt++;
      tmp /= 3;
    }
    ll dis = 0;
    rep(j,n) {
      dis += min(costx[j][xx],costy[j][yy]);
    }
    ans[k] = min(ans[k],dis);  
  }
  rep(i,n+1) cout << ans[i] << endl;

  return 0;
}

```