#include <bits/stdc++.h>
 
using namespace std;
 
int main() {
  ios_base::sync_with_stdio(false);

  int n;
  cin >> n;
  vector<int> x(n), y(n), w(n);
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i] >> w[i];
  }

  vector<vector<int>> dis_x(n, vector<int>(1 << n));
  vector<vector<int>> dis_y(n, vector<int>(1 << n));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < (1 << n); j++) {
      dis_x[i][j] = abs(x[i]);
      dis_y[i][j] = abs(y[i]);
      for (int k = 0; k < n; k++) {
        if (j & (1 << k)) {
          dis_x[i][j] = min(dis_x[i][j], abs(x[i] - x[k]));
          dis_y[i][j] = min(dis_y[i][j], abs(y[i] - y[k]));
        }
      }
    }
  }

  
  vector<long long> ans(n + 1, 5e16);
  auto solve = [&](int mx, int my, int mz) {
    assert((mx | my | mz) == (1 << n) - 1); 
    long long res = 0;
    for (int i = 0; i < n; i++) {
      res += 1LL * w[i] * min(dis_x[i][mx], dis_y[i][my]);
    }
    int used = __builtin_popcount(mx | my);
    ans[used] = min(ans[used], res);
  };
  
  for (int i = 0; i < (1 << n); i++) { // mask x
    int mask = ((1 << n) - 1) ^ i;
    for (int j = mask; j > 0; j = (j - 1) & mask) { // mask y
      int k = ((1 << n) - 1) ^ i ^ j;
      solve(i, j, k);
    }
    solve(i, 0, mask);
  }

  for (auto i : ans) {
    cout << i << '\n';
  }

  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;
 
int main() {
  ios_base::sync_with_stdio(false);

  int n;
  cin >> n;
  vector<int> x(n), y(n), w(n);
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i] >> w[i];
  }

  vector<vector<int>> dis_x(n, vector<int>(1 << n));
  vector<vector<int>> dis_y(n, vector<int>(1 << n));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < (1 << n); j++) {
      dis_x[i][j] = abs(x[i]);
      dis_y[i][j] = abs(y[i]);
      for (int k = 0; k < n; k++) {
        if (j & (1 << k)) {
          dis_x[i][j] = min(dis_x[i][j], abs(x[i] - x[k]));
          dis_y[i][j] = min(dis_y[i][j], abs(y[i] - y[k]));
        }
      }
    }
  }

  
  vector<long long> ans(n + 1, 5e16);
  auto solve = [&](int mx, int my, int mz) {
    assert((mx | my | mz) == (1 << n) - 1); 
    long long res = 0;
    for (int i = 0; i < n; i++) {
      res += 1LL * w[i] * min(dis_x[i][mx], dis_y[i][my]);
    }
    int used = __builtin_popcount(mx | my);
    ans[used] = min(ans[used], res);
  };
  
  for (int i = 0; i < (1 << n); i++) { // mask x
    int mask = ((1 << n) - 1) ^ i;
    for (int j = mask; j > 0; j = (j - 1) & mask) { // mask y
      int k = ((1 << n) - 1) ^ i ^ j;
      solve(i, j, k);
    }
    solve(i, 0, mask);
  }

  for (auto i : ans) {
    cout << i << '\n';
  }

  return 0;
}

```