#include <bits/stdc++.h>

using namespace std;
using int64 = long long;

constexpr int DEBUG = 0;

template<class T> inline bool UpdateMin(T& a, T b) {
  if (a > b) { a = b; return 1; } return 0;
}

template<class T> inline bool UpdateMax(T& a, T b) {
  if (a < b) { a = b; return 1; } return 0;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int n;
  cin >> n;
  vector<int64> xs(n), ys(n), ps(n);
  for (int i = 0; i < n; i++) {
    cin >> xs[i] >> ys[i] >> ps[i];
  }

  vector<int64> ans(n + 1, INT64_MAX);
  for (int city_pattern = 0; city_pattern < 1 << n; city_pattern++) {
    vector<int> cs;
    for (int c = 0; c < n; c++) {
      if (city_pattern & (1 << c)) {
        cs.push_back(c);
      }
    }

    int num_roads = cs.size();

    vector<int64> r_xs;
    vector<int64> r_ys;
    for (int d_pattern = 0; d_pattern < 1 << num_roads; d_pattern++) {
      r_xs.clear();
      r_ys.clear();
      r_xs.push_back(0);
      r_ys.push_back(0);

      for (int r = 0; r < num_roads; r++) {
        if (d_pattern & (1 << r)) {
          r_ys.push_back(ys[cs[r]]);
        } else {
          r_xs.push_back(xs[cs[r]]);
        }
      }

      int64 cost = 0;
      for (int c = 0; c < n; c++) {
        if (city_pattern & (1 << c)) continue;
        int64 min_d = INT64_MAX;
        for (int64 r_x : r_xs) {
          UpdateMin(min_d, abs(xs[c] - r_x));
        }
        for (int64 r_y : r_ys) {
          UpdateMin(min_d, abs(ys[c] - r_y));
        }
        cost += min_d * ps[c];
      }

      UpdateMin(ans[num_roads], cost);
    }
  }

  for (int i = 0; i <= n; i++) {
    cout << ans[i] << endl;
  }
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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;
using int64 = long long;

constexpr int DEBUG = 0;

template<class T> inline bool UpdateMin(T& a, T b) {
  if (a > b) { a = b; return 1; } return 0;
}

template<class T> inline bool UpdateMax(T& a, T b) {
  if (a < b) { a = b; return 1; } return 0;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int n;
  cin >> n;
  vector<int64> xs(n), ys(n), ps(n);
  for (int i = 0; i < n; i++) {
    cin >> xs[i] >> ys[i] >> ps[i];
  }

  vector<int64> ans(n + 1, INT64_MAX);
  for (int city_pattern = 0; city_pattern < 1 << n; city_pattern++) {
    vector<int> cs;
    for (int c = 0; c < n; c++) {
      if (city_pattern & (1 << c)) {
        cs.push_back(c);
      }
    }

    int num_roads = cs.size();

    vector<int64> r_xs;
    vector<int64> r_ys;
    for (int d_pattern = 0; d_pattern < 1 << num_roads; d_pattern++) {
      r_xs.clear();
      r_ys.clear();
      r_xs.push_back(0);
      r_ys.push_back(0);

      for (int r = 0; r < num_roads; r++) {
        if (d_pattern & (1 << r)) {
          r_ys.push_back(ys[cs[r]]);
        } else {
          r_xs.push_back(xs[cs[r]]);
        }
      }

      int64 cost = 0;
      for (int c = 0; c < n; c++) {
        if (city_pattern & (1 << c)) continue;
        int64 min_d = INT64_MAX;
        for (int64 r_x : r_xs) {
          UpdateMin(min_d, abs(xs[c] - r_x));
        }
        for (int64 r_y : r_ys) {
          UpdateMin(min_d, abs(ys[c] - r_y));
        }
        cost += min_d * ps[c];
      }

      UpdateMin(ans[num_roads], cost);
    }
  }

  for (int i = 0; i <= n; i++) {
    cout << ans[i] << endl;
  }
}
```