#include <bits/stdc++.h>

using namespace std;

const int64_t INF = INT64_C(1) << 60;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N;
  cin >> N;
  vector<int64_t> X(N), Y(N), P(N);
  for (int i = 0; i < N; i++) {
    cin >> X[i] >> Y[i] >> P[i];
  }
  vector<vector<int64_t>> SX(1 << N, vector<int64_t>(N, INF)), SY(1 << N, vector<int64_t>(N, INF));
  for (int bit = 0; bit < (1 << N); bit++) {
    for (int i = 0; i < N; i++) {
      SX[bit][i] = min(SX[bit][i], abs(X[i]));
      SY[bit][i] = min(SY[bit][i], abs(Y[i]));
      if ((bit >> i) & 1) {
        for (int j = 0; j < N; j++) {
          SX[bit][j] = min(SX[bit][j], abs(X[i] - X[j]));
          SY[bit][j] = min(SY[bit][j], abs(Y[i] - Y[j]));
        }
      }
    }
  }
  vector<int64_t> res(N + 1, INF);
  for (int bitx = 0; bitx < (1 << N); bitx++) {
    int m = ((1 << N) - 1) ^ bitx;
    for (int bity = m;; bity = (bity - 1) & m) {
      int64_t sum = 0;
      for (int i = 0; i < N; i++) {
        sum += P[i] * min(SX[bitx][i], SY[bity][i]);
      }
      int b = __builtin_popcount(bitx) + __builtin_popcount(bity);
      res[b] = min(res[b], sum);
      if (bity == 0) break;
    }
  }
  for (auto r : res) {
    cout << r << '\n';
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;

const int64_t INF = INT64_C(1) << 60;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N;
  cin >> N;
  vector<int64_t> X(N), Y(N), P(N);
  for (int i = 0; i < N; i++) {
    cin >> X[i] >> Y[i] >> P[i];
  }
  vector<vector<int64_t>> SX(1 << N, vector<int64_t>(N, INF)), SY(1 << N, vector<int64_t>(N, INF));
  for (int bit = 0; bit < (1 << N); bit++) {
    for (int i = 0; i < N; i++) {
      SX[bit][i] = min(SX[bit][i], abs(X[i]));
      SY[bit][i] = min(SY[bit][i], abs(Y[i]));
      if ((bit >> i) & 1) {
        for (int j = 0; j < N; j++) {
          SX[bit][j] = min(SX[bit][j], abs(X[i] - X[j]));
          SY[bit][j] = min(SY[bit][j], abs(Y[i] - Y[j]));
        }
      }
    }
  }
  vector<int64_t> res(N + 1, INF);
  for (int bitx = 0; bitx < (1 << N); bitx++) {
    int m = ((1 << N) - 1) ^ bitx;
    for (int bity = m;; bity = (bity - 1) & m) {
      int64_t sum = 0;
      for (int i = 0; i < N; i++) {
        sum += P[i] * min(SX[bitx][i], SY[bity][i]);
      }
      int b = __builtin_popcount(bitx) + __builtin_popcount(bity);
      res[b] = min(res[b], sum);
      if (bity == 0) break;
    }
  }
  for (auto r : res) {
    cout << r << '\n';
  }
  return 0;
}

```