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 debug(x) cerr << #x << ": " << x << endl
#define debugArray(x, n)                           \
  for (long long hoge = 0; (hoge) < (n); ++(hoge)) \
  cerr << #x << "[" << hoge << "]: " << x[hoge] << endl
using namespace std;

signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  long long X[N], Y[N], P[N];
  for (int i = 0; i < N; i++) {
    cin >> X[i] >> Y[i] >> P[i];
  }
  long long x_pre[1 << N][N], y_pre[1 << N][N];
  for (int s = 0; s < (1 << N); s++) {
    for (int i = 0; i < N; i++) {
      x_pre[s][i] = abs(X[i]);
      y_pre[s][i] = abs(Y[i]);
      for (int j = 0; j < N; j++)
        if ((s >> j) & 1) {
          x_pre[s][i] = min(x_pre[s][i], abs(X[i] - X[j]));
          y_pre[s][i] = min(y_pre[s][i], abs(Y[i] - Y[j]));
        }
    }
  }
  long long ans[N + 1];
  fill(ans, ans + N + 1, LLONG_MAX / 100);
  for (int s = 0; s < (1 << N); s++) {
    int n = __builtin_popcount(s);
    for (int t = s;; t = (t - 1) & s) {
      long long sum = 0;
      for (int i = 0; i < N; i++)
        if (!((s >> i) & 1)) {
          sum += min(x_pre[t][i], y_pre[s - t][i]) * P[i];
        }
      ans[n] = min(ans[n], sum);
      if (!t) break;
    }
  }
  for (int K = 0; K <= N; K++) {
    cout << ans[K] << endl;
  }
  return 0;
}