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 double long double
#define ff first
#define ss second
#define endl '\n'
#define ii pair<int, int>
#define mp make_pair
#define mt make_tuple
#define DESYNC                      \
  ios_base::sync_with_stdio(false); \
  cin.tie(0);                       \
  cout.tie(0)
#define pb push_back
#define vi vector<int>
#define vii vector<ii>
#define all(x) x.begin(), x.end()
#define EPS 1e-9
#define INF 1e18
#define ROOT 1
#define M 1000000007
#define curtime chrono::steady_clock::now().time_since_epoch().count
#define rep(i, beg, n, s) for (int i = beg; i < n; i += s)
const double PI = acos(-1);

using namespace std;

int x[20], y[20], p[20];

long long best[20];

int32_t main() {
  DESYNC;
  int n;
  cin >> n;

  for (int i = 0; i < n; i++) cin >> x[i] >> y[i] >> p[i], best[i] = INF;

  // brute people
  int d[n];
  for (int msk = 0; msk < (1 << n); msk++) {
    int cnt = __builtin_popcount(msk);
    for (int dir = msk;; dir = (dir - 1) & msk) {
      for (int i = 0; i < n; i++) d[i] = min(abs(x[i]), abs(y[i]));
      for (int i = 0; i < n; i++) {
        if (msk >> i & 1) {
          if (dir >> i & 1)
            for (int j = 0; j < n; j++) {
              d[j] = min(d[j], abs(x[i] - x[j]));
            }
          else
            for (int j = 0; j < n; j++) {
              d[j] = min(d[j], abs(y[i] - y[j]));
            }
        }
      }

      long long ans = 0;
      for (int i = 0; i < n; i++) {
        ans += (1LL * d[i] * p[i]);
      }
      best[cnt] = min(best[cnt], ans);
      if (dir == 0) break;
    }
  }

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