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>
using namespace std;
using ll = long long;
using V = vector<int>;
using Vll = vector<ll>;
using VV = vector<V>;
using VVV = vector<VV>;
#define rep2(i, m, n) for (int i = (m); i < (n); ++i)
#define drep2(i, m, n) for (int i = (m)-1; i >= (n); --i)
#define rep(i, n) rep2(i, 0, n)
#define drep(i, n) drep2(i, n, 0)
#define all(a) (a).begin(), (a).end()
template<typename T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true; } return false; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &v) { for (auto &e : v) os << e << '\n'; return os; }
inline int popcount(const int x) { return __builtin_popcount(x); }
const int INF  = 1<<30;
const ll INFll = 1ll<<62;


int main() {
  int n; cin >> n;
  VV xy(2, V(n));
  Vll p(n);
  rep(i, n) cin >> xy[0][i] >> xy[1][i] >> p[i];

  int N = 1<<n;
  VVV dst(2, VV(n, V(N)));

  rep(i, 2) rep(S, N) {
    VV res(n, V(n+1, INF));
    rep(j, n) rep(k, n) if (S>>k&1) res[j][k] = abs(xy[i][j]-xy[i][k]);
    rep(j, n) res[j][n] = abs(xy[i][j]);
    rep(j, n) dst[i][j][S] = *min_element(all(res[j]));
  }

  Vll ans(n+1, INFll);

  rep(S, N) drep(T, S+1) {
    T &= S;
    int U = S^T;
    ll res = 0;
    rep(i, n) res += p[i] * min(dst[0][i][T], dst[1][i][U]);
    int k = popcount(S);
    chmin(ans[k], res);
  }

  cout << ans;
  return 0;
}