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;
#define rep(i, n) for(ll i = 0, i##_len = (n); i < i##_len; ++i)
#define rep2(i, x, n) for(ll i = x, i##_len = (n); i < i##_len; ++i)
#define all(n) begin(n), end(n)
using ll = long long;
using P = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vs = vector<string>;
using vc = vector<char>;
using vb = vector<bool>;
using vd = vector<double>;
vi dir = {-1, 0, 1, 0, -1, -1, 1, 1, -1};

int main() {
  ll n;
  cin >> n;
  vl x(n), y(n), p(n);
  rep(i, n) cin >> x[i] >> y[i] >> p[i];
  vector<vl> dx(1 << n, vl(n)), dy(1 << n, vl(n));
  vl ans(n + 1, 1e18);
  rep(i, 1 << n) {
    rep(j, n) dx[i][j] = abs(x[j]);
    rep(j, n) dy[i][j] = abs(y[j]);
    vl v;
    rep(j, n) if(i >> j & 1) v.push_back(j);
    for(auto u : v) {
      rep(j, n) {
        dx[i][j] = min(dx[i][j], abs(x[u] - x[j]));
        dy[i][j] = min(dy[i][j], abs(y[u] - y[j]));
      }
    }
  }
  ll m = pow(3, n);
  rep(i, m) {
    ll ii = i, ax = 0, ay = 0, cnt = 0;
    rep(j, n) {
      ll r = ii % 3;
      ii /= 3;
      if(r == 1)
        ax += 1 << j;
      else if(r == 2)
        ay += 1 << j;
      if(r) cnt++;
    }
    ll now = 0;
    rep(j, n) now += min(dx[ax][j], dy[ay][j]) * p[j];
    ans[cnt] = min(ans[cnt], now);
  }
  rep(i, n + 1) cout << ans[i] << '\n';
}